Question

Identical series capacitors are installed at both ends of the line in Problem 5.16, providing 30%total series compensation. (a) Determine the equivalent ABCDparameters for this compensated line. (b) Determine the theoretical maximum real power that this series-compensated line candeliver when ${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$. Compare your result with that of Problem 5.40.

Short answer

(a)Therefore, the equivalent ABCD parameters are$\left[\begin{array}{c}0.9492\angle 0.2553°71.45\angle 80.5°\\ 1.405\times {10}^{-3}\angle 90.09°0.9492\angle 0.2553°\end{array}\right]$and reactance of each capacitance is$-\mathrm{j}14.71\mathrm{\Omega }.$.

(b)Therefore, the maximum power is $2936\mathrm{MW}$ and is more than the power obtained in problem 5.40.

Step-by-step solution

Given data.

From problems 5.16,

The ABCD parameters are,

$$\begin{gathered} \mathrm{A}=0.9285\angle 0.{258}^0\mathrm{pu} \\ \mathrm{D}=0.9285\angle 0.{258}^0\mathrm{pu} \\ \mathrm{C}=1.405\times {10}^{-3}\angle 90.{09}^0\mathrm{S} \\ \mathrm{B}=98.25\angle 86.{69}^0\mathrm{\Omega } \end{gathered}$$

The 30% series compensation is mentioned.

The impedance from problem 5.16 is,

$\mathrm{Z}=(5.673+\mathrm{j}98.09)\mathrm{\Omega }$

Determine the formula of ABCD parameters and maximum power.

Write the formula of equivalent ABCD parameters.

$\left[\begin{array}{c}{A}_{\mathrm{eq}}{B}_{\mathrm{eq}}\\ C_{\mathrm{eq}}{D}_{\mathrm{eq}}\end{array}\right]\mathbf{=}\left[\begin{array}{c}{A}_S{B}_S\\ C_S{D}_S\end{array}\right]\mathbf{}\left[\begin{array}{c}AB\\ CD\end{array}\right]\left[\begin{array}{c}{A}_R{B}_R\\ C_R{D}_R\end{array}\right]$ .................(1)

Here, eq is representing equivalent, S is representing sending end, and R is representing receiving end.

Write the formula of maximum power.

${\mathbf{P}}_{\mathbf{max}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{\mathbf{Z}}\mathbf{-}\frac{{\mathbf{AV}}_{\mathbf{R}}^{\mathbf{2}}}{\mathbf{Z}}\mathbf{cos}\mathbf{(}{\mathbf{\theta }}_{\mathbf{B}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{A}}\mathbf{)}$ ...........…… (2)

Here, ${\mathrm{V}}_{\mathrm{R}}$ and ${\mathrm{V}}_{\mathrm{S}}$ are receiving and sending end voltages.

Determine the formula of ABCD parameters and capacitance reactance.

(a) The reactance of each series capacitor with compensation is,

$$\begin{gathered} \mathrm{X}=-\left(\mathrm{j}\frac{98.09}{2}\right)\left(0.3\right)\mathrm{\Omega } \\ =-\mathrm{j}14.71\mathrm{\Omega } \end{gathered}$$

The ABCD parameters at sending and receiving end will be,

Determine the maximum power.

(b)

Substitute 500 kV for ${\mathrm{V}}_{\mathrm{S}}$, ${\mathrm{V}}_{\mathrm{R}}$, localid="1655281050243" $0.9492\angle 0.{2553}^0$, for $\mathrm{A}\angle {\mathrm{\theta }}_{\mathrm{A}}$ and $71.25\angle 80.5^0$ for $\mathrm{Z}\angle {\mathrm{\theta }}_{\mathrm{B}}$ in equation (2).

$$\begin{gathered} \mathrm{P}=\frac{500\times 500}{71.45}-\frac{0.9492{\left(500\right)}^2}{71.45}\cos \left(80.5^0-0.{2553}^0\right) \\ =2936\mathrm{MW} \end{gathered}$$

The maximum power in problem 5.40 is $2397.5\mathrm{MW}$and is less than $2936\mathrm{MW}$ .