Question

What is the minimum amount of series capacitive compensation ${\mathbf{N}}_{\mathbf{C}}$in percent of the positive-sequence line reactance needed to reduce the number ofrole="math" localid="1655273957159" $\mathbf{765}\mathbf{}\mathbf{kV}$lines in Example 5.8 from five to four? Assume two intermediate substations with one line section out of service. Also, neglect line losses and assume that the series compensation is sufficiently distributed along the line so as to effectively reduce the series reactance of the equivalent$\mathbf{\pi }$circuit to${\mathbf{X}}^{\mathbf{\text{'}}}\left(1-\frac{N_C}{100}\right)$.

Short answer

Therefore, the series capacitive compensation is$18.4\%$ .

Step-by-step solution

Given data.

From problems 5.8, the given data is,

The power transferred through the line is $\mathrm{P}=9000\mathrm{MW}$.

The line-to-line voltage is$765\mathrm{kV}$ .

The per-unit receiving end voltage is ${\mathrm{V}}_{\mathrm{R}}=0.95\mathrm{pu}$.

The load angle is $\mathrm{\delta }={35}^0$and ${\mathrm{X}}^{\text{'}}=156.35\mathrm{\Omega }$.

Determine the formula of power.

Write the formula of power transferred through the transmission line.

$\mathbf{P}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{\mathbf{X}}\mathbf{sin\delta }$ …… (1)

Here, ${\mathrm{V}}_{\mathrm{R}}$ and ${\mathrm{V}}_{\mathrm{S}}$ are receiving and sending end voltages, $\mathrm{\delta }$ is load angle and X is transmission line reactance.

Determine the reactance of line.

For four lines with two intermediate substations, and one line out of service, the reactance of line will be$0.2778{\mathrm{X}}_{\mathrm{eq}}$.

Substitute 9000 MVA for P , 765 kV for ${\mathrm{V}}_{\mathrm{S}}$, $(0.95\times 765)\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{R}}$, ${35}^0$ for $\mathrm{\delta }$and $0.2778{\mathrm{X}}_{\mathrm{eq}}$for X in equation (1).

$$\begin{gathered} 9000\mathrm{MVA}=\frac{(765\mathrm{kV})(0.95\times 765)\mathrm{kV}}{0.2778{\mathrm{X}}_{\mathrm{eq}}}\sin {35}^0 \\ {\mathrm{X}}_{\mathrm{eq}}=127.54\mathrm{\Omega } \end{gathered}$$

The percent of compensation is calculated as,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{eq}}={\mathrm{X}}^{\text{'}}\left(1-\frac{{\mathrm{N}}_{\mathrm{C}}}{100}\right) \\ 127.54\mathrm{\Omega }=(156.35\mathrm{\Omega })\left(1-\frac{{\mathrm{N}}_{\mathrm{C}}}{100}\right) \\ {\mathrm{N}}_{\mathrm{C}}=18.4\% \end{gathered}$$

Therefore, the series capacitive compensation is $18.4\%$.