Question

Recalculate the percent voltage regulation in Problem 5.15 when identical shunt reactors are installed at both ends of the line during light loads, providing 65% total shunt compensation. The reactors are removed at full load. Also calculate the impedance of each shunt reactor.

Short answer

Therefore, the voltage regulation is 15.7% and the shunt reactor is$1713\mathrm{\Omega }$ .

Step-by-step solution

Given data.

From problems, 5.14,5.15 and 5.23,

The shunt admittance and series impedance of equivalent$\mathrm{\pi }$circuit is,

role="math" localid="1655288258435" $\frac{\mathrm{Y}}{2}=(1.57\times {10}^{-6}+\mathrm{j}8.981\times {10}^{-4})\mathrm{S}$ and$\mathrm{Z}=134.8\angle 85.3^0\mathrm{\Omega }$ .

Source end voltage is ${\mathrm{V}}_{\mathrm{S}}=526.14\mathrm{kV}$.

The full-load receiving end voltage is${\mathrm{V}}_{\mathrm{RFL}}=475\mathrm{kV}$ .

The 65% shunt compensation at both sides are given.

Determine the formulas of A parameter and voltage regulation.

Write the formula of A parameter

$\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{Y}}_{\mathbf{eq}}\mathbf{Z}}{\mathbf{2}}$ …… (1)

Here, ${\mathrm{Y}}_{\mathrm{eq}}$ is equivalent shunt admittance and Z is series impedance.

Write the formula of voltage regulation.

$\mathbf{VR}\mathbf{\%}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{V}}_{\mathbf{S}}}{\mathbf{A}}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\times }\mathbf{100}$ …… (2)

Here, ${\mathrm{V}}_{\mathrm{S}}$ is sending end voltage and ${\mathrm{V}}_{\mathrm{RFL}}$ is full-load receiving end voltage.

Determine the A parameter.

The total shunt admittance is,

$$\begin{gathered} \frac{\mathrm{Y}}{2}=(1.57\times {10}^{-6}+\mathrm{j}8.981\times {10}^{-4})\mathrm{S} \\ \mathrm{Y}=(3.14\times {10}^{-6}+\mathrm{j}1.796\times {10}^{-3})\mathrm{S} \end{gathered}$$

The equivalent shunt admittance after the 65% shunt compensation is,

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{eq}}=(3.14\times {10}^{-6}+\mathrm{j}1.796\times {10}^{-3}(1-0.65\left)\right)\mathrm{S} \\ =(3.14\times {10}^{-6}+\mathrm{j}1.796\times {10}^{-3})\mathrm{S} \end{gathered}$$

Substitute $(3.14\times {10}^{-6}+\mathrm{j}1.796\times {10}^{-3})\mathrm{S}$for ${\mathrm{Y}}_{\mathrm{eq}}$and $134.8\angle 85.3^0\mathrm{\Omega }$ for Z in equation (1).

$$\begin{gathered} \mathrm{A}=1+\frac{(3.14\times {10}^{-6}+\mathrm{j}6.287\times {10}^{-4})(314.8\angle 85.3^0)}{2} \\ =0.9578\angle 0.{22}^0\mathrm{pu} \end{gathered}$$

Derive the voltage regulation.

Substitute 0.9578 pu for A , $526.14\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{S}}$ and 475 kV for ${\mathrm{V}}_{\mathrm{RFL}}$ in equation (2).

$$\begin{gathered} \mathrm{VR}\%=\frac{{\displaystyle \frac{526.14\mathrm{kV}}{0.9578}}-475\mathrm{kV}}{475\mathrm{kV}}\times 100 \\ =15.7\% \end{gathered}$$

Therefore, the voltage regulation is $15.7\%$.

Determine the shunt reactor.

The impedance of shunt reactor with 65% compensation is,

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{reactor}}=\frac{1}{{\displaystyle \frac{1}{2}}\left(1.796\times {10}^{-3}\right)(1-0.65))\mathrm{S}} \\ =1713\mathrm{\Omega } \end{gathered}$$

Therefore, the shunt reactor is $1713\mathrm{\Omega }$.