Question

Rework Problem 5.17 when identical shunt reactors are installed at both ends of the line, providing 50%total shunt compensation. The reactors are removed at full load.

Short answer

Therefore, the voltage regulation is (a) 41.2% , (b) 20.3% and (c) -6% .

Step-by-step solution

Given data.

From problems, 5.17,

The shunt admittance and series impedance is,$\frac{\mathrm{Y}}{2}=(6.37\times {10}^{-7}+\mathrm{j}7.294\times {10}^{-4})\mathrm{S}$

and $\mathrm{Z}=98.25\angle 86.{69}^0\mathrm{\Omega }$ .

Source end voltage is (a) ${\mathrm{V}}_{\mathrm{S}}=653.7\mathrm{kV}$, (b) ${\mathrm{V}}_{\mathrm{S}}=556.7\mathrm{kV}$and (c) ${\mathrm{V}}_{\mathrm{S}}=435.1\mathrm{kV}$.

The full-load receiving end voltage is ${\mathrm{V}}_{\mathrm{RFL}}=480\mathrm{kV}$.

The 50% 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,

$$\begin{gathered} \frac{\mathrm{Y}}{2}=(6.37\times {10}^{-7}+\mathrm{j}7.294\times {10}^{-4})\mathrm{S} \\ \mathrm{Y}=(1.274\times {10}^{-6}+\mathrm{j}1.4588\times {10}^{-3})\mathrm{S} \end{gathered}$$

The equivalent shunt admittance after the 50% shunt compensation is,

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{eq}}=(1.274\times {10}^{-6}+\mathrm{j}1.4588\times {10}^{-3}(1-0.5\left)\right)\mathrm{S} \\ =(1.274\times {10}^{-6}+\mathrm{j}1.729\times {10}^{-4})\mathrm{S} \end{gathered}$$

Substitute $(1.274\times {10}^{-6}+\mathrm{j}1.729\times {10}^{-4})\mathrm{S}$ for ${\mathrm{Y}}_{\mathrm{eq}}$and $98.25\angle 86.{69}^0\mathrm{\Omega }$ for Z in equation (1).

$$\begin{gathered} \mathrm{A}=1+\frac{(1.274\times {10}^{-6}+\mathrm{j}1.729\times {10}^{-4})(98.25\angle 86.{69}^0)}{2} \\ =0.9644\angle 0.{21}^0\mathrm{pu} \end{gathered}$$

Derive the voltage regulation for (a).

(a)

Substitute $0.9644\mathrm{pu}$for A , $653.7\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{S}}$ and 480 kV for ${\mathrm{V}}_{\mathrm{RFL}}$in equation (2).

role="math" localid="1655292962783" $$\begin{gathered} VR\%=\frac{{\displaystyle \frac{653.7kV}{0.9644}}-480kV}{480kV}\times 100 \\ =41.2\% \end{gathered}$$

Therefore, the voltage regulation is $41.2\%$.

Derive the voltage regulation for (b).

(b)

Substitute $0.9644\mathrm{pu}$for A , 556.7 kV for ${\mathrm{V}}_{\mathrm{S}}$and 480 kV for ${\mathrm{V}}_{\mathrm{RFL}}$ in equation (2).

role="math" localid="1655293284518" $$\begin{gathered} VR\%=\frac{{\displaystyle \frac{556.7kV}{0.9644}-480kV}}{480kV}\times 100 \\ =20.3\% \end{gathered}$$

Therefore, the voltage regulation is $20.3\%$ .

Derive the voltage regulation for (c).

(c)

Substitute $0.9644\mathrm{pu}$for A , 435.1 kV for ${\mathrm{V}}_{\mathrm{S}}$ and 480 kV for ${\mathrm{V}}_{\mathrm{RFL}}$ in equation (2).

$$\begin{gathered} VR\%=\frac{{\displaystyle \frac{435.1kV}{0.9644}-480kV}}{480kV}\times 100 \\ =-6\% \end{gathered}$$

Therefore, the voltage regulation is $-6\%$.