Question

For the system from Problem 6.30, assume that a $\mathbf{75}\mathbf{Mvar}$shunt capacitance (three phase assuming one per unit bus voltage) is added at bus

4. Calculate the new value of${\mathbf{Y}}_{\mathbf{44}}$.

Short answer

The value of the element$Y_{44}$ is$9.88-j35.65\mathrm{pu}$.

Step-by-step solution

Write the given data.

The base MVA,$S_{\mathrm{base}}=1000\mathrm{MVA}$

The capacitance of$75\mathrm{Mvar}$ is added at bus 4.

Determine the formula to calculate the admittance matrix for threephase power system.

The equation to calculate the diagonal element of the$Y_{\mathrm{bus}}$matrix of the system is

given as follows.

${\mathbf{Y}}_{\mathbf{kk}}$ =Sum of the admittance connected to bus. ......( 1 )

The equation to calculate the non-diagonal element of the localid="1655275377909" $Y_{\mathrm{bus}}$matrix of the system

is given as follows.

${\mathbf{Y}}_{\mathbf{kk}}$=(-sum of admittance connected between bus and n) ......(2)

The equation to calculate the per unit value is given as follows.

$\mathbf{Per}\mathbf{}\mathbf{unit}\mathbf{=}\frac{\mathbf{Actual}\mathbf{}\mathbf{value}}{\mathbf{Base}\mathbf{}\mathbf{value}}$ ......(3)

The equation to calculate per unit admittance of capacitor is given as follows.

${\mathbf{Y}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{Mvar}}_{\mathbf{pu}}}{{\mathbf{V}}^{\mathbf{2}}}$ ......(4)

Here,$\mathbf{V}$is the per unit voltage at specific bus.

Calculate the value ofY44.

Calculate the $\mathrm{Mvar}$per unit value..

Substitute $1000\mathrm{MVA}$for base value and $75\mathrm{Mvar}$for actual value into equation $$\begin{gathered} {\mathrm{Mvar}}_{pu}=\frac{75}{100} \\ {\mathrm{Mvar}}_{\mathrm{pu}}=0.75\mathrm{pu} \end{gathered}$$

Calculate the admittance of capacitor.

Substitute $0.75\mathrm{pu}$for ${\mathrm{Mvar}}_{\mathrm{pu}}$and $1\mathrm{pu}$for $V$into equation (4).

$$\begin{gathered} Y_{\mathit{c}}\mathit{=}\frac{\mathit{0}\mathit{.}\mathit{75}}{{\mathit{1}}^{\mathit{2}}} \\ {Y}_{\mathit{c}}=0.75\mathrm{pu} \end{gathered}$$

Calculate the element$Y_{44}$.

$$\begin{gathered} Y_{44}=\frac{1}{0.08+j0.24}+\frac{1}{0.01+\mathrm{j}0.24}+\frac{1}{0.03+\mathrm{j}0.10}+j\frac{0.05}{2}\mathit{+}j\frac{0.04}{2}\mathit{+}j\frac{0.010}{2}\mathit{+}j\mathit{0}\mathit{.}\mathit{75} \\ {Y}_{\mathit{44}}=1.25-j3.75+5.8823-j23.5294+2.7522+j0.025+j0.02+j0.005+j0.75 \\ {Y}_{44}=9.88-j35.65\mathrm{pu} \end{gathered}$$

Hence the element $Y_{44}$is$9.88-j35.65\mathrm{pu}$.