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Z-Bus Building Algorithm

Consider the ...

Question

Consider the network shown. Bus numbers and impedances are marked. An element having an impedance of $0.8 Ω$ is connected between bus 2 and reference bus. $Z_{22 n e w}$ becomes ___pu.
0.218

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Solution

The correct option is A 0.218
Without line:
$Z_{b u s} = ⎡ ⎢ ⎣ \begin{matrix} 0.1 & 0.1 & 0.1 0.1 & 0.3 & 0.3 0.1 & 0.3 & 0.7 \end{matrix} ⎤ ⎥ ⎦$

With line:
$Z_{b u s} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0.1 & 0.1 & 0.1 & - 0.1 0.1 & 0.3 & 0.3 & - 0.3 0.1 & 0.3 & 0.7 & - 0.3 - 0.1 & - 0.3 & - 0.3 & [- (- 0.3) + 0.8) = 1.1] \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$Z_{b u s m o d i f i e d} = Z_{b u s o l d} - \frac{⌊ Z_{l i} ⌊ Z_{l j} ⌋ ⌋}{Z_{l l}}$

$= ⎡ ⎢ ⎣ \begin{matrix} 0.1 & 0.1 & 0.1 0.1 & 0.3 & 0.3 0.1 & 0.3 & 0.7 \end{matrix} ⎤ ⎥ ⎦ - \frac{1}{1.1} ⎡ ⎢ ⎣ \begin{matrix} - 0.1 - 0.3 - 0.3 \end{matrix} ⎤ ⎥ ⎦ [- 0.1 - 0.3 - 0.3]$

$Z_{22 n e w} = 0.3 - \frac{1}{1.1} (- 0.3) (- 0.3)$

$= 0.3 - \frac{0.09}{1.1}$

$= 0.218$

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