Question

Compute the $\mathbf{5}\mathbf{\times }\mathbf{5}$per-units zero-, positive-, and negative-sequence bus impedance matrices for the power system given in Problem 9.8. Use a base of $\mathbf{100}\mathbf{MVA}$and $\mathbf{15}\mathbf{kV}$in the zone of generator.

Short answer

The value of positive-sequence bus impedance matrix is,

$Z_{\text{bus1}}=-j\left[\begin{array}{ccccc}1.6397& 1.362& 1.0975& 0.8331& 0.2777\\ 1.362& 5.2846& 4.2584& 3.2323& 1.0774\\ 1.0975& 4.2584& 7.2688& 5.5173& 1.8391\\ 0.8331& 3.2323& 5.5173& 7.8023& 2.6008\\ 0.2777& 1.0774& 1.8391& 2.6008& 4.2003\end{array}\right]\text{p.u.}$

The value of negative-sequence bus impedance matrix is,

$Z_{\text{bus2}}=-j\left[\begin{array}{ccccc}1.6376& 1.354& 1.0839& 0.8138& 0.2466\\ 1.354& 5.2536& 4.2056& 3.1576& 0.9569\\ 1.0839& 4.2056& 7.1787& 5.3899& 1.6333\\ 0.8138& 3.1576& 5.3899& 7.6221& 2.3097\\ 0.2466& 0.9569& 1.6333& 2.3097& 3.7302\end{array}\right]\text{\hspace{0.33em}p.u.}$

The value of zero-sequence bus impedance matrix is,

$Z_{\text{bus0}}=-j\left[\begin{array}{ccccc}2.9287& 2.146& 1.8479& 1.5498& 0\\ 2.146& 5.2363& 4.509& 3.7816& 0\\ 1.8479& 4.509& 5.5225& 4.6316& 0\\ 1.5498& 3.7816& 4.6316& 5.4817& 0\\ 0& 0& 0& 0& 10\end{array}\right]\text{\hspace{0.33em}p.u.}$

Step-by-step solution

Write the given data from the question.

As reference of problem 9.8.

Thevalue of base MVA is.$\text{100\hspace{0.33em}MVA}$

The value of voltage base zone of generator$\text{100\hspace{0.33em}MVA}$ is$15\text{\hspace{0.33em}kV}$.

Determine the formula of sequence impedance bus matrix.

Write the formula of positive-sequence bus impedance matrix

${\mathbf{Z}}_{\mathrm{bus}1}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}1}}$ …… (1)

Here, ${\mathrm{Y}}_{\mathrm{bus}1}$is per-unit positive-sequence bus admittance matrix.

Write the formula of negative-sequence bus impedance matrix

${\mathbf{Z}}_{\mathrm{bus}2}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}2}}$ …… (2)

Here,role="math" localid="1656476644681" ${\mathbf{Y}}_{\mathrm{bus}2}$is per-unit negative-sequence bus admittance matrix.

Write the formula of zero-sequence bus impedance matrix

${\mathbf{Z}}_{\mathrm{bus}0}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}0}}$ …… (3)

Here,${\mathbf{Y}}_{\mathrm{bus}0}$ is per-unit zero-sequence bus admittance matrix.

(a)Determine thevalue of positive sequence, negative sequence and zero sequence bus impedance matrix of the system.

Determine the per unit positive sequence reactance for generator$1$.

$\begin{array}{c}{X}_1={X^{\text{'}\text{'}}}_{\text{d}}\\ =0.2{\left(\frac{12}{10}\right)}^2\left(\frac{100}{50}\right)\\ =0.576\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per-unit negative sequence reactance for generator.localid="1656480615528" $1$

$X_2=0.576\text{\hspace{0.33em}p.u.}$

Determine the per-unit zero sequence reactance for generator.localid="1656480597938" $1$

width="172">X0=0.11210210050=0.288 p.u.

Determine the per-unit positive sequence reactance for generatorlocalid="1656480541504" $2$.

localid="1656480524282" $\begin{array}{c}{X}_1={X^{\text{'}}}_{\text{d}}\\ =0.2\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per-unit negative sequence reactance for generatorlocalid="1656480644009" $2$.

localid="1656480625284" $X_2=0.23\text{\hspace{0.33em}p.u.}$

Determine the per-unit zero sequence reactance for generatorlocalid="1656480656079" $2$.

localid="1656480666096" $X_0=0.1\text{\hspace{0.33em}p.u.}$

Determine the per-unit reactance for transformerlocalid="1656480677563" $\text{T1}$.

localid="1656480688195" $\begin{array}{c}{X}_{\text{T1}}=0.10\left(\frac{100}{50}\right)\\ =0.2\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit reactance for transformer.localid="1656480698312" $\text{T2}$

localid="1656480710583" $X_{\text{T2}}=0.10\text{\hspace{0.33em}p.u.}$ --

Determine the base impedance value.

localid="1656480943690" $\begin{array}{c}{Z}_{\text{base}}=\frac{{\left(138\times {10}^3\right)}^2}{1000\times {10}^6}\\ =190.44\Omega \end{array}$

Determine the per-unit positive and negative quantities for lines.

localid="1656480953551" $\begin{array}{c}{X}_1=X_2\\ =\left(\frac{40}{190.44}\right)\\ =0.210\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per-unit zero sequence reactance for line.

localid="1656480962740" $\begin{array}{c}{X}_0=\left(\frac{100}{190.44}\right)\\ =0.5251\text{\hspace{0.33em}p.u.}\end{array}$

Draw the circuit diagram of positive sequence network of the system.

The circuit's positive-sequence bus admittance matrix is shown below.

localid="1656480970935" $Y_{\text{bus1}}=j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.3\end{array}\right]\text{\hspace{0.17em}\hspace{0.33em}p.u.}$

Determine the per unit positive sequence bus impedance matrix of the system.

localid="1656480981802" $Z_{\text{bus1}}=\frac{1}{Y_{\text{bus1}}}$

Here,localid="1656480991001" $Y_{\text{bus1}}$is positive-sequence admittance matrix.

Substitute localid="1656481000247" $j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.3\end{array}\right]\text{\hspace{0.17em}\hspace{0.33em}p.u.}$forlocalid="1656481011500" $Y_{\text{bus1}}$into above equation.

localid="1656481023177" $\begin{array}{c}{Z}_{\text{bus1}}=\frac{1}{j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.3\end{array}\right]\text{\hspace{0.17em}\hspace{0.33em}p.u.}}\\ =-j\left[\begin{array}{ccccc}1.6397& 1.362& 1.0975& 0.8331& 0.2777\\ 1.362& 5.2846& 4.2584& 3.2323& 1.0774\\ 1.0975& 4.2584& 7.2688& 5.5173& 1.8391\\ 0.8331& 3.2323& 5.5173& 7.8023& 2.6008\\ 0.2777& 1.0774& 1.8391& 2.6008& 4.2003\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Therefore, the value of per unit positive-sequence bus impedance matrix is,

localid="1656481033610" $Z_{\text{bus1}}=-j\left[\begin{array}{ccccc}1.6397& 1.362& 1.0975& 0.8331& 0.2777\\ 1.362& 5.2846& 4.2584& 3.2323& 1.0774\\ 1.0975& 4.2584& 7.2688& 5.5173& 1.8391\\ 0.8331& 3.2323& 5.5173& 7.8023& 2.6008\\ 0.2777& 1.0774& 1.8391& 2.6008& 4.2003\end{array}\right]\text{\hspace{0.33em}p.u.}$

Determine the per unit negative sequence bus impedance matrix of the system.

localid="1656481043179" $Z_{\text{bus2}}=\frac{1}{Y_{\text{bus2}}}$

The circuit's negative-sequence bus admittance matrix is shown below.

localid="1656481052033" $Y_{\text{bus2}}=j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.33\end{array}\right]\text{\hspace{0.33em}p.u.}$

localid="1656481065986" $\begin{array}{c}{Z}_{\text{bus2}}=\frac{1}{j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.33\end{array}\right]}\\ =-j\left[\begin{array}{ccccc}1.6376& 1.354& 1.0839& 0.8138& 0.2466\\ 1.354& 5.2536& 4.2056& 3.1576& 0.9569\\ 1.0839& 4.2056& 7.1787& 5.3899& 1.6333\\ 0.8138& 3.1576& 5.3899& 7.6221& 2.3097\\ 0.2466& 0.9569& 1.6333& 2.3097& 3.7302\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Therefore, the value of per unit negative-sequence bus impedance matrix is,

localid="1656481078881" $Z_{\text{bus2}}=-j\left[\begin{array}{ccccc}1.6376& 1.354& 1.0839& 0.8138& 0.2466\\ 1.354& 5.2536& 4.2056& 3.1576& 0.9569\\ 1.0839& 4.2056& 7.1787& 5.3899& 1.6333\\ 0.8138& 3.1576& 5.3899& 7.6221& 2.3097\\ 0.2466& 0.9569& 1.6333& 2.3097& 3.7302\end{array}\right]\text{\hspace{0.33em}p.u.}$

The circuit's zero-sequence bus admittance matrix is shown below.

$Y_{\text{bus0}}=\frac{1}{j}\left[\begin{array}{ccccc}0.488& -0.2& 0& 0& 0\\ -0.2& 0.7251& -0.5251& 0& 0\\ 0& -0.5251& 1.0502& -0.5251& 0\\ 0& 0& -0.5251& 0.6261& 0\\ 0& 0& 0& 0& 0.1\end{array}\right]\text{\hspace{0.33em}p.u.}$

$Z_{\text{bus0}}=\frac{1}{Y_{\text{bus0}}}$

Here,$Y_{\text{bus0}}$is zero-sequence bus admittance matrix.

Substitute$\frac{1}{j}\left[\begin{array}{ccccc}0.488& -0.2& 0& 0& 0\\ -0.2& 0.7251& -0.5251& 0& 0\\ 0& -0.5251& 1.0502& -0.5251& 0\\ 0& 0& -0.5251& 0.6261& 0\\ 0& 0& 0& 0& 0.1\end{array}\right]\text{\hspace{0.33em}p.u.}$for $Y_{\text{bus0}}$into above equation.

$\begin{array}{c}{Z}_{\text{bus0}}=\frac{1}{\frac{1}{j}\left[\begin{array}{ccccc}0.488& -0.2& 0& 0& 0\\ -0.2& 0.7251& -0.5251& 0& 0\\ 0& -0.5251& 1.0502& -0.5251& 0\\ 0& 0& -0.5251& 0.6261& 0\\ 0& 0& 0& 0& 0.1\end{array}\right]\text{\hspace{0.33em}p.u.}}\\ =-j\left[\begin{array}{ccccc}2.9287& 2.146& 1.8479& 1.5498& 0\\ 2.146& 5.2363& 4.509& 3.7816& 0\\ 1.8479& 4.509& 5.5225& 4.6316& 0\\ 1.5498& 3.7816& 4.6316& 5.4817& 0\\ 0& 0& 0& 0& 10\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Therefore, the value of per unit zero-sequence bus impedance matrix is,

$Z_{\text{bus0}}=-j\left[\begin{array}{ccccc}2.9287& 2.146& 1.8479& 1.5498& 0\\ 2.146& 5.2363& 4.509& 3.7816& 0\\ 1.8479& 4.509& 5.5225& 4.6316& 0\\ 1.5498& 3.7816& 4.6316& 5.4817& 0\\ 0& 0& 0& 0& 10\end{array}\right]\text{\hspace{0.33em}p.u.}$