Question

For the five-bus network shown in Figure 9.25, a bolted single-line-toground fault occurs at the bus$\mathbf{2}$ end of the transmission line between buses $\mathbf{1}$and. role="math" localid="1656410079694" $\mathbf{2}$The fault causes the circuit breaker at the bus end of the line to open, but all other breakers remain closed. The fault is shown in Figure 9.26. Compute the subtransient fault current with the circuit breaker at the bus- $\mathbf{2}$end of the faulted line open. Neglect prefault current and assume a prefault voltage ofrole="math" localid="1656410247948" $\mathbf{1}\mathbf{.}\mathbf{0}$ per unit.

Short answer

The value of subtransient fault current at the end of the faulted line $I_f^{*}$is$-j3.912\text{\hspace{0.17em}per\hspace{0.33em}unit}$

Step-by-step solution

Write the given data from the question.

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

Assume a prefault voltage of.$\text{1.0perunit}$

Determine the formula ofsub transient fault current.

Write the formula of sub transient fault current at the end of the faulted line.

${\mathbf{I}}_f^{*}=\frac{{\mathbf{V}}_F}{{\mathbf{Z}}_{\mathrm{kk},\mathrm{th}}}$ …… (1)

Here,${\mathbf{V}}_F$is fault voltage and${\mathbf{Z}}_{\mathrm{kk},\mathrm{th}}$ is Thevenin equivalent impedance.

Step 3:Determine thevalue ofsubtransient fault current at the end of the faulted line.

Refer to five bus network in figure 9.25 in the textbook.

Determine the bus admittance matrix.

$Y_{\text{bus1}}=\left[\begin{array}{ccccc}{Y}_{11}& Y_{12}& Y_{13}& Y_{14}& Y_{15}\\ Y_{12}& Y_{22}& Y_{23}& Y_{24}& Y_{25}\\ Y_{13}& Y_{23}& Y_{33}& Y_{34}& Y_{35}\\ Y_{14}& Y_{24}& Y_{43}& Y_{44}& Y_{45}\\ Y_{15}& Y_{25}& Y_{53}& Y_{54}& Y_{55}\end{array}\right]$

Determine the value of admittance$Y_{11}$.

$\begin{array}{c}{Y}_{11}=\frac{1}{Z_{01}}+\frac{1}{Z_{12}}+\frac{1}{Z_{15}}\\ =\frac{1}{j0.111}+\frac{1}{j0.168}+\frac{1}{j0.126}\\ =-j22.89\end{array}$

Determine the value of admittance$Y_{12}$.

$\begin{array}{c}{Y}_{12}=\frac{-1}{Z_{12}}\\ =\frac{-1}{j0.168}\\ =j5.95\end{array}$

Determine the value of the admittance$Y_{23}$.

$\begin{array}{c}{Y}_{23}=\frac{-1}{Z_{23}}\\ =\frac{-1}{j0.126}\\ =j7.94\end{array}$

Determine the value of the admittance$Y_{24}$.

$\begin{array}{c}{Y}_{24}=\frac{-1}{Z_{24}}\\ =0\end{array}$

Determine the value of admittance.$Y_{25}$

$\begin{array}{c}{Y}_{25}=\frac{-1}{Z_{25}}\\ =0\end{array}$

Determine the value of admittance$Y_{34}$.

$\begin{array}{c}{Y}_{34}=\frac{-1}{Z_{34}}\\ =\frac{-1}{j0.336}\\ =j2.976\end{array}$

Determine the value of admittance$Y_{35}$.

$\begin{array}{c}{Y}_{35}=\frac{-1}{Z_{35}}\\ =\frac{-1}{j0.210}\\ =j4.76\end{array}$

Determine the value of admittance.localid="1656413893891" $Y_{45}$

localid="1656420801300" $\begin{array}{c}{Y}_{45}=\frac{-1}{Z_{45}}\\ =\frac{-1}{j0.252}\\ =j3.97\end{array}$

Determine the value of admittance.localid="1656420807213" $Y_{13}$$Y_{45}$

localid="1656420814703" $\begin{array}{c}{Y}_{13}=\frac{-1}{Z_{13}}\\ =0\end{array}$

Determine the value of admittance.localid="1656420823609" $Y_{14}$

localid="1656420831604" $\begin{array}{c}{Y}_{14}=\frac{-1}{Z_{14}}\\ =0\end{array}$

Determine the value of admittancelocalid="1656420840106" $Y_{44}$

localid="1656420848493" $\begin{array}{c}{Y}_{44}=\frac{1}{Z_{43}}+\frac{1}{Z_{45}}\\ =\frac{1}{j0.336}+\frac{1}{j0.252}\\ =-j6.94\end{array}$

Determine the value of admittance.localid="1656420857502" $Y_{55}$

.localid="1656420864877" $\begin{array}{c}{Y}_{55}=\frac{1}{Z_{51}}+\frac{1}{Z_{53}}+\frac{1}{Z_{54}}\\ =\frac{1}{j1.26}+\frac{1}{j0.21}+\frac{1}{j0.252}\\ =-j16.67\end{array}$

Determine the value of admittancelocalid="1656420875726" $Y_{15}$.

localid="1656420883411" $\begin{array}{c}{Y}_{15}=\frac{-1}{Z_{15}}\\ =\frac{-1}{j0.126}\\ =j7.94\end{array}$

Determine the value of admittancelocalid="1656420892016" $Y_{22}$.

localid="1656420899185" $\begin{array}{c}{Y}_{22}=\frac{1}{Z_{21}}+\frac{1}{Z_{23}}\\ =\frac{1}{j0.168}+\frac{1}{j0.126}\\ =-j13.89\end{array}$

Determine the value of admittancelocalid="1656420910344" $Y_{33}$.

localid="1656420919667" $\begin{array}{c}{Y}_{33}=\frac{1}{Z_{03}}+\frac{1}{Z_{23}}+\frac{1}{Z_{34}}+\frac{1}{Z_{53}}\\ =\frac{1}{j0.1333}+\frac{1}{j0.126}+\frac{1}{j0.336}+\frac{1}{j0.210}\\ =-j23.19\end{array}$

The positive sequence bus admittance matrix is,

localid="1656420927051" $Y_{\text{bus1}}=\left[\begin{array}{ccccc}-j22.89& j5.95& 0& 0& j7.94\\ j5.95& -j13.89& j7.94& 0& 0\\ 0& j7.94& -j23.19& j2.976& j4.76\\ 0& 0& j2.976& -j6.94& j3.97\\ j7.94& 0& j4.76& j3.97& -j16.67\end{array}\right]$

Determine the$Z$-bus matrix of positive sequence bus admittance matrix.

localid="1656420938006" $\begin{array}{c}{Z}_{\text{bus1}}=\frac{1}{Y_{\text{bus1}}}\\ =\frac{1}{\left[\begin{array}{ccccc}-j22.89& j5.95& 0& 0& j7.94\\ j5.95& -j13.89& j7.94& 0& 0\\ 0& j7.94& -j23.19& j2.976& j4.76\\ 0& 0& j2.976& -j6.94& j3.97\\ j7.94& 0& j4.76& j3.97& -j16.67\end{array}\right]}\\ =\left[\begin{array}{ccccc}j0.0793& j0.0558& j0.0382& j0.0512& j0.0609\\ j0.0558& j0.1338& j0.0664& j0.0631& j0.0606\\ j0.0382& j0.0664& j0.0875& j0.0721& j0.0603\\ j0.0512& j0.0631& j0.0721& j0.2324& j0.1003\\ j0.0609& j0.0606& j0.0603& j0.1003& j0.1301\end{array}\right]\end{array}$

Therefore, the value of positive sequence bus impedance matrixlocalid="1656420948026" $Z_{\text{bus1}}$is,

localid="1656420957462" $\left[\begin{array}{ccccc}j0.0793& j0.0558& j0.0382& j0.0512& j0.0609\\ j0.0558& j0.1338& j0.0664& j0.0631& j0.0606\\ j0.0382& j0.0664& j0.0875& j0.0721& j0.0603\\ j0.0512& j0.0631& j0.0721& j0.2324& j0.1003\\ j0.0609& j0.0606& j0.0603& j0.1003& j0.1301\end{array}\right]$

Assume the impedance of the line.

localid="1656420968306" $Z_{\text{b}}=j0.168$

Figure 1 illustrates the circuit of the bolted single-line-to-ground fault on the transmission line connecting buses 1 and 2 at the bus 2 ends.

Figure 1

Draw the thevenin equivalent circuit of the bolted single-line-to-ground fault.

Figure 2

Determine the Thevenin equivalent impedance.

$\begin{array}{c}{Z}_{\text{kk,th}}=Z_{12}+Z_{\text{b}}-\left(Z_{11}-Z_{12}\right)\parallel \left(Z_{22}-Z_{21}-Z_{\text{b}}\right)\\ =j0.0558+j0.168+\frac{\left(j0.0235\right)\left(-j0.09\right)}{j0.0235-j0.09}\\ =j0.0558+j0.168+j0.0318\\ =j0.2556\end{array}$

Now, determine the sub transient fault current at the end of the faulted line.

Substitute $1$for$V_{\text{F}}$and$j0.2556$for$Z_{\text{kk,th}}$into equation (1).

$\begin{array}{c}{I}_f^{*}=\frac{1}{j0.2556}\\ =-j3.912\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the value of subtransient fault current at the end of the faulted line$I_f^{*}$is$-j3.912\text{\hspace{0.17em}per\hspace{0.33em}unit}$.