Question

Repeat Problem 9.24 for a bolted line-to-line fault.

Short answer

The value of sub transient fault current is$\left[\begin{array}{c}0\\ 2.43\angle {180}^0\\ 2.43\angle 0^0\end{array}\right]\mathrm{per}\mathrm{unit}$ and $\left[\begin{array}{c}0\\ 14.03\angle {180}^0\\ 14.03\angle 0^0\end{array}\right]\mathrm{kA}$.

The value of contributions to the fault current from each generator is$\left[\begin{array}{c}\mathrm{j}0.0092\\ 1.502\angle -179.8^0\\ 1.502\angle -0.2^0\end{array}\right]\mathrm{per}\mathrm{unit}$ and $\left[\begin{array}{c}3.463\angle -{90}^0\\ 1.731\angle 89.4^0\\ 1.731\angle 90.6^0\end{array}\right]\mathrm{kA}$.

The value of contributions to the fault current from each transformer is$\left[\begin{array}{c}-\mathrm{j}0.011\\ 0.925\angle 179.{66}^0\\ 0.925\angle 0.{34}^0\end{array}\right]\mathrm{per}\mathrm{unit}$ and $\left[\begin{array}{c}4.932\angle -{90}^0\\ 1.731\angle -90.6^0\\ 1.731\angle -89.4^0\end{array}\right]\mathrm{kA}$.

Step-by-step solution

Write the given data from the question.

The circuit zero sequence impedance is ${\mathrm{Z}}_0$.

The positive sequence impedance of the circuit is ${\mathrm{Z}}_1$.

The negative sequence impedance of the circuit is ${\mathrm{Z}}_2$.

${\mathrm{I}}_0$Is the zero sequence current flowing through zero sequence impedance ${\mathrm{Z}}_0$.

${\mathrm{I}}_1$Is the positive sequence current flowing through positive impedance ${\mathrm{Z}}_1$.

${\mathrm{I}}_2$Is the negative sequence current flowing through negative sequence impedance ${\mathrm{Z}}_2$.

${\mathrm{I}}_{\mathrm{a}}$,${\mathrm{I}}_{\mathrm{b}}$ and${\mathrm{I}}_{\mathrm{c}}$ are phase currents.

${\mathrm{V}}_{\mathrm{F}}$Is the pre-fault bus voltage in the positive-sequence network.

Determine the formula of per unit reactance of generator, motor and neutral reactor.

Write the formula of sub transient fault current.

$\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{a}}\\ {\mathbf{I}}_{\mathbf{b}}\\ {\mathbf{I}}_{\mathbf{c}}\end{array}\right]=\left[\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{2}}\end{array}\right]$ …… (1)

Here,${\mathbf{I}}_{\mathbf{0}}$ is a zero sequence current,${\mathbf{I}}_{\mathbf{1}}$ is the current flowing through impedance${\mathbf{Z}}_{\mathbf{1}}$ and${\mathbf{I}}_{\mathbf{2}}$ is the current flowing through impedance ${\mathbf{Z}}_{\mathbf{2}}$.

Write the formula ofcontributions to the fault current from each generator.

$\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{a}}\\ {\mathbf{I}}_{\mathbf{G}\mathbf{2}\mathbf{-}\mathbf{b}}\\ {\mathbf{I}}_{\mathbf{G}\mathbf{3}\mathbf{-}\mathbf{c}}\end{array}\right]=\left[\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{2}}\end{array}\right]$ …… (2)

Here, ${\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{0}}$is the fault current by zero sequence generator $\mathbf{G}\mathbf{1}$, role="math" localid="1656392539847" ${\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{1}}$is fault current by positive sequence generator role="math" localid="1656392563995" $\mathbf{G}\mathbf{1}$and role="math" localid="1656392579832" ${\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{-}\mathbf{2}}$is fault current by negative sequence generator $\mathbf{G}\mathbf{1}$.

Write the formula ofcontributions to the fault current from each transformer.

$\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{a}}\\ {\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{b}}\\ {\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{c}}\end{array}\right]=\left[\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{2}}\end{array}\right]$ …… (3)

Here, ${\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{0}}$is the fault current by zero sequence transform, ${\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{1}}$is the fault current by positive sequence transformer $\mathbf{T}\mathbf{1}$, ${\mathbf{I}}_{\mathbf{T}\mathbf{1}\mathbf{-}\mathbf{2}}$is the fault current by negative sequence transformer $\mathbf{T}\mathbf{1}$.

 Determine the sub transient fault current, contributions to the fault current from each generator and contributions to the fault current from each transformer.

Refer to figure 7.15 in the textbook of five bus power system.

Determine the per-unit positive sequence reactances for generator 1.

$\begin{array}{rcl}{\mathrm{X}}_1& =& {\mathrm{X}}_{\mathrm{d}}^{*}\\ & =& 0.2{\left(\frac{12}{10}\right)}^2\left(\frac{100}{50}\right)\\ & =& 0.576\mathrm{p}.\mathrm{u}.\end{array}$

Determine the per-unit negative sequence reactances for generator 1.

${\mathrm{X}}_2=0.576\mathrm{p}.\mathrm{u}.$

Determine the per-unit zero sequence reactances for generator 1.

$\begin{array}{rcl}{\mathrm{X}}_0& =& 0.1{\left(\frac{12}{10}\right)}^2\left(\frac{100}{50}\right)\\ & =& 0.288\mathrm{p}.\mathrm{u}.\end{array}$

Determine the per-unit positive sequence reactances for generator.

$\begin{array}{rcl}{\mathrm{X}}_1& =& {\mathrm{X}}_{\mathrm{d}}^{*}\\ & =& 0.2\mathrm{p}.\mathrm{u}.\end{array}$

Determine the per-unit negative sequence reactances for generator.

${\mathrm{X}}_2=0.23\mathrm{p}.\mathrm{u}.$

Determine the per-unit zero sequence reactances for generator.

${\mathrm{X}}_0=0.1$

Determine the per-unit positive sequence reactances for transformer.

localid="1656394643848" $\begin{array}{rcl}{\mathrm{X}}_{\mathrm{T}1}& =& 0.10\left(\frac{100}{50}\right)\\ & =& 0.2\mathrm{p}.\mathrm{u}.\\ & & \end{array}$

${\mathrm{X}}_{\mathrm{T}2}=0.10\mathrm{p}.\mathrm{u}.$

Determine the base impedance value.

$\begin{array}{rcl}{\mathrm{Z}}_{\mathrm{base}}& =& \frac{{\left(1.38\times {10}^3\right)}^2}{100\times {10}^6}\\ & =& 190.44\mathrm{\Omega }\end{array}$

Determine the per unit positive sequence line reactance equal to negative sequence line reactances.

$\begin{array}{rcl}{\mathrm{X}}_1& =& {\mathrm{X}}_2\\ & =& \left(\frac{40}{190.44}\right)\\ & =& 0.210\mathrm{p}.\mathrm{u}.\\ & & \end{array}$

Determine the zero sequence reactance is.

$\begin{array}{rcl}{\mathrm{X}}_0& =& \left(\frac{100}{190.44}\right)\\ & =& 0.5251\mathrm{p}.\mathrm{u}.\end{array}$

Draw the per unit zero-sequence network.

Figure 1

Draw the per unit positive sequence network.

Figure 2

Draw the per unit negative sequence network.

Figure 3

Now assume bus 1 is faulty bus.

Calculate the zero sequence equivalent impedance from figure 1.

$\begin{array}{rcl}{\mathrm{Z}}_0& =& \mathrm{j}0.2+\mathrm{j}0.5251+\mathrm{j}0.5251+\mathrm{j}0.1\\ & =& \mathrm{j}1.3502\mathrm{p}.\mathrm{u}.\end{array}$

Draw zero sequence Thévenin equivalent circuit is shown in figure.

Figure 4

Draw the Thévenin equivalent circuit for zero sequence is shown in figur

Figure 5

Now assume bus 1 is faulty bus.

Calculate the negative sequence equivalent impedance from figure 3.

$\begin{array}{rcl}{\mathrm{Z}}_2& =& \left(\mathrm{j}0.576\right)\mathrm{P}\left(\mathrm{j}0.2+\mathrm{j}0.21+\mathrm{j}0.21+\mathrm{j}0.1+0.23\right)\\ & =& \left(\mathrm{j}0.576\right)\mathrm{P}\left(\mathrm{j}0.95\right)\\ & =& \frac{\left(\mathrm{j}0.576\right)\left(\mathrm{j}0.95\right)}{\mathrm{j}0.576+\mathrm{j}0.95}\\ & =& \mathrm{j}0.3586\mathrm{p}.\mathrm{u}.\end{array}$

Draw the Thévenin equivalent circuit for negative sequence is shown in figure 5

Figure 6

Draw the Thévenin equivalent circuit for negative sequence is shown in figure

Figure 7

Calculate the positive sequence equivalent impedance from figure 2.

$\begin{array}{rcl}{\mathrm{Z}}_1& =& \left(\mathrm{j}0.576\right)\mathrm{P}\left(\mathrm{j}0.2+\mathrm{j}0.21+\mathrm{j}0.21+\mathrm{j}0.1+0.2\right)\\ & =& \left(\mathrm{j}0.576\right)\mathrm{P}\left(\mathrm{j}0.92\right)\\ & =& \frac{\left(\mathrm{j}0.576\right)\left(\mathrm{j}0.92\right)}{\mathrm{j}0.576+\mathrm{j}0.92}\\ & =& \mathrm{j}0.3542\mathrm{p}.\mathrm{u}.\end{array}$

Draw the Thévenin equivalent circuit for positive sequence for bus is shown in figure.

Figure 8

Draw the Thévenin equivalent circuit for positive sequence is shown in figure.

Figure 9

Draw the bolted line-to-line fault values of impedance bus and is shown in figure 10.

Figure 10

Determine the sequence current through each impedance.

$\begin{array}{rcl}{\mathrm{I}}_1& =& -{\mathrm{I}}_2\\ & =& \frac{{\mathrm{V}}_{\mathrm{F}}}{{\mathrm{Z}}_1+{\mathrm{Z}}_2}\\ & =& \frac{1\angle 0^0}{\mathrm{j}0.3542+\mathrm{j}0.3586}\\ & =& -\mathrm{j}1.1403\mathrm{p}.\mathrm{u}.\end{array}$

Determine the sub transient fault current.

Substitute $0$for ${\mathrm{I}}_0$, $-\mathrm{j}1.403$for ${\mathrm{I}}_1$and $\mathrm{j}1.403$for ${\mathrm{I}}_2$into equation (1).

$\begin{array}{rcl}\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{c}}\end{array}\right]& =& \left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}0\\ -\mathrm{j}1.403\\ \mathrm{j}1.403\end{array}\right]\\ & =& \left[\begin{array}{c}0\\ 2.43\angle {180}^0\\ 243\angle 0^0\end{array}\right]\mathrm{per}\mathrm{unit}\end{array}$

Change in $\mathrm{kA}$.

localid="1656396808984" $\begin{array}{rcl}\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{c}}\end{array}\right]& =& \left[\begin{array}{c}0\\ 2.43\angle {180}^0\\ 2.43\angle 0^0\end{array}\right]\left(\frac{100}{10\sqrt{3}}\right)\mathrm{kA}\\ & =& \left[\begin{array}{c}0\\ 14.03\angle {180}^0\\ 14.03\angle 0^0\end{array}\right]\mathrm{kA}\end{array}$

Therefore, the sub transient fault current $\left[\begin{array}{c}0\\ 2.43\angle {180}^0\\ 2.43\angle 0^0\end{array}\right]\mathrm{per}\mathrm{unit}$and $\left[\begin{array}{c}0\\ 14.03\angle {180}^0\\ 14.03\angle 0^0\end{array}\right]\mathrm{kA}$.

Determine the contribution of fault current by zero sequence for generator$\mathrm{G}1$is ${\mathrm{I}}_{\mathrm{G}1-0}=0$

Determine the contribution of fault current by zero sequence for transformer $\mathrm{T}1$is ${\mathrm{I}}_{\mathrm{T}1-0}=0\mathrm{per}\mathrm{unit}$.

Draw the circuit diagram of positive sequence network.

Figure 11

Determine the currents from Generator$\mathrm{G}1$as.

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{G}1-1}& =& \left(-\mathrm{j}1.403\right)\left(\frac{0.92}{0.92+.576}\right)\\ & =& \left(-\mathrm{j}1.403\right)\left(0.615\right)\\ & =& -\mathrm{j}0.8628\mathrm{per}\mathrm{unit}\end{array}$

Determine the currents from Transformer $\mathrm{T}1$as.

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{T}1-1}& =& \left(-\mathrm{j}1.403\right)\left(\frac{.576}{0.92+.576}\right)\\ & =& -\mathrm{j}0.54\mathrm{per}\mathrm{unit}\end{array}$

Draw the circuit diagram of negative sequence network.

Figure 12

Determine the current of Generator $\mathrm{G}1$.

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{G}1-2}& =& -\mathrm{j}1.403\left(\frac{0.95}{0.95+.76}\right)\\ & =& \left(\mathrm{j}1.403\right)\left(0.622\right)\\ & =& \mathrm{j}0.872\mathrm{per}\mathrm{unit}\end{array}$

Determine the current of Transformer $\mathrm{T}1$.

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{T}1-2}& =& -\mathrm{j}0.4847\left(\frac{.576}{.576+0.95}\right)\\ & =& \left(\mathrm{j}1.403\right)\left(0.377\right)\\ & =& \mathrm{j}0.529\mathrm{per}\mathrm{unit}\end{array}$

Determine the contribution to the fault current from each generator.

Substitute $0$for ${\mathrm{I}}_{\mathrm{G}1-0}$, $-\mathrm{j}0.8628$for ${\mathrm{I}}_{\mathrm{G}1-1}$and $\mathrm{j}0.872$for ${\mathrm{I}}_{\mathrm{G}1-2}$into equation (2).

$\begin{array}{rcl}\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{G}1-\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{G}1-\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{G}1-\mathrm{c}}\end{array}\right]& =& \left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}0\\ -\mathrm{j}0.8628\\ \mathrm{j}0.872\end{array}\right]\\ & =& \left[\begin{array}{c}\mathrm{j}0.0092\\ 1.502\angle -179.8^0\\ 1.502\angle -0.2^0\end{array}\right]\mathrm{per}\mathrm{unit}\end{array}$

Change in $\mathrm{kA}$.

$\begin{array}{rcl}\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{G}1-\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{G}1-\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{G}1-\mathrm{c}}\end{array}\right]& =& \left[\begin{array}{c}\mathrm{j}0.0092\\ 1.502\angle -179.8^0\\ 1.502\angle -0.2^0\end{array}\right]\left(\frac{100}{10\sqrt{3}}\right)\mathrm{kA}\\ & =& \left[\begin{array}{c}0.053\angle {90}^0\\ 8.67\angle -179.8^0\\ 8.67\angle -0.2^0\end{array}\right]\mathrm{kA}\end{array}$

Therefore, the value of contributions to the fault current from each generator is $\left[\begin{array}{c}3.463\angle -{90}^0\\ 1.731\angle 89.4^0\\ 1.731\angle 90.6^0\end{array}\right]\mathrm{kA}$.

Determine the contribution to the fault current from each transformer.

Substitute $0$for ${\mathrm{I}}_{\mathrm{T}1-0}$, $-\mathrm{j}0.54$for ${\mathrm{I}}_{\mathrm{T}1-1}$and $\mathrm{j}0.529$for ${\mathrm{I}}_{\mathrm{T}1-2}$into equation (3).

$\begin{array}{rcl}\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{T}1-\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{T}1-\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{T}1-\mathrm{c}}\end{array}\right]& =& \left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}0\\ -\mathrm{j}0.54\\ \mathrm{j}0.529\end{array}\right]\\ & =& \left[\begin{array}{c}-\mathrm{j}0.011\\ 0.925\angle 179.{66}^0\\ 0.925\angle 0.{34}^0\end{array}\right]\mathrm{per}\mathrm{unit}\end{array}$

Change in $\mathrm{kA}$.

$\begin{array}{rcl}\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{T}1-\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{T}1-\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{T}1-\mathrm{c}}\end{array}\right]& =& \left[\begin{array}{c}-\mathrm{j}0.011\\ 0.925\angle 179.{66}^0\\ 0.925\angle 0.{34}^0\end{array}\right]\left(\frac{100}{10\sqrt{3}}\right)\mathrm{kA}\\ & =& \left[\begin{array}{c}0.063\angle -{90}^0\\ 5.34\angle 179.{66}^0\\ 5.34\angle 0.{34}^0\end{array}\right]\mathrm{kA}\end{array}$.

Therefore, the value of contributions to the fault current from each transformer is $\left[\begin{array}{c}0.063\angle -{90}^0\\ 5.34\angle 179.{66}^0\\ 5.34\angle 0.{34}^0\end{array}\right]\mathrm{kA}$.