Question

For the system of Problem 9.12, compute the fault current and voltages at the fault for the following faults at bus 3: (a) a bolted single line-to ground fault, (b) a bolted line-to-line fault, (c) a bolted double line-to ground fault. Also, for the single line-to-ground fault at bus 3, determine the currents and voltages at the terminals of generators G1 and G2.

Short answer

(a)

The value of sub-transient fault current from a bolted single line-to-ground fault is.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}-j5.475\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$

The value of fault voltages from single line-to-line voltages at the faulted bus is.

$\left[\begin{array}{c}{V^{\text{'}\text{'}}}_a\\ {V^{\text{'}\text{'}}}_b\\ {V^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}0\\ 0.5906\\ 0.5906\end{array}\right]\text{\hspace{0.33em}pu}$

(b)

The value of sub-transient fault current from a bolted single line-to-line fault is.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}0\\ 4.962\angle 180°\\ 4.962\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}$

The value of fault voltages from a single line-to-ground fault system is.

$\left[\begin{array}{c}{V}_{ag}\\ V_{bg}\\ V_{cg}\end{array}\right]=\left[\begin{array}{c}1\\ 0.5\angle 180°\\ 0.5\angle 180°\end{array}\right]\text{\hspace{0.33em}pu}$

(c)

The value of sub-transient fault current from a bolted double line-to-ground fault system is$5.239\angle 90°\text{\hspace{0.33em}pu}$.

The value of fault voltages from a double line to ground fault $\left[\begin{array}{c}{V}_{\text{ag}}\\ V_{\text{bg}}\\ V_{\text{cg}}\end{array}\right]=\left[\begin{array}{c}0.041\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$.

Step-by-step solution

Write the given data from the question.

As reference Problem 9.12

A given value of Synchronous generators is:

$\begin{array}{l}G1100\text{MVA}25\text{kV}{X}_1=X_2=0.2X_0=0.05\\ G2100\text{MVA}13.8\text{kV}{X}_1=X_2=0.2X_0=0.05\end{array}$

A given value of Transformers is:

$\begin{array}{l}\text{T1100MVA25/230kV}{X}_1=X_2=X_0=0.05\\ \text{T2100MVA13.8/230kV}{X}_1=X_2=X_0=0.05\end{array}$

Transmission lines:

$\begin{array}{l}\text{TL12100MVA230kV}{X}_1=X_2=0.1X_0=0.3\\ \text{TL13100MVA230kV}{X}_1=X_2=0.1X_0=0.3\\ \text{TL23100MVA230kV}{X}_1=X_2=0.1X_0=0.3\end{array}$

The base MVA value of transmission line is 100.

The base voltage of transmission line is 230-kV.

Determine the formula of line-to-line fault and line-to-ground fault.

Write the formula of sub-transient fault current from a bolted single line-to-ground fault.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (1)

Here, ${\mathbf{I}}_0\mathbf{,}{\mathbf{I}}_1\&{\mathbf{I}}_2$ and are phase sequence currents.

Write the formula of fault voltages from line-to-line fault system.

${V^{\text{'}\text{'}}}_a=\frac{{V^{\text{'}\text{'}}}_{\mathrm{ag}}}{\sqrt{3}}$ …… (2)

Here, ${V^{\text{'}\text{'}}}_{\mathrm{ag}}$is a single line-to-ground voltage.

${V^{\text{'}\text{'}}}_b=\frac{{V^{\text{'}\text{'}}}_{\mathrm{bg}}}{\sqrt{3}}$ …… (3)

Here, ${V^{\text{'}\text{'}}}_{\mathrm{bg}}$is a single line-to-ground voltage.

${V^{\text{'}\text{'}}}_c=\frac{{V^{\text{'}\text{'}}}_{\mathrm{cg}}}{\sqrt{3}}$ …… (4)

Here, ${V^{\text{'}\text{'}}}_{\mathrm{cg}}$is a single line-to-ground voltage.

Write the formula of fault voltages from a bolted single line to ground voltages at the faulted bus.

${\mathbf{V}}_{\mathrm{ag}}={\mathbf{V}}_0\mathbf{+}{\mathbf{V}}_1\mathbf{+}{\mathbf{V}}_2$ …… (5)

Here, ${\mathbf{V}}_0\mathbf{,}\mathbf{}{\mathbf{V}}_1\mathbf{}\&\mathbf{}{\mathbf{V}}_2$are sequence voltages.

${\mathbf{V}}_{\mathrm{bg}}={\mathbf{V}}_0\mathbf{+}{\mathbf{a}}^2{\mathbf{V}}_1\mathbf{+}{\mathbf{aV}}_2$ …… (6)

Here, ${\mathbf{V}}_0\mathbf{,}\mathbf{}{\mathbf{V}}_1\mathbf{}\&\mathbf{}{\mathbf{V}}_2$are sequence voltages.

role="math" localid="1656146730750" ${\mathbf{V}}_{\mathrm{cg}}={\mathbf{V}}_0\mathbf{+}{\mathbf{aV}}_1\mathbf{+}{\mathbf{a}}^2{\mathbf{V}}_2$ …… (7)

Here, ${\mathbf{V}}_0\mathbf{,}\mathbf{}{\mathbf{V}}_1\mathbf{}\&\mathbf{}{\mathbf{V}}_2$ are sequence voltages.

Write the formula of fault current a bolted double line-to-ground fault.

${I^{\text{'}\text{'}}}_a=\mathbf{3}{\mathbf{I}}_0$ …… (8)

Here,${\mathbf{I}}_0$ is sequence current.

(a) Determine the sub-transient fault current and fault voltages from a bolted single line to ground fault.

Reference as a Problem 9.12 from the text book.

Draw a circuit diagram of per unit positive-sequence network.

Convert delta 123 into star, the equivalent star reactance is,

$\begin{array}{c}{X}_1=\frac{\left(j0.1\right)\left(j0.1\right)}{j0.1+j0.1+j0.1}\\ =j0.033\end{array}$

Draw a circuit diagram of delta-to-star transformation of positive-sequence network.

Calculate the equivalent impedance of the positive sequence network between the terminals 3 and ground.

$\begin{array}{c}{Z}_1=\left[\left(j0.25+j0.033\right)\parallel \left(j0.25+j0.033\right)\right]+j0.033\\ =\frac{\left(j0.283\right)\left(j0.283\right)}{\left(j0.283\right)\left(j0.283\right)}+j0.033\\ =j0.1415+j0.033\\ =j0.1745\end{array}$

Draw a circuit diagram of negative sequence network is equal to positive sequence network; they not present any source so they show negative sequence reactances.

Draw a circuit diagram of equivalent impedance of negative sequence network is similar to positive sequence network.

${\text{Z}}_{\text{2}}=j0.1745$

Draw a circuit diagram of zero-sequence network from generators G1 and G2 are grounded through a neutral reactance, which is three times the neutral reactance $j0.03$, zero-sequence generator reactance is also included as $j0.09$.

Convert delta 123 into star, the equivalent star reactance is,

$\begin{array}{c}{X}_1=\frac{\left(j0.3\right)\left(j0.3\right)}{j0.3+j0.3+j0.3}\\ =j0.1\end{array}$

Draw a circuit diagram of zero-sequence network after delta-to-star transformation.

Calculate the equivalent impedance of the zero sequence network between the terminals 3 and ground.

$\begin{array}{c}{Z}_0=\left[\left(j0.09+j0.05+j0.05+j0.1\right)\parallel \left(j0.05+j0.1\right)\right]+j0.1\\ =\left[\left(j0.29\right)\parallel \left(j0.15\right)\right]+j0.1\\ =\frac{\left(j0.29\right)\left(j0.15\right)}{j0.29+j0.15}+j0.1\\ =j0.099+j0.1\\ =j0.199\end{array}$

A single line-to-ground fault occurs in a system through fault impedance at F.

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1=-I_2\\ =\frac{E_{\text{g}}}{\left({\text{Z}}_0+{\text{Z}}_1+{\text{Z}}_2\right)}\\ =\frac{1}{j\left(0.199+0.1745+0.1745\right)}\\ =\frac{1}{j0.548}\end{array}$

As further solve as

$\begin{array}{c}{I}_0=I_1=I_2\\ =-j1.825\end{array}$

Determine the value of sub-transient fault current from a bolted single line-to-ground fault.

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}-j1.825\\ -j1.825\\ -j1.825\end{array}\right]\\ =\left[\begin{array}{c}\left(-j1.825\right)\left(3\right)\\ \left(1+1\angle 120°+1\angle 120°\right)\left(-j1.825\right)\\ \left(1+1\angle 120°+1\angle 120°\right)\left(-j1.825\right)\end{array}\right]\\ =\left[\begin{array}{c}-j5.475\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Since, the sub-transient fault current from a bolted single line-to-ground fault in $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$pu is $\left[\begin{array}{c}-j5.475\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$.

Determine the zero-sequence component of voltage at fault current.

$\begin{array}{c}{V}_0=-I_0{Z}_0\\ =-\left(-j1.825\right)\left(j0.199\right)\\ =-0.363\text{\hspace{0.33em}pu}\end{array}$

Determine the positive-sequence component of voltage at fault current.

$\begin{array}{c}{V}_1=1-I_1{Z}_1\\ =1-\left(-j1.825\right)\left(j0.1745\right)\\ =0.682\text{\hspace{0.33em}pu}\end{array}$

Determine the negative-sequence component of voltage at fault current.

$\begin{array}{c}{V}_2=-I_2{\text{Z}}_{\text{2}}\\ =-\left(j1.825\right)\left(j0.1745\right)\\ =-0.318\text{\hspace{0.33em}pu}\end{array}$

Determine thevalue of fault voltages from a bolted single line to ground voltages at the faulted bus.

Substitute for -0.363 $V_0$, 0.682 for $V_1$and 0.318 for $V_2$into equation (5).

$\begin{array}{c}{V}_{\text{ag}}=-0.363+0.682-0.318\\ =0\text{\hspace{0.33em}pu}\end{array}$

Substitute -0.363 for V_o, 0.682 for V₁, $1\angle -120°$for $a^2$, $1\angle 120°$ for a and -0.318 for V₂into equation (6).

$\begin{array}{c}{V}_{\text{bg}}=-0.363+\left(0.682\right)\left(1\angle -120°\right)+\left(-0.318\right)\left(1\angle 120°\right)\\ =1.023\angle -122.2°\text{\hspace{0.33em}pu}\end{array}$

Substitute -0.363 for V_o, 0.682 for V₁, $1\angle -120°$ for a², $1\angle 120°$ for a and -0.318 for V₂into equation (7).

$\begin{array}{c}{V}_{\text{cg}}=-0.363+\left(0.682\right)\left(1\angle 120°\right)+\left(-0.318\right)\left(1\angle -120°\right)\\ =0.023\angle 122.2°\end{array}$

Therefore, the line to ground voltages$\left[\begin{array}{c}{V}_{\text{ag}}\\ V_{\text{bg}}\\ V_{\text{cg}}\end{array}\right]$at the fault bus is $\left[\begin{array}{c}0\\ 1.023\angle -122.2°\\ 1.023\angle 122.2°\end{array}\right]\text{\hspace{0.33em}pu}$.

Now, determine thevalue of fault voltages from a bolted single line-to-line fault system.

Substitute 0 for ${V^{\text{'}\text{'}}}_{ag}$into equation (2).

$\begin{array}{c}{V^{\text{'}\text{'}}}_a=\frac{0}{\sqrt{3}}\\ =0\text{\hspace{0.33em}pu}\end{array}$

Substitute 1.023 for ${V^{\text{'}\text{'}}}_{bg}$into equation (3).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{bg}=\frac{1.023}{\sqrt{3}}\\ =0.5906\text{\hspace{0.33em}pu}\end{array}$

Substitute 1.023 for ${V^{\text{'}\text{'}}}_{cg}$into equation (8).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{cg}=\frac{1.023}{\sqrt{3}}\\ =0.5906\text{\hspace{0.33em}pu}\end{array}$

Hence, the line-to-line voltages$V_a\text{'}\text{'},V_b\text{'}\text{'}\&V_c\text{'}\text{'}$are 0 pu, 0.5906 pu and 0.5906 pu.

(b) Determine the sub-transient fault current and fault voltages from bolted line to line fault.

A bolted line-to-line fault occurs in a system.

The fault current in a zero-sequence component for line to line fault is zero.

Determine the fault current in the positive and negative sequence domain.

$\begin{array}{c}{I}_1=I_2\\ =\frac{E_{\text{g}}}{j\left({\text{Z}}_{\text{1}}+{\text{Z}}_{\text{2}}\right)}\\ =\frac{1}{j\left(0.1745+0.1745\right)}\\ =-j2.865\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current from line to line fault in the system.

Substitute $-j2.865$for $I_1$and $j2.865$for $I_2$into equation (1)

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j2.865\\ j2.865\end{array}\right]\\ =\left[\begin{array}{c}0-j2.865+j2.865\\ 0+\left(-j2.865\right)\left(1\angle -120\right)+\left(j2.865\right)\left(1\angle 120\right)\\ 0+\left(-j2.865\right)\left(1\angle 120\right)+\left(j2.865\right)\left(1\angle -120\right)\end{array}\right]\\ =\left[\begin{array}{c}0\\ 4.962\angle 180°\\ 4.962\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Hence, the sub-transient fault currents in phase of a system $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$is $\left[\begin{array}{c}0\\ 4.962\angle 180°\\ 4.962\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}$

Determine the zero-sequence component of voltage at fault current.

$\begin{array}{c}{V}_0=-I_0{Z}_0\\ =-\left(0\right)\left(j0.199\right)\\ =0\text{\hspace{0.33em}pu}\end{array}$

Determine the positive-sequence component of voltage at fault current.

$\begin{array}{c}{V}_1=1-I_1{Z}_1\\ =1-\left(-j2.865\right)\left(j0.1745\right)\\ =0.5\text{\hspace{0.17em}pu}\end{array}$

Determine the negative-sequence component of voltage at fault current.

$\begin{array}{c}{V}_2=-I_2{\text{Z}}_{\text{2}}\\ =-\left(-j2.865\right)\left(j0.1745\right)\\ =-0.5\text{\hspace{0.33em}pu}\end{array}$

Determine the value of fault voltages from line to line voltage at the faulted bus.

Substitute 0 for V_o, 0.5 for V₁ and 0.5 for V₂ into equation (5).

$\begin{array}{c}{V}_{\text{ag}}=0+0.5+0.5\\ =1\text{\hspace{0.33em}pu}\end{array}$

Substitute 0 for V_o, 0.5 for V₁, $1\angle -120°$ for a², $1\angle 120°$ for a and 0.5 for V₂ into equation (6).

$\begin{array}{c}{V}_{\text{bg}}=0+\left(0.5\right)\left(1\angle -120°\right)+\left(0.5\right)\left(1\angle 120°\right)\\ =0.5\angle 180°\text{\hspace{0.33em}pu}\end{array}$

Substitute 0 for V_o, 0.5 for V₁, $1\angle -120°$for a², $1\angle 120°$for a and 0.5 for V₂ into equation (6).

$\begin{array}{c}{V}_{\text{cg}}=0+\left(0.5\right)\left(1\angle 120°\right)+\left(0.5\right)\left(1\angle -120°\right)\\ =0.5\angle 180°\text{\hspace{0.33em}pu}\end{array}$

Hence, the line to ground voltages $\left[\begin{array}{c}{V}_{\text{ag}}\\ V_{\text{bg}}\\ V_{\text{cg}}\end{array}\right]$at the fault bus is $\left[\begin{array}{c}1\\ 0.5\angle 180°\\ 0.5\angle 180°\end{array}\right]\text{\hspace{0.33em}pu}$.

(c) Determine the sub-transient fault current and fault voltages from a bolted double line to ground fault.

When a bolted double line-to-ground fault occurs in a system.

Draw a circuit diagram for the system.

Determine the positive-sequence current component in a system.

$I_1=\frac{E_R}{j\left(Z_1+\left(\frac{Z_2{Z}_0}{Z_2+Z_0}\right)\right)}$

Here, E_R is required voltage, Z₁, Z₂ and Z₀ is sequence impedance.

Substitute 1 for E_R , 0.1745 for Z₁, 0.1745 for Z₂ and 0.199 for Z₀ into above equation.

$\begin{array}{c}{I}_1=\frac{1}{j\left(0.1745+\left(\frac{\left(0.1745\right)\left(0.199\right)}{0.1745+0.199}\right)\right)}\\ =\frac{1}{j0.2675}\\ =-j3.738\text{\hspace{0.33em}pu}\end{array}$

Determine the negative sequence current domain in the network.

$\begin{array}{c}{I}_2=\left(-I_1\right)\frac{Z_0}{Z_2+Z_0}\\ =\left(j3.738\text{\hspace{0.33em}pu}\right)\left(\frac{j0.199}{j0.1745+j0.199}\right)\\ =\left(j3.738\text{\hspace{0.33em}pu}\right)\left(0.533\right)\end{array}$

Determine the zero-sequence current domain in the network.

$\begin{array}{c}{I}_0=\left(-I_1\right)\frac{Z_2}{Z_2+Z_0}\\ =\left(j3.738\text{\hspace{0.33em}pu}\right)\left(\frac{j0.1745}{j0.1745+j0.199}\right)\\ =\left(j3.738\text{\hspace{0.33em}pu}\right)\left(0.467\right)\\ =j1.746\text{\hspace{0.17em}pu}\end{array}$

Determine thesub-transient fault current.

Substitute $j1.746$ for I₀ into equation (8).

$\begin{array}{c}{I}_n=3\left(j1.746\right)\\ =5.239\angle 90°\text{\hspace{0.33em}pu}\end{array}$

Since, the sub-transient fault current in the system is $I_n=5.239\angle 90°\text{\hspace{0.33em}pu}$.

Determine the value of fault voltages from zero-sequence component at fault current.

$V_0=-I_0{Z}_0$ …… (9)

Here,$I_0$ is zero sequence current and Z₀ is zero sequence impedance.

Substitute $j1.746$for $I_0$and $j0.199$for $Z_0$into equation (9).

$\begin{array}{c}{V}_0=-\left(j1.746\right)\left(j0.199\right)\\ =0.347\text{\hspace{0.33em}pu}\end{array}$

Determine the value of fault voltages from positive-sequence component at fault current.

$V_1=1-I_1{Z}_1$ …… (10)

Here, $I_1$is positive sequence current and $Z_1$is positive sequence impedance.

Substitute $-j3.738$for $I_1$and $j0.1745$for $Z_1$into equation (10).

$\begin{array}{c}{V}_1=1-\left(-j3.738\right)\left(j0.1745\right)\\ =0.347\text{\hspace{0.33em}pu}\end{array}$

Determine the value of fault voltages from negative-sequence component at fault current.

$V_2=-I_2{Z}_2$ …… (11)

Here, $I_2$is negative sequence current and $Z_2$is negative sequence impedance.

Substitute role="math" localid="1656153786412" $j1.99$for $I_2$and $j0.1745$for $Z_2$into equation (11)

$\begin{array}{c}{V}_2=-\left(j1.99\right)\left(j0.1745\right)\\ =0.347\text{\hspace{0.33em}pu}\end{array}$

Now, determine thevalue of fault voltages from double line-to-ground fault system.

Substitute 0.347 for V_o, 0.347 for V₁ and 0.347 for V₂ into equation (5)

$\begin{array}{c}{V}_{\text{ag}}=0.347+0.347+0.347\\ =1.041\text{\hspace{0.33em}pu}\end{array}$

Substitute 0.347 for V_o, 0.347 for V₁, $1\angle -120°$ for a², $1\angle 120°$ for a and 0.347 for V₂ into equation (6).

$\begin{array}{c}{V}_{\text{bg}}=0.347+\left(0.347\right)\left(1\angle -120°\right)+\left(0.347\right)\left(1\angle 120°\right)\\ =0°\text{\hspace{0.33em}pu}\end{array}$

Substitute 0.347 for V_o, 0.347 forV₁, $1\angle -120°$for a², $1\angle 120°$for a and 0.347 forV₂ into equation (7).

$\begin{array}{c}{V}_{\text{cg}}=0.347+\left(0.347\right)\left(1\angle 120°\right)+\left(0.347\right)\left(1\angle -120°\right)\\ =0°\text{\hspace{0.33em}pu}\end{array}$

Hence, the line to ground voltages$\left[\begin{array}{c}{V}_{\text{ag}}\\ V_{\text{bg}}\\ V_{\text{cg}}\end{array}\right]$ at the fault bus is $\left[\begin{array}{c}1.041\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$.