Question

Question: Determine the subtransient fault current in per-unit and in kA during a bolted three-phase fault at the fault bus selected in Problem 9.9.

Short answer

Answer

The sub transient fault current in per unit and kA are $\left[\begin{array}{c}2.823\angle -90°\\ 2.823\angle 150°\\ 2.823\angle 30°\end{array}\right]pu$and $\left[\begin{array}{c}16.30\angle -90°\\ 16.30\angle 150°\\ 16.30\angle 30°\end{array}\right]kA$respectively.

Step-by-step solution

Write the given data from the question:

Write the rating of the generators.

The power rating of generatorG₁ is 50 MVA, the voltage rating is 15 KV, the reactance and are X_d''equal X₂ to and the value of X₀ is 0.10 unit.

The power rating of generatorG₂ is 100 MVA , the voltage rating is 12 KV, the reactance is ,is and X_d''equal X₂ to and the value of X₀is 0.10 unit. .

The power rating of transformer T₁is 100 MVA, the voltage rating is 10KV Y/138 KV Y, and the value X₀is 0.10 per unit. .

The power rating of transformer T₂is 50 MVA, the voltage rating is 15KV Y/138 KV Y, and the value X₀is 0.10 per unit. .

The reactance of each line,X₁ is 40$\Omega$ andX₀ is 100$\Omega$.

The base MVA, S_base=100 MVA

The base voltage for generator side, V_base=15 KV

The prefault fault voltage,V_F = 1 per unit

Determine the formulas to calculate the reactance of the zero, positive and negative sequence network.

The equation to calculate the reactance on the new base value is given as follows.

${\left(X_{pu}\right)}_{new}={\left(X_{pu}\right)}_{old}\frac{{\left(K{V}_{base}\right)}_{old}}{{\left(K{V}_{base}\right)}_{new}}\frac{{\left(MV{A}_{base}\right)}_{new}}{{\left(MV{A}_{base}\right)}_{old}}$ …… (1)

Here (X_pu)_new is the per unit value on new base values, (X_pu)_old is the per unit value on old base values,(KV_base)_old it the old voltage rating,(KV_base)_new is the new voltage rating,(MVA_base)_old is the new power rating and (MVA_base)_new is the old power rating.

The equation to calculate the base impedance is given as follows.

$Z_{base}=\frac{K{V}^2}{MVA}$ …… (2)

Here, Z_base is the base value, is the power rating, is the voltage rating.

The equation to calculate the new per unit impedance of the transmission line is given as follows.

$I_1=\frac{V_F}{Z_1}$ …… (3)

The equation to calculate current in the positive sequence circuit is given as follows.

${\left(X\right)}_{new}=\frac{X_{old}}{Z_{base}}$ …… (4)

Here is the positive sequence impedance and is the fault voltage.

The equation to calculate the base current is given as follows.

$I_{base}=\frac{S_{base}}{\sqrt{3}{V}_{base}}$ …… (5)

The equation to calculate sub transient current is given as follows.

$\left[\begin{array}{c}{I\text{'}\text{'}}_a\\ {I\text{'}\text{'}}_b\\ {I\text{'}\text{'}}_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (6)

Here and are the sub transient current, and are the sequence current.

The equation to calculate per quantities is given as follows.

$I_{pu}=\frac{I_{base}}{I\text{'}\text{'}}$ …… (7)

Here is the sub transient current.

 Step 3: Calculate the reactance of the zero, positive and negative sequence network.

Calculate the new value of the reactance of generator 1.

For positive and negative sequence reactance.

Substitute 50 MVA for (MVA_base)_old , 12 KV for(KV_base)_old , 0.20pu for (X₁)_odd, 100 MVA for (MVA_base)_new , 10 KV for(KV_base)_new into equation (1).

$$\begin{gathered} {\left(X_1\right)}_{new}={\left(X_2\right)}_{new} \\ {\left(X_1\right)}_{new}={\left({X\text{'}\text{'}}_d\right)}_{new} \\ {\left(X_1\right)}_{new}=0.20\times \frac{{12}^2}{{10}^2}\times \frac{100}{50} \\ {\left(X_1\right)}_{new}=0.576\text{pu} \end{gathered}$$

For zero-sequence reactance.

Substitute 50 MVA for (MVA_base)_old , 12 KV for(KV_base)_old , 0.10pu for (X₁)_odd, 100 MVA for (MVA_base)_new , 10 KV for(KV_base)_new into equation (1).

$$\begin{gathered} {\left(X_0\right)}_{new}=0.10\times \frac{{12}^2}{{10}^2}\times \frac{100}{50} \\ {\left(X_0\right)}_{new}=0.288\text{pu} \end{gathered}$$

As the base old MVA and KV values of generator 2 is same as the new value, therefore the reactance of generator remains the same.

$$\begin{gathered} X_1={X\text{'}\text{'}}_d \\ {X}_1=0.2\text{pu} \\ {\text{X}}_2=0.23\text{pu} \\ {\text{X}}_0=0.1\text{pu} \end{gathered}$$

Calculate the new value of the reactance of transformer 1.

Substitute 50 MVA for (MVA_base)_old , 12 KV for(KV_base)_old , 0.10pu for (X₁)_odd, 100 MVA for (MVA_base)_new , 10 KV for(KV_base)_new into equation (1).

$$\begin{gathered} {\left(X_{T1}\right)}_{new}=0.10\times \frac{{10}^2}{{10}^2}\times \frac{100}{50} \\ {\left(X_{T1}\right)}_{new}=0.20\text{pu} \end{gathered}$$

Calculate the new value of the reactance of transformer 2.

Substitute 50 MVA for (MVA_base)_old , 12 KV for(KV_base)_old , 0.10pu for (X₁)_odd, 100 MVA for (MVA_base)_new , 10 KV for(KV_base)_new into equation (1).

$$\begin{gathered} {\left(X_{T2}\right)}_{new}=0.10\times \frac{{10}^2}{{10}^2}\times \frac{100}{100} \\ {\left(X_{T2}\right)}_{new}=0.10\text{pu} \end{gathered}$$

Calculate the base impedance.

Substitute 1000 MVA for S_base and 138 KV for V_base into equation (2).

$$\begin{gathered} Z_{base}=\frac{{138}^2}{1000} \\ {Z}_{base}=190.44\Omega \end{gathered}$$

Calculate the positive and negative sequence reactance for line.

Substitute 190.44 for Z_base and 40 for X_old into equation (3)

$$\begin{gathered} X_1=X_2 \\ {X}_1=\frac{40}{190.44} \\ {X}_1=0.210\text{pu} \end{gathered}$$

Calculate the zero-sequence reactance for line.

Substitute 190.44 for Z_base and 40 for X_old into equation (3)

$$\begin{gathered} X_0=\frac{100}{190.44} \\ {X}_0=0.5251\text{pu} \end{gathered}$$

Draw the positive sequence network.

Calculate the positive-sequence equivalent impedance.

Calculate current in the positive sequence circuit.

Substitute j=0.3542 for Z₁ and 1.0< 0⁰ for V_Finto equation (4).

$$\begin{gathered} X_1=\left(0.576\right)\parallel \left(0.2+0.21+0.1+0.2\right) \\ {X}_1=\left(0.576\right)\parallel \left(0.92\right)\backslash \\ {X}_1=0.3542\text{pu}\backslash \end{gathered}$$

Calculate the base current.

Substitute 100 MVA for S_base and 10 KV for V_base into equation (5)

localid="1656421248875" $$\begin{gathered} I_{base}=\frac{100}{\sqrt{3}\times 10} \\ {I}_{base}=5.774\text{kA} \end{gathered}$$

Since the fault is symmetrical, therefore, zero and negative sequence currents are zero.

$$\begin{gathered} I_0=I_2 \\ {I}_0=0\text{A} \end{gathered}$$

Calculate the sub transient fault current in per unit.

Substitute 0.A for I₀ ,I _{2 ,}and -j 2.823 for I₁ into equation (6).

localid="1656422035328" $$\begin{gathered} \left[\begin{array}{c}{I\text{'}\text{'}}_a\\ {I\text{'}\text{'}}_b\\ {I\text{'}\text{'}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j2.823\\ 0\end{array}\right] \\ \left[\begin{array}{c}{I\text{'}\text{'}}_a\\ {I\text{'}\text{'}}_b\\ {I\text{'}\text{'}}_c\end{array}\right]=\left[\begin{array}{ccc}0& -j2.823& 0\\ 0& -j2.823a^2& 0\\ 0& -j2.823a& +1\end{array}\right] \\ \left[\begin{array}{c}{I\text{'}\text{'}}_a\\ {I\text{'}\text{'}}_b\\ {I\text{'}\text{'}}_c\end{array}\right]=\left[\begin{array}{c}2.823\angle -90°\\ 2.823\angle 150°\\ 2.823\angle 30°\end{array}\right]pu \end{gathered}$$

Calculate the sub transient current in kA.

Substitute $\left[\begin{array}{c}2.823\angle -90°\\ 2.823\angle 150°\\ 2.823\angle 30°\end{array}\right]pu$for I_d '' and 5.774 k for I_base into equation (7).

$$\begin{gathered} \left[\begin{array}{c}{I\text{'}\text{'}}_a\\ {I\text{'}\text{'}}_b\\ {I\text{'}\text{'}}_c\end{array}\right]=\left[\begin{array}{c}2.823\angle -90°\\ 2.823\angle 150°\\ 2.823\angle 30°\end{array}\right]\times 5.774 \\ \left[\begin{array}{c}{I\text{'}\text{'}}_a\\ {I\text{'}\text{'}}_b\\ {I\text{'}\text{'}}_c\end{array}\right]=\left[\begin{array}{c}16.30\angle -90°\\ 16.30\angle 150°\\ 16.30\angle 30°\end{array}\right]KA \end{gathered}$$

Hence the sub transient fault current in per unit and kA are width="121">2.823∠-90°2.823∠150°2.823∠30°and $\left[\begin{array}{c}16.30\angle -90°\\ 16.30\angle 150°\\ 16.30\angle 30°\end{array}\right]$respectively.