Question

A Y-connected synchronous generator rated $\mathbf{20}\mathbf{}\mathbf{MVA}$at $\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{kV}$has a positive-sequence reactance of $\mathbf{j}\mathbf{2}\mathbf{.}\mathbf{38}\mathbf{V}$, negative-sequence reactance of $\mathbf{j}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{V}$, and zero-sequence reactance of $\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{V}$. The generator neutral is solidly grounded. With the generator operating unloaded at rated voltage, a so-called single line-to-ground fault occurs at the machine terminals. During this fault, the line-to-ground voltages at the generator terminals are ${\mathbf{V}}_{\mathrm{ag}}=\mathbf{0}$,${\mathbf{V}}_{\mathrm{bg}}=\mathbf{8}\mathbf{.}\mathbf{071}\angle -\mathbf{102}\mathbf{.}\mathbf{25}\mathbf{°}$ and ${\mathbf{V}}_{\mathrm{cg}}=\mathbf{8}\mathbf{.}\mathbf{071}\angle \mathbf{102}\mathbf{.}\mathbf{25}\mathbf{°}$. Determine the sequence components of the generator fault currents and the generator fault currents. Draw a phasor diagram of the prefault and postfault generator terminal voltages. (Note: For this fault, the sequence components of the generator fault currents are all equal to each other.)

Short answer

The sequence components of the generator fault currents are:

$\begin{array}{l}\overline{I_{a0}}=-j1.43\text{\hspace{0.17em}pu}\\ \overline{I_{a1}}=-j1.43\text{\hspace{0.17em}pu}\\ \overline{I_{a2}}=-j1.43\text{\hspace{0.17em}pu}\end{array}$.

The generator fault current is $3590A$.

Draw the phasor diagram for the phase pre-fault voltage:

Draw the phasor diagram for the post fault generator voltage.

Step-by-step solution

Write the given data from the question.

A synchronous generator rated at $20MVA$.

A synchronous voltage rated at $13.8\mathrm{Kv}$.

A positive-sequence reactance of $\text{j2.38\hspace{0.17em}V}$.

A negative-sequence reactance of $\mathrm{j}3.33\text{\hspace{0.17em}V}$.

A zero-sequence reactance of $\mathrm{j}0.95\text{\hspace{0.17em}V}$.

The line-to-ground voltage $a$at the generator terminals are $V_{ag}=0$.

The line-to-ground voltage $b$at the generator terminals are $V_{bg}=8.071\angle -102.25°$.

The line-to-ground voltage $c$at the generator terminals are $V_{cg}=8.071\angle 102.25°$.

Step 2: Determine the formula of sequence components of the generator fault currents and the generator fault currents.

Write the formula of zero sequence components of fault currents.

$\overline{{\mathbf{I}}_{a0}}=-\frac{\overline{{\mathbf{V}}_{a0}}}{{\mathbf{Z}}_0}$ …… (1)

Here, $\overline{{\mathbf{V}}_{a0}}$is zero sequence component voltage and ${\mathbf{Z}}_0$ is sequence impedance.

Write the formula of positive sequence components of fault currents.

$\overline{{\mathbf{I}}_{\mathbf{a}1}}=\frac{{\mathbf{E}}_{\mathrm{an}}-\overline{{\mathbf{V}}_{a1}}}{{\mathbf{Z}}_1}$ …… (2)

Here, ${\mathbf{E}}_{\mathrm{an}}$ is internal phase voltage with neutral connected, $\overline{{\mathbf{V}}_{a1}}$is positive sequence phase voltage and ${\mathbf{Z}}_{\mathbf{1}}$is positive sequence impedance.

Write the formula of negative sequence components of fault currents.

role="math" localid="1656152859342" $\overline{{\mathbf{I}}_{a2}}=-\frac{\overline{{\mathbf{V}}_{a1}}}{{\mathbf{Z}}_1}$ …… (3)

Here, $\overline{{\mathbf{V}}_{a1}}$is positive sequence phase voltage and ${\mathbf{Z}}_{\mathbf{1}}$is positive sequence impedance.

Write the formula of generator fault currents.

${\mathbf{I}}_a={\mathbf{I}}_{a0}\mathbf{+}{\mathbf{I}}_{a1}\mathbf{+}{\mathbf{I}}_{a2}$ …… (4)

Here,${\mathbf{I}}_{a0}$ ,${\mathbf{I}}_{a1}$ and${\mathbf{I}}_{a2}$ is of zero, positive and negative sequence components of fault currents.

 Determine the sequence components of the generator fault currents.

Determine the base voltage

Determine the base impedance

$\begin{array}{c}{Z}_{base}=\frac{{\left(13.8\text{\hspace{0.17em}kV}\right)}^2}{20\text{\hspace{0.17em}MVA}}\\ =9.52\Omega \end{array}$

Determine the per unit generator phase voltage $\overline{V_a}$.

$\overline{V_a}=0$

Determine the per unit generator phase voltage $\overline{V_b}$.

$\begin{array}{c}\overline{V_b}=\frac{8.071\angle -102.25\text{\hspace{0.17em}kV}}{7.967\text{\hspace{0.17em}kV}}\\ =1.013\angle -102.25°\\ =-0.215-j0.99\text{\hspace{0.17em}pu}\end{array}$

Determine the per unit generator phase voltage $\overline{V_c}$.

$\begin{array}{c}\overline{V_c}=\frac{8.071\angle 102.25\text{\hspace{0.17em}kV}}{7.967\text{\hspace{0.17em}kV}}\\ =1.013\angle 102.25°\\ =-0.215+j0.99\text{\hspace{0.17em}pu}\end{array}$

Determine the sequence impedance in per unit.

$Z_{\text{0,pu}}=\frac{Z_0}{Z_{base}}$ …… (5)

Here,$Z_0$ is zero sequence impedance and$Z_{base}$ is base impedance.

Substitute$j0.95$ for$Z_0$ and$9.52$ for$Z_{base}$ into equation (5).

$\begin{array}{c}{Z}_{\text{0,pu}}=\frac{j0.95}{9.52}\\ =j0.1\end{array}$

Determine the positive sequence impedance in per unit.

$Z_{\text{1,pu}}=\frac{Z_1}{Z_{base}}$ …… (6)

Here,$Z_1$ is positive sequence impedance and$Z_{base}$ is base impedance.

Substitute $j2.38$ for$Z_1$ and$9.52$ for$Z_{base}$ into equation (6).

$\begin{array}{c}{Z}_{\text{1,pu}}=\frac{j2.38}{9.52}\\ =j0.25\end{array}$

Determine the negative sequence impedance in per unit.

$Z_{\text{2,pu}}=\frac{Z_2}{Z_{base}}$ …… (7)

Here,$Z_2$ is negative sequence impedance and$Z_{base}$ is base impedance.

Substitute$j3.33$ for$Z_2$ and $9.52$for$Z_{base}$ into equation (7).

$\begin{array}{c}{Z}_{\text{2,pu}}=\frac{j3.33}{9.52}\\ =j0.35\end{array}$

At various fault point determine the symmetrical components of voltages.

$\left[\begin{array}{c}\overline{V_{a0}}\\ \overline{V_{a1}}\\ \overline{V_{a2}}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}\overline{V_a}\\ \overline{V_b}\\ \overline{V_c}\end{array}\right]$ …… (8)

Here,$V_a.V_b$and $V_c$is generator phase voltage.

$\begin{array}{l}a=1\angle 120°\\ a^2=1\angle -120°\end{array}$

Substitute $0$for $\overline{V_a}$, $-0.215-j0.99$for $\overline{v_b}$and $-0.215+j0.99$for $\overline{V_C}$into equation (8).

role="math" localid="1656154846479"

Now determine the zero sequence components of the generator fault currents.

Substitute$-0.143$ forrole="math" localid="1656154555761" $\overline{V_{a0}}$ and$j0.1$ for$Z_0$ into equation (1).

$\begin{array}{c}\overline{I_{a0}}=-\left(\frac{-0.143}{j0.1}\right)\\ =-j1.43\text{\hspace{0.17em}pu}\end{array}$

Determine the positive sequence components of the generator fault currents.

Substitute$1$ for $E_{an}$,$0.643$ for$\overline{V_{a1}}$ and$j0.25$ for$Z_1$ into equation (2).

Determine the negative sequence components of the generator fault currents.

Substitute$-0.5$ for$\overline{V_{a1}}$ and$j0.35$ for$Z_1$ into equation (3).

Hence, the sequence components of generator fault currents are.

$\begin{array}{l}\overline{I_{a0}}=-j1.43\text{\hspace{0.17em}pu}\\ \overline{I_{a1}}=-j1.43\text{\hspace{0.17em}pu}\\ \overline{I_{a2}}=-j1.43\text{\hspace{0.17em}pu}\end{array}$

Determine the fault currents.

…… (9)

Substitute equation 9 into equation 4.

$\begin{array}{c}{I}_a=3I_{a0}\\ =3(-j1.43)\\ =-j4.29\text{\hspace{0.17em}pu}\end{array}$

Determine the base current.

$\begin{array}{c}{I}_{base}=\frac{20\text{\hspace{0.17em}MVA}}{\sqrt{3}\times 13.8\text{\hspace{0.17em}kV}}\\ =837\text{\hspace{0.17em}A}\end{array}$

Determine the sub-transient current of line $a$.

$\begin{array}{c}{I}_a=4.29\times 837\\ =3590\text{\hspace{0.17em}A}\end{array}$

Hence, the generator fault current is $3590A$.

Determine the line-to-line voltage at fault present in phase $ab$.

Determine the line-to-line voltage at fault present in phase $bc$.

Determine the line-to-line voltage at fault present in phase $ca$.

The prefault generator terminal line voltages are,

$\begin{array}{l}{V}_{ab}=13.8\angle 30°\text{\hspace{0.17em}kV}\\ V_{bc}=13.8\angle 270°\text{\hspace{0.17em}kV}\\ V_{ca}=13.8150°\text{\hspace{0.17em}kV}\end{array}$

Draw the phasor diagram of prefault generator terminal voltages.

Figure 1

The postfault generator terminal line voltages are,

role="math" localid="1656153549722" $\begin{array}{l}{V}_{ab}=8.05\angle 77.7°\\ V_{bc}=15.78\angle 270°\\ V_{ca}=8.05\angle 102.3°\end{array}$

Draw the phasor diagram of postfault generator terminal voltages.

Figure 2