Question

A 500 MVA, 13.8 kV synchronous generator with${\mathit{X}}_{\mathbf{d}}^{\mathbf{"}}\mathbf{=}{\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}$and ${\mathit{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$per unit is connected to a$\mathbf{500}\mathbf{}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{}\mathit{k}\mathit{V}\mathbf{∆}\mathbf{/}\mathbf{500}\mathbf{}\mathit{k}\mathit{V}\mathbf{}\mathit{Y}$ transformer with 0.10 per-unit leakage reactance. The generator and transformer neutrals are solidly grounded. The generator is operated at no-load and rated voltage, and the high-voltage side of the transformer is disconnected from the power system. Compare the sub transient fault currents for the following bolted faults at the transformer high-voltage terminals: three-phase fault, single line-to-ground fault, line-to-line fault, and double line-to-ground fault.

Short answer

Answer

The sub transient fault current for three phase fault, line to ground fault, single line to line fault and double line to line fault are $1.923\angle -{90}^{°}kA$, $2.473\angle -{90}^{°}kA$, $1.666\angle {180}^{°}kA$and $2.401\angle 133.{89}^{°}kA$respectively.

Step-by-step solution

write the given data from the question.

Write the generator reactance.

$$\begin{gathered} X_d^{"}=X_2=0.20pu \\ {X}_0=0.05pu \end{gathered}$$

The base MVA rating of generator, $S_{baseG}=500MVA$.

The base voltage of generator, $V_{baseG}=13.8kV$.

The base MVA rating of transformer, $S_{base}=500MVA$.

The base voltage rating of transformer is $V_{base}=13.8kV∆/500kVY$.

The transformer reactance, $X_{Tr}=0.10pu$.

Determine the equations to calculated the sub transient current of three-phase fault, single line-to-ground fault, line-to-line fault, and double line-to-ground fault.

The equation for three phase faults.

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

${\mathit{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ …… (1)

The equation to calculate the sequence current for the three-phase fault is given as follows.

$$\begin{gathered} {\mathit{I}}_{\mathbf{0}}\mathbf{=}{\mathit{I}}_{\mathbf{2}} \\ {\mathit{I}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{}\mathit{p}\mathit{u} \\ {\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{1}}} \end{gathered}$$ …… (2)

Here, ${\mathit{V}}_{\mathbf{F}}$is the fault voltage, ${\mathit{Z}}_{\mathbf{1}}$is the positive sequence impedance, ${\mathit{I}}_{\mathbf{0}}{\mathit{I}}_{\mathbf{1}}$and ${\mathit{I}}_{\mathbf{2}}$ are the zero, positive and negative sequence component.

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

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{=}{\mathit{I}}_{\mathbf{1}}$ …… (3)

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

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{kA}\mathbf{\right)}\mathbf{=}{\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}{\mathit{I}}_{\mathbf{base}}$ …… (4)

The equation for single line to ground fault.

The equation to calculate the fault current in the sequence component is given as follows.

$$\begin{gathered} {\mathit{I}}_{\mathbf{0}}\mathbf{=}{\mathit{I}}_{\mathbf{1}} \\ {\mathit{I}}_{\mathbf{0}}\mathbf{=}{\mathit{I}}_{\mathbf{2}} \\ {\mathit{I}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{1}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}} \end{gathered}$$ …… (5)

Here, ${\mathit{V}}_{\mathbf{F}}$is the fault voltage, ${\mathit{Z}}_{\mathbf{0}}\mathbf{,}{\mathit{Z}}_{\mathbf{1}}$and ${\mathit{Z}}_{\mathbf{2}}$ are the zero, positive and negative sequence impedance respectively.

The equation to calculate the sub transient fault current in per unit is given as follows.

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}\mathbf{=}\mathbf{3}{\mathit{I}}_{\mathbf{0}}$ …… (6)

The equation to calculate the sub transient current in kA .

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{kA}\mathbf{\right)}\mathbf{=}{\mathit{I}}_{\mathbf{base}}{\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}$ …… (7)

The equations for line to line fault occur in the system.

The equation to calculate the fault current in the sequence component is given as follows.

$$\begin{gathered} {\mathit{I}}_{\mathbf{0}}\mathbf{=}\mathbf{0} \\ {\mathit{I}}_{\mathbf{1}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{2}} \\ {\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{R}}}{{\mathbf{Z}}_{\mathbf{1}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}} \end{gathered}$$ …… (8)

The equation to calculate the sub transient currentis given as follows.

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{=}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}\mathbf{a}\mathbf{)}{\mathit{I}}_{\mathbf{1}}$ ……. (9)

The equation to calculate the sub transient current in kA.

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{kA}\mathbf{\right)}\mathbf{=}{\mathit{I}}_{\mathbf{base}}{\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}$ …… (10)

The equations for double line to ground fault.

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

${\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{R}}}{{\mathbf{Z}}_{\mathbf{1}}\mathbf{+}{\displaystyle \frac{{\mathbf{Z}}_{\mathbf{0}}{\mathbf{Z}}_{\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}}}}$ …… (11)

The equation to calculate the negative sequence fault current is given as follows.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}}\mathbf{\right)}$ …… (12)

The equation to calculate the zero-sequence fault current is given as follows.

${\mathit{I}}_{\mathbf{0}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{Z}}_{\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}}\mathbf{\right)}$ …… (13)

The equation to calculate the sub transient currentis given as follows.

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathit{I}}_{\mathbf{1}}\mathbf{+}\mathit{a}{\mathit{I}}_{\mathbf{2}}$ …… (14)

The equation to calculate the sub transient current in kA for double line to line faults.

${\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{kA}\mathbf{\right)}\mathbf{=}{\mathit{I}}_{\mathbf{base}}{\mathit{I}}_{\mathbf{a}}^{\mathbf{"}}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}$ …… (15)

Calculated the sub transient current of three-phase fault, single line-to-ground fault, line-to-line fault, and double line-to-ground fault.

Draw the sequence network of the given system as,

Calculate the equivalent impedance of the zero-sequence network.

$Z_0=j0.10\text{pu}$

Calculate the equivalent impedance of the positive-sequence network.

$$\begin{gathered} Z_1=j0.2+j0.1 \\ {Z}_1=j0.3\text{pu} \end{gathered}$$

Calculate the equivalent impedance of the negative-sequence network.

$$\begin{gathered} Z_2=j0.2+j0.1 \\ {Z}_2=j0.3\text{pu} \end{gathered}$$

Calculate the base current.

Substitute $500\text{MVA}$ for $S_{base}$and $500\text{kV}$for $V_{base}$into equation (1).

$$\begin{gathered} I_{base}=\frac{500\times {10}^6}{\sqrt{3}\times 500\times {10}^3} \\ {I}_{base}=0.577\text{kA} \end{gathered}$$

The zero and negative sequence current for three phase faults are zero.

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

Calculate the positive sequence current for thee phase fault.

Substitute $1\angle 0°\text{pu}$for $V_F$and $j0.3\text{pu}$for $Z_1$into equation (2).

$$\begin{gathered} I_1=\frac{1\angle 0°}{j0.3} \\ {I}_1=-j3.33\text{pu} \end{gathered}$$

Calculate the sub transient current for three phase faults.

Substitute $-j3.33\text{pu}$for $I_1$into equation (3).

$I\text{'}{\text{'}}_a=-j3.33\text{pu}$

Calculate the sub transient current for three phase faults in kA.

Substitute $-j3.33\text{pu}$for $I\text{'}{\text{'}}_a$and $0.577\text{kA}$for $I_{base}$into equation (4).

$$\begin{gathered} {I\text{'}\text{'}}_a=\left(-j3.33\right)\left(0577\right) \\ {I\text{'}\text{'}}_a=j1.923\text{kA} \\ {I\text{'}\text{'}}_a=1.923\angle -90°\text{kA} \end{gathered}$$

Calculate the fault current is sequence component for line to ground faults.

Substitute $1\angle 0°\text{pu}$for $V_F$, $j0.3\text{pu}$for $Z_1$, $j0.3\text{pu}$for $Z_2$and $j0.1\text{pu}$for $Z_0$into equation (5).

$$\begin{gathered} I_0=\frac{1\angle 0°}{j0.1+j0.3+j0.3} \\ {I}_0=\frac{1\angle 0°}{j0.7} \\ {I}_0=-j1.428\text{pu} \end{gathered}$$

Calculate the sub-transient fault current in per unit for line to ground faults.

Substitute $-j1.428\text{pu}$for $I_0$into equation (6).

$$\begin{gathered} {I\text{'}\text{'}}_a=3\left(-j1.428\right) \\ {I\text{'}\text{'}}_a=j4.286\text{pu} \end{gathered}$$

Calculate the sub-transient fault current in kA.

Substitute $0.577\text{kA}$for $I_{base}$and $-j4.286\text{pu}$for $I\text{'}{\text{'}}_a\left(\text{pu}\right)$into equation (7).

$$\begin{gathered} {I\text{'}\text{'}}_a=0.577\times \left(-j4.286\right) \\ {I\text{'}\text{'}}_a=2.473\angle -90°\text{kA} \end{gathered}$$

The zero-sequence current for line to line fault is zero.

$I_0=0\text{pu}$

Calculate the positive and negative sequence fault current for line to line faults.

Substitute $1\angle 0°\text{pu}$for $V_F$, $j0.3\text{pu}$for $Z_1$and $j0.3\text{pu}$for $Z_2$into equation (8).

$$\begin{gathered} I_1=-I_2 \\ {I}_1=\frac{1\angle 0°}{j\left(0.3+0.3\right)} \\ {I}_1=\frac{1\angle 0°}{j0.6} \\ {I}_1=-j1.667\text{pu} \end{gathered}$$

Calculate the sub transient current $I\text{'}{\text{'}}_b$for line to line faults.

Substitute $-j1.667$for $I\text{'}{\text{'}}_a$into equation (9).

$$\begin{gathered} {I\text{'}\text{'}}_b=\left[\left(-0.5-j0.866\right)-\left(-0.5+j0.866\right)\right]\left(-j1.667\right) \\ {I\text{'}\text{'}}_b=\left(-j1.732\right)\left(-j1.667\right) \\ {I\text{'}\text{'}}_b=-2.887\text{pu} \end{gathered}$$

Calculate the sub-transient fault current in kA.

Substitute $0.577\text{kA}$for$I_{base}$and $-2.887\text{pu}$for $I\text{'}{\text{'}}_a\left(\text{pu}\right)$into equation (10).

$$\begin{gathered} {I\text{'}\text{'}}_b=0.577\times \left(-2.887\right) \\ {I\text{'}\text{'}}_b=1.666\text{kA} \\ {I\text{'}\text{'}}_b=1.666\angle 180°\text{kA} \end{gathered}$$

Calculate the positive sequence component of fault current for double line to line faults.

Substitute $1\angle 0°\text{pu}$for $V_F$, $j0.3\text{pu}$for $Z_1$, $j0.3\text{pu}$for $Z_2$and $j0.1\text{pu}$for $Z_0$into equation (11).

$$\begin{gathered} I_1=\frac{1\angle 0°}{j\left(0.3+\frac{\left(0.3\right)\left(0.1\right)}{0.3+0.1}\right)} \\ {I}_1=\frac{1\angle 0°}{j0.375} \\ {I}_1=-j2.667\text{pu} \end{gathered}$$

Calculate the negative sequence component of fault current for double line to line faults.

Substitute $-j2.667\text{pu}$for $I_1$, $j0.3\text{pu}$for $Z_2$and$j0.1\text{pu}$for $Z_0$into equation (12).

$$\begin{gathered} I_2=-\left(-j2.887\right)\times \frac{j0.1}{j0.1+j0.3} \\ {I}_2=j0.667\text{pu} \end{gathered}$$

Calculate the zero-sequence component of fault current for double line to line faults.

Substitute $-j2.667\text{pu}$for $I_1$, $j0.3\text{pu}$for $Z_2$and$j0.1\text{pu}$for $Z_0$into equation (13).

$$\begin{gathered} I_0=-\left(-j2.667\right)\times \frac{j0.3}{j0.1+j0.3} \\ {I}_0=j2\text{pu} \end{gathered}$$

Calculate the sub transient current $I\text{'}{\text{'}}_b$for double line to line faults.

Substitute $-j2.887\text{pu}$for $I_1$, $j0.667\text{pu}$for $I_2$and $j2\text{pu}$for$I_0$into equation (14).

$$\begin{gathered} {I\text{'}\text{'}}_b=j2.0+\left(-0.5-j0.866\right)\left(-j2.667\right)+\left(-0.5+j0.866\right)\left(j0.667\right) \\ {I\text{'}\text{'}}_b=-j2+j1.333-2.309-j0.333-0.577 \\ {I\text{'}\text{'}}_b=-2.886+j3 \\ {I\text{'}\text{'}}_b=4.162\angle 133.89° \end{gathered}$$

Calculate the sub-transient fault current in kA.

Substitute $0.577\text{kA}$for$I_{base}$and $4.162\angle 133.89°\text{pu}$for $I\text{'}{\text{'}}_a\left(\text{pu}\right)$into equation (15).

$$\begin{gathered} {I\text{'}\text{'}}_b=0.577\times \left(4.162\angle 133.89°\right) \\ {I\text{'}\text{'}}_b=2.401\angle 133.89°\text{kA} \end{gathered}$$

Hence, the sub transient fault current for three phase fault, line to ground fault, single line to line fault and double line to line fault are$1.923\angle -90°\text{kA}$,$2.473\angle -90°\text{kA}$,$1.666\angle 180°\text{kA}$and $2.401\angle 133.89°\text{kA}$respectively.