Question

For Problem 7.6, determine (a) the instantaneous symmetrical fault current in kA in phase a of the generator as a function of time, assuming maximum dc offset occurs in this generator phase, and (b) the maximum dc offset current in kA as a function of time that can occur in any one generator phase.

Short answer

(a) The instantaneous symmetrical fault current in kA in phase a of the generator as a function of time is $40.824[1.203{\mathrm{e}}^{-\frac{\mathrm{t}}{0.05}}+1.875{\mathrm{e}}^{-\mathrm{t}}+9.625]sin(\mathrm{\omega }t-\frac{\mathrm{\pi }}{2})\mathrm{kA}$.

(b) The maximum dc offset current in kA that can occur in any one generator phase $151.202e^{-\frac{t}{0.1}}\mathrm{kA}$.

Step-by-step solution

Write the given data from the question.

Base MVA of generator,$S_{base}=1000\mathrm{MVA}$

Base voltage of generator,$V_{base}=20\mathrm{kV}$

Base MVA of transformer rating;$S_{base}=1000\mathrm{MVA}$

Base voltage of transformer,$V_{base}=20\mathrm{kV}∆/345\mathrm{kV}\mathrm{Y}$

Generator reactances:

$$\begin{gathered} X_d^{\text{'}\text{'}}=0.17\mathrm{pu} \\ {X}_d^{\text{'}}=0.30\mathrm{pu} \\ {X}_d=1.5\mathrm{pu} \end{gathered}$$

Time constant for generator;

$$\begin{gathered} T_d^{\text{'}\text{'}}=0.05\mathrm{s} \\ {T}_d^{\text{'}}=1.0\mathrm{s} \\ {T}_d=0.1\mathrm{s} \end{gathered}$$

Transformer series reactance,$X_{tr}=0.1\mathrm{pu}$

Determine the formula to calculate the instantaneous fault current per phase as function of time and maximum dc offset current as function of time.

The equation to calculate instantaneous symmetrical fault current is given as follows.

${\mathbf{\left[}{\mathbf{I}}_{\mathbf{ac}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right]}}_{\mathbf{pu}}\mathbf{=}\mathbf{\left\{}\begin{array}{c}\sqrt{\mathbf{2}}{\mathbf{E}}_{\mathbf{g}}\mathbf{[}\left(\frac{1}{X_d^{\text{'}\text{'}}+X_{\mathrm{tr}}}‐\frac{1}{X_d^{\text{'}}+X_{\mathrm{tr}}}\right){\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{d}}^{\mathbf{\text{'}\text{'}}}}}\mathbf{+}\left(\frac{1}{X_d^{\text{'}}+X_{\mathrm{tr}}}‐\frac{1}{X_d+X_{\mathrm{tr}}}\right){\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{d}}^{\mathbf{\text{'}}}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{X}}_{\mathbf{d}}\mathbf{+}{\mathbf{X}}_{\mathbf{tr}}}\mathbf{]}\\ \mathbf{sin}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{\alpha }\mathbf{‐}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}\mathbf{\right\}}$…… (1)

Here ${\mathbf{E}}_{\mathbf{g}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{pu}$is the rms line to neutral pre fault voltage.

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

${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ …… (2)

The equation to calculate the instantaneous symmetrical fault current is given as follows.

role="math" localid="1655895014946" $\mathbf{I}\mathbf{=}{\mathbf{\left[}{\mathbf{I}}_{\mathbf{ac}}\left(t\right)\mathbf{\right]}}_{\mathbf{pu}}{\mathbf{I}}_{\mathbf{base}}$ …… (3)

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

$\mathbf{I}\mathbf{\text{'}\text{'}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{g}}}{{\mathbf{X}}_{\mathbf{d}}^{\mathbf{\text{'}\text{'}}}\mathbf{+}{\mathbf{X}}_{\mathbf{tr}}}$ …… (4)

The equation to calculate the maximum dc offset current is given as follows.

${\left[I_{\mathrm{dc}}\left(t\right)\right]}_{\mathbf{pu}}\mathbf{=}\sqrt{\mathbf{2}}\mathbf{I}\mathbf{\text{'}\text{'}}{\mathbf{e}}^{\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{A}}}}$ …… (5)

The equation to calculate the current from per unit to Amperes is given as follows.

$\mathbf{Per}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{=}\frac{\mathbf{Actual}\mathbf{}\mathbf{value}}{\mathbf{Base}\mathbf{}\mathbf{value}}$ …… (6)

Calculate the instantaneous symmetrical fault current in kA in phase a of the generator as a function of time.

(a)

Calculate the instantaneous symmetrical fault current.

Substitute 0 for $\alpha$, 0.05 for $t$, 1 pu for $E_g$, 0.17 pu for $X_d^{\text{'}\text{'}}$, 0.30 pu for $X_d^{\text{'}}$, 1.5 pu for $X_d$, 0.1 pu for $X_{tr}$, 0.05 s for $T_d^{\text{'}\text{'}}$, 1.0 s for $T_d^{\text{'}}$and 0.1 s for $T_A$into equation (1).

$$\begin{gathered} I_{ac}\left(0.05\right)=\left\{\begin{array}{c}\sqrt{2}\times 1\left[\begin{array}{c}\left(\frac{1}{0.17+0.1}-\frac{1}{0.3+0.1}\right)e^{-\frac{t}{0.05}}\\ +\left(\frac{1}{0.3+0.1}-\frac{1}{1.5+0.1}\right)e^{-\frac{t}{1}}+\frac{1}{1.5+0.1}\end{array}\right]\\ \sin \left(\omega t-0-\frac{\mathrm{\pi }}{2}\right)\end{array}\right\} \\ {I}_{ac}\left(0.05\right)=\sqrt{2}\left[\left(\frac{1}{0.27}-\frac{1}{0.4}\right)e^{-\frac{t}{0.05}}+\left(\frac{1}{0.4}-\frac{1}{1.6}\right)e^{-t}+\frac{1}{1.6}\right]\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right) \\ {I}_{ac}\left(0.05\right)=1.414\left[1.203e^{-\frac{t}{0.05}}+1.875e^{-t}+9.625\right]\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right)\mathrm{pu} \end{gathered}$$

calculate the generator base current.

Substitute 1000 MVA for$S_{base}$ and 20 kV for $V_{base}$into equation (2).

$$\begin{gathered} I_{base}=\frac{1000\times {10}^6}{\sqrt{3}\times 20\times {10}^3} \\ {I}_{base}=28.867\mathrm{kA} \end{gathered}$$

The instantaneous fault current in kA.

Substitute role="math" localid="1655896358740" $1.414\left[\begin{array}{c}1.203e^{-\frac{t}{0.05}}+1.875e^{-t}+9.625\end{array}\right]\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right)\mathrm{pu}$for $I_{ac}\left(0.05\right)$and$28.867\mathrm{kA}$ for $I_{base}$into equation (3).

$$\begin{gathered} I=1.414\left[\begin{array}{c}1.203e^{-\frac{t}{0.05}}+1.875e^{-t}+9.625\end{array}\right]\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right)\times 28.867 \\ I=40.824\left[\begin{array}{c}1.203e^{-\frac{t}{0.05}}+1.875e^{-t}+9.625\end{array}\right]\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right)\mathrm{kA} \end{gathered}$$

Hence, the instantaneous symmetrical fault current in kA in phase a of the generator as a function of time is role="math" localid="1655896500448" $40.824\left[\begin{array}{c}1.203e^{-\frac{t}{0.05}}+1.875e^{-t}+9.625\end{array}\right]\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right)\mathrm{kA}$.

Calculate maximum dc offset current in kA as a function of time.

(b)

Calculate the sub transient current.

Substitute 0.17 pu for$X_d^{\text{'}\text{'}}$ , 0.1 pu for$X_{tr}$ and 1 pu for$E_g$ into equation (4).

$$\begin{gathered} I\text{'}\text{'}=\frac{1}{0.17+0.1} \\ I\text{'}\text{'}=\frac{1}{0.27} \\ I\text{'}\text{'}=3.703\mathrm{pu} \end{gathered}$$

Calculate the maximum dc offset current.

Substitute 3.703 pu for$I\text{'}\text{'}$ and 0.10 s for$T_A$ into equation (5).

$$\begin{gathered} {\left[I_{dc}\left(t\right)\right]}_{pu}=\sqrt{2}\times 3.703e^{-\frac{t}{0.1}} \\ {\left[I_{dc}\left(t\right)\right]}_{pu}=5.237e^{-\frac{t}{0.1}}\mathrm{pu} \end{gathered}$$

Calculate the dc offset current in kA.

Substitute $5.237e^{-\frac{t}{0.1}}\mathrm{pu}\mathrm{for}{\left[I_{dc}\left(t\right)\right]}_{pu}$and 28.867 kA for $I_{base}$into equation (6).

role="math" localid="1655897083635" $$\begin{gathered} I_{dc}\left(t\right)=5.237e^{-\frac{t}{0.1}}\times 28.867 \\ {I}_{dc}=151.202e^{-\frac{t}{0.1}}\mathrm{kA} \end{gathered}$$

Hence, the maximum dc offset current in kA that can occur in any one generator phase $151.202e^{-\frac{t}{0.1}}\mathrm{kA}$.