Question

A 300 MVA, 13.8 kV three-phase, 60 Hz, Y-connected synchronous generator is adjusted to produce rated voltage on open circuit. A balanced three-phase fault is applied to the terminals at t=0 . After analysing the raw data, the symmetrical transient current is obtained as

${\mathbf{i}}_{\mathbf{ac}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{10}}^{\mathbf{4}}\mathbf{\left(}\begin{array}{c}\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{1}}}}\mathbf{+}\mathbf{6}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{2}}}}\end{array}\mathbf{\right)}\mathbf{A}$

where ${\mathbf{T}}_{\mathbf{1}}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{ms}$and ${\mathbf{T}}_{\mathbf{1}}\mathbf{=}\mathbf{15}\mathbf{}\mathbf{ms}$. (a) Sketch${\mathbf{i}}_{\mathbf{ac}}\mathbf{}\left(t\right)$as a function of time for $\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{500}\mathbf{}\mathbf{ms}$. (b) Determine ${\mathbf{X}}_{\mathbf{d}}^{\mathbf{\text{'}\text{'}}}$and${\mathbf{X}}_{\mathbf{d}}^{\mathbf{\text{'}}}$in per unit based on the machine ratings.

Short answer

(a) Therefore the $i_{ac}\left(t\right)$can be drawn as,

(b) The reactance $X_d^{\text{'}\text{'}}$is 0.157 pu and$X_d$is 1.255 pu .

Step-by-step solution

write the given data from the question.

The power,$S_{base}=300\mathrm{MVA}$

The voltage,$V_{base}=13.8\mathrm{kV}$

Time constant,

$$\begin{gathered} {\tau }_1=200\mathrm{ms} \\ {\tau }_2=15\mathrm{ms} \end{gathered}$$

The equation for symmetrical current,

$i\left(t\right)={10}^4\left(\begin{array}{c}1+e^{-\frac{t}{{\tau }_1}}+6e^{-\frac{t}{{\tau }_2}}\end{array}\right)\mathrm{A}$

Determine the formula to calculate the symmetrical transient current and Xd'' & Xd' .

The equation to calculate base current is given as follows.

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

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

${\mathbf{Z}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{base}}^{\mathbf{2}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$

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

localid="1655892814813" ${\mathbf{i}}_{\mathbf{ac}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{10}}^{\mathbf{4}}\mathbf{\left(}\begin{array}{c}\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{1}}}}\mathbf{+}\mathbf{6}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{T}}_{\mathbf{2}}}}\end{array}\mathbf{\right)}\mathbf{A}$

Draw the graph of asymmetrical transient current.

(a)

The equation for asymmetrical transient current is $i_{ac}\left(t\right)={10}^4\left(\begin{array}{c}1+e^{-\frac{t}{T_1}}+6e^{-\frac{t}{T_2}}\end{array}\right)\mathrm{A}$.

The graph for $i_{ac}\left(t\right)$can be drawn as,

Calculate the value of Xd'' & Xd' .

(b)

Calculate the base current.

Substitute 300 MVA for $S_{base}$and 13.8 kV for $V_{base}$into equation (1).

$$\begin{gathered} I_{base}=\frac{300\times {10}^6}{\sqrt{3}\times 13.8\times {10}^3} \\ {I}_{base}=12551\mathrm{A} \end{gathered}$$

Calculate the base impedance.

Substitute 300 MVA for$S_{base}$and 13.8 kV for $V_{base}$into equation (2).

$$\begin{gathered} Z_{base}=\frac{{\left(13.8\times {10}^3\right)}^2}{300\times {10}^6} \\ {Z}_{base}=\frac{190.44\times {10}^6}{300\times {10}^6} \\ {Z}_{base}=0.635\mathrm{\Omega } \end{gathered}$$

Calculate the transient current.

role="math" localid="1655892783845" $$\begin{gathered} i_{ac}\left(t\right)=\frac{{10}^4}{12552}\left(\begin{array}{c}1+e^{-\frac{t}{{\tau }_1}}+6e^{-\frac{t}{{\tau }_2}}\end{array}\right) \\ {i}_{ac}\left(t\right)=0.797\left(\begin{array}{c}1+e^{-\frac{t}{{\tau }_1}}+6e^{-\frac{t}{{\tau }_2}}\end{array}\right)\mathrm{pu} \end{gathered}$$

When time approaches to zero, then obtained impedance is $X_d^{\text{'}}$and when time approaches to infinite then obtained impedance is $X_d^{\text{'}}$.

When time approaches to zero.

role="math" localid="1655892794589" $$\begin{gathered} i_{ac}\left(0\right)=\underset{t\to 0}{\lim 0.797}\left(\begin{array}{c}1+e^{-\frac{t}{{\tau }_1}}+6e^{-\frac{t}{{\tau }_2}}\end{array}\right) \\ {i}_{ac}\left(0\right)=0.797\left(1+1+6\right) \\ {i}_{ac}\left(0\right)=6.376\mathrm{pu} \end{gathered}$$

When time approaches to infinity.

$$\begin{gathered} i_{ac}(\infty )=\underset{t\to 0}{\lim 0.797}\left(\begin{array}{c}1+e^{-\frac{t}{{\tau }_1}}+6e^{-\frac{t}{{\tau }_2}}\end{array}\right) \\ {i}_{ac}(\infty )=0.797\left(1+0+0\right) \\ {i}_{ac}(\infty )=0.797\mathrm{pu} \end{gathered}$$

Calculate the reactance $X_d^{\text{'}\text{'}}$.

$$\begin{gathered} X_d^{\text{'}\text{'}}=\frac{1}{6.397} \\ {X}_d^{\text{'}\text{'}}=0.157\mathrm{pu} \end{gathered}$$

Calculate the reactance $X_d$.

$$\begin{gathered} X_d=\frac{1}{0.797} \\ {X}_d=1.255\mathrm{pu} \end{gathered}$$

Hence, the reactance $X_d^{\text{'}\text{'}}$is 0.157 pu and $X_d$is 1.255 pu .