Question

As shown in Figure 7.21, a$\mathbf{25}\mathbf{-}\mathbf{MVA}\mathbf{}\mathbf{,}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{kV}\mathbf{}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{-}\mathbf{Hz}$, synchronous generator with ${\mathbf{X}}_{\mathbf{d}}^{\mathbf{"}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{per}\mathbf{}\mathbf{units}$is connected through a transformer to a bus that supplies four identical motors. The rating of the three-phase transformer is $\mathbf{25}\mathbf{MVA}\mathbf{}\mathbf{and}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{/}\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{kV}$with a leakage reactance of$\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$. Each motor has a sub-transient reactance ${\mathbf{X}}_{\mathbf{d}}^{\mathbf{"}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{per}\mathbf{}\mathbf{units}$on a base of $\mathbf{5}\mathbf{}\mathbf{MVA}\mathbf{}\mathbf{and}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{kV}$.A three-phase fault occurs at point P, when the bus voltage at the motors is$\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{kV}$.

(a) the sub-transient fault current,

(b) the sub-transient current through breaker A, and

(c) the symmetrical short-circuit interrupting current (as defined for circuit breaker applications) in the fault and in breaker A.

Short answer

(a)The value of the sub-transient fault current is $I_{\mathrm{sc}}^{\text{'}}=16.734\mathrm{kA}$.

(b)The value of the sub-transient fault current through circuit breaker A is $I_{\mathrm{sc}}^{\text{'}}=16.734\mathrm{kA}$.

(c)The value of the symmetrical short circuit fault current is$-j6.4\mathrm{per}\mathrm{units}$ and the value of symmetrical short circuit interrupting fault current in breaker A is$14.726\mathrm{kA}$.

Step-by-step solution

Given data

The base power rating, $S_b=25\mathrm{MVA}$.

The base voltage rating, $V_b=13.8kV$.

The frequency of $60\mathrm{Hz}$.

The synchronous generator reactance, $X_{dg}^{"}=0.15\mathrm{per}\mathrm{units}$.

The rating of the three phase transformer is $25\mathrm{MVA}$.

The step up and step down voltage of the transformer is $13.8/6\mathrm{kV}$with a leakage reactance, $X_{\mathrm{L}}0\mathrm{f}0.1\mathrm{per}\mathrm{unit}$.

Sub-transient reactance ofeach motor is${\mathrm{X}}_{\mathrm{dm}}^{"}=0.2\mathrm{per}\mathrm{units}$on a base of$5\mathrm{MVA}\mathrm{and}6.9\mathrm{kV}$.

Determine the sub-transient fault current

(a)

The general formula to calculate the sub-transient fault current is

${\mathbf{l}}_{\mathbf{sc}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{l}}_{\mathbf{sc}}{\mathbf{l}}_{\mathbf{b}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here,

$$\begin{gathered} {\mathbf{l}}_{\mathbf{sc}}\mathbf{=}\frac{\mathbf{Per}\mathbf{-}\mathbf{unit}\mathbf{}\mathbf{voltage}}{\mathbf{Reactance}\mathbf{}\mathbf{of}\mathbf{}\mathbf{system}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{in}\mathbf{}\mathbf{parallel}}{\mathbf{Motor}\mathbf{}\mathbf{sub}\mathbf{-}\mathbf{transient}\mathbf{}\mathbf{reactances}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{Total}\mathbf{}\mathbf{system}\mathbf{}\mathbf{reactance}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{in}\mathbf{}\mathbf{parallel}}{{\mathbf{X}}_{\mathbf{m}}^{\mathbf{"}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{X}}_{\mathbf{total}}} \end{gathered}$$

Calculate the motor sub-transient reactance,

$X_m^{"}=\frac{j\left(X_{dm}^{"}\right)\left(baseMVA\right)}{MotorMVA}$

Substitute the value $25\mathrm{MVA}$for $\mathrm{base}\mathrm{MVA}\mathrm{and}5\mathrm{MVA}\mathrm{for}\mathrm{motor}\mathrm{MVA}$, also $0.2\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{X}}_{\mathrm{dm}}^{"}$in the above equation,

$X_m^{"}=\frac{j\left(0.2\right)\left(25\right)}{\left(5\right)}$

Calculate the total reactance of the system

$X_{total}=j{X}_d^{"}+j{X}_L$

Substitute the value $0.2\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{X}}_{\mathrm{d}}^{"}\mathrm{and}j0.1\mathrm{for}{\mathrm{X}}_{\mathrm{L}}$in the above equation,

$$\begin{gathered} X_{total}=j{X}_d^{"}+j{X}_L \\ =\mathrm{j}0.15+\mathrm{j}0.1.....\left(3\right) \\ =\mathrm{j}0.25\mathrm{per}\mathrm{units} \end{gathered}$$

As there are 4 motors in parallel, therefore the value for $\left(\mathrm{Number}\mathrm{of}\mathrm{reactance}\mathrm{in}\mathrm{parallel}\right)$will be 4 in equation (2)

Thus, substituting all the derived values such as 4 for $\left(\mathrm{Number}\mathrm{of}\mathrm{reactance}\mathrm{in}\mathrm{parallel}\right)$, $j1\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{X}}_{\mathrm{m}}^{"}\mathrm{and}j0.25\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{X}}_{\mathrm{total}}$in equation (2),

$$\begin{gathered} i=\frac{4}{j1}+\frac{1}{j0.25} \\ =-j8perunit \end{gathered}$$

The general formula to calculate the base current in the circuit,

${\mathbf{l}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{Base}\mathbf{}\mathbf{MVA}}{\sqrt{\mathbf{3}}\mathbf{\times }\mathbf{Base}\mathbf{}\mathbf{voltage}}$......(4)

Substitute the value $25\mathrm{MVA}$for $\mathrm{baseMVA}\mathrm{and}6.9\mathrm{kV}\mathrm{for}Basevoltage$in the above equation,

$$\begin{gathered} l_b=\frac{25MVA}{\sqrt{3}\times 6.9kV} \\ =2.0918kA \end{gathered}$$

Thus, substitute the value $2.918kA\mathrm{for}-j8perunit\mathrm{for}{l}_{sc}$in the equation (1)

$$\begin{gathered} l_{sc}^{\text{'}}=\left(-\mathrm{j}8\mathrm{per}\mathrm{unit}\right)\left(2.918\mathrm{kA}\right) \\ =16.734\mathrm{kA} \end{gathered}$$

So the value of the sub-transient fault current is$l_{sc}^{\text{'}}=16.734\mathrm{kA}$.

Determine the sub-transient current through breaker A

(b)

To calculate short circuit current for circuit breaker A, substitute the value 3 for$\mathrm{Number}\mathrm{of}\mathrm{reactance}\mathrm{in}\mathrm{parallel}$, $j1\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{X}}_{\mathrm{m}}^{"}\mathrm{and}\mathrm{j}0.25\mathrm{per}\mathrm{unit}$for $totalreac\tan ceofthesystem$in the equation (2)

$$\begin{gathered} l_{sc}=\frac{3}{j1}+\frac{1}{j0.25} \\ =j7\mathrm{per}\mathrm{unit} \end{gathered}$$

Now, calculate the sub-transient current through circuit breaker A by substituting the value $-j7\mathrm{per}\mathrm{unit}\mathrm{for}{l}_{sc}and2.0918kA\mathrm{for}{l}_b$in equation (1)

$$\begin{gathered} l_{sc}^{\text{'}}=\left(-j7\mathrm{per}\mathrm{unit}\right)\left(2.918\mathrm{kA}\right) \\ =14.642\mathrm{kA} \end{gathered}$$

So the value of the sub-transient fault current through circuit breaker A is$l_{sc}^{\text{'}}=16.734\mathrm{kA}$.

Determine the symmetrical short-circuit interrupting current in the fault and in breaker A

(c)

During a fault in a breaker dc-offset current is multiplied by factor 1.1.

Therefore, the general formula to calculate the symmetrical short circuit interrupting current in the fault is,

$\left(\begin{array}{c}\mathrm{Symmetrical}\mathrm{short}-\mathrm{circuit}\\ \mathrm{interrupting}\mathrm{fault}\mathrm{current}\end{array}\right)=1.1\left(\mathrm{short}-\mathrm{circuit}\right){\mathrm{l}}_{\mathrm{b}......\left(4\right)}$

Here, localid="1655380433995" $l_b$is the base current.

Calculate the symmetrical short circuit current.

Substitute 3 for because there are three motors other than circuit breaker A also put the value obtained from equation (3) in equation (2),

$$\begin{gathered} l_{sc}=\frac{Numberofrec\tan cesinparallel}{Motorsub-transientreac\tan ce}+\frac{1}{Totalsystemreac\tan ce} \\ =\frac{3}{X_d^{\text{'}}}+\frac{1}{j0.25} \end{gathered}$$......(5)

Now, the formula to calculate the motor transient reactance is

$X_d^{\text{'}}=\frac{\mathrm{j}\left(\mathrm{Transient}\mathrm{reactance}\mathrm{of}\mathrm{each}\mathrm{motor}\right)\times \left(\mathrm{BaseMVA}\right)}{\mathrm{MotorMVA}}$

localid="1655381304092" $$\begin{gathered} \mathrm{Substitute}\mathrm{the}\mathrm{value}0.2\mathrm{per}\mathrm{unit}\mathrm{for}\mathrm{Transient}\mathrm{reactance}\mathrm{of}\mathrm{each}\mathrm{motor}\left({\mathrm{X}}_{\mathrm{dm}}^{\text{'}}\right) \\ ,25\mathrm{MVA}\mathrm{for}\mathrm{base}\mathrm{MVA}\mathrm{and}5\mathrm{MVA}\mathrm{for}\mathrm{motor}\mathrm{MVA}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equatio}. \end{gathered}$$

$$\begin{gathered} x_{\mathrm{d}}^{\text{'}}=\frac{\mathrm{j}\left(0.2\mathrm{per}\mathrm{unit}\right)\times \left(25\mathrm{MVA}\right)}{5\mathrm{MVA}} \\ =\mathrm{j}1.25\mathrm{per}\mathrm{unit} \end{gathered}$$

Now replace role="math" localid="1655381482761" $X_d^{\text{'}}$by $j1.25\mathrm{per}\mathrm{unit}$in equation (5)

$$\begin{gathered} l_{sc}=\frac{3}{j1.25}+\frac{1}{j0.25} \\ =j6.4\mathrm{per}\mathrm{unit} \end{gathered}$$

Hence, the symmetrical short circuit fault current is $-j6.4\mathrm{per}\mathrm{unit}$

Thus, calculate the symmetrical short circuit current,

Substitute the value$-6.4\mathrm{per}\mathrm{unit}\mathrm{for}\mathrm{short}-\mathrm{circuit}\mathrm{current}\mathrm{and}2.0918\mathrm{kA}\mathrm{for}{\mathrm{l}}_{\mathrm{b}}$in equation (4)

$$\begin{gathered} \left(\begin{array}{c}\mathrm{Symmetrical}\mathrm{short}-\mathrm{circuit}\\ \mathrm{interrupting}\mathrm{fault}\mathrm{current}\end{array}\right)=1.1\left(\mathrm{short}-\mathrm{circuit}\right){\mathrm{l}}_{\mathrm{b}} \\ =1.1\times 6.4\times 2091.84 \\ =14.726\mathrm{kA} \end{gathered}$$

Hence, the value of symmetrical short circuit interrupting fault current is$14.726\mathrm{kA}$