Question

Figure 8.26 shows a single-line diagram of a three-phase, interconnected generator-reactor system, in which the given per-unit reactances are based on the ratings of the individual pieces of equipment. If a three phase short-circuit occurs at fault point $\mathbf{F}$, obtain the fault$\mathbf{MVA}$ and fault current in$\mathbf{kA}$ if the prefault busbar line-to-line voltage is $\mathbf{13}\mathbf{.}\mathbf{2}\mathbf{kV}$. Choose $\mathbf{100}\mathbf{MVA}$as the base$\mathbf{MVA}$ for the system.

Short answer

The fault$\text{MVA}$ of a three-phase interconnected generator-reactor system is $\text{406.504\hspace{0.17em}MVA}$.

The fault current of a three-phase interconnected generator-reactor system is $\text{17.764\hspace{0.17em}kA}$.

Step-by-step solution

Write the given data from the question.

Refer to Figure 8.26 in the textbook.

The pre-fault bus-bar line-to-line voltage is $\text{13.2\hspace{0.17em}kV}$.

Base$\text{MVA}$ for the system is $\text{100\hspace{0.17em}MVA}$.

Step 2: Determine the fault of a three-phase interconnected generator-reactor system.

Write the formula of fault current.

${\mathbf{I}}_{\mathrm{fault}}=\frac{I}{{\mathbf{Z}}_{\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}}$ …… (1)

Here, $\mathbf{I}$line current and ${\mathbf{Z}}_{\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$ is total impedance per-unit.

Write the formula of fault $\mathrm{MVA}$.

$\mathbf{Fault}\mathbf{MVA}=\frac{\mathbf{Generator}\mathbf{MVA}}{{\mathbf{Z}}_{\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}}}$ …… (2)

Here, ${\mathbf{Z}}_{\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$ is total impedance per-unit.

 Determine the fault current of a three-phase interconnected generator-reactor system

Determine the per-unit impedance for generator.

$G_{1\text{p.u.}}=\left(G_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right){\left(\frac{V_{\text{baseold}}}{V_{\text{basenew}}}\right)}^2$

Here,$G_{\text{p.u.old}}$is per-unit old impedance of generator,$S_{\text{basenew}}$is sequence component of new base quantity,$S_{\text{baseold}}$is sequence component of old base quantity,$V_{\text{baseold}}$is old voltage of base quantity and$V_{\text{basenew}}$is new voltage of base quantity.

Substitute$0.1$for $G_{\text{p.u.old}}$, 100 for $S_{\text{basenew}}$, 20 for $S_{\text{baseold}}$,$1$for$V_{\text{baseold}}$and$1$for$V_{\text{basenew}}$into above equation.

$\begin{array}{c}{G}_{\text{1p.u.}}=0.1\left(\frac{100}{20}\right){\left(1\right)}^2\\ =0.5\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Determine the per-unit impedance for generator $2$.

$G_{1\text{p.u.}}=\left(G_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right){\left(\frac{V_{\text{baseold}}}{V_{\text{basenew}}}\right)}^2$

Here, $G_{\text{p.u.old}}$is per-unit old impedance of generator, $S_{\text{basenew}}$is sequence component of new base quantity, $S_{\text{baseold}}$is sequence component of old base quantity, $V_{\text{baseold}}$is old voltage of base quantity and $V_{\text{basenew}}$is new voltage of base quantity.

Substitute 0.15 for $G_{\text{p.u.old}}$,$100$for $S_{\text{basenew}}$, 40 for $S_{\text{baseold}}$,1 for $V_{\text{baseold}}$and$1$for$V_{\text{basenew}}$into above equation.

$\begin{array}{c}{G}_{\text{1p.u.}}=0.15\left(\frac{100}{40}\right){\left(1\right)}^2\\ =0.375\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Determine the per-unit reactance for reactor 1.

$X_{1p.u.}=\left(X_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right)$

Here,$X_{\text{p.u.old}}$is per-unit old reactance, $S_{\text{basenew}}$is sequence component of new base quantity and $S_{\text{baseold}}$ is sequence component of old base quantity.

Substitute$0.05$for $X_{\text{p.u.old}}$,$100$for $S_{\text{basenew}}$and$20$for$S_{\text{baseold}}$into above equation.

$\begin{array}{c}{X}_{1p.u.}=0.05\left(\frac{100}{20}\right)\\ =0.25\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Determine the per-unit reactance for reactor 2.

$X_{2p.u.}=\left(X_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right)$

Here, $X_{\text{p.u.old}}$is per-unit old reactance, $S_{\text{basenew}}$is sequence component of new base quantity and $S_{\text{baseold}}$ is sequence component of old base quantity.

Substitute$0.04$for $X_{\text{p.u.old}}$,$100$for$S_{\text{basenew}}$and$16$for$S_{\text{baseold}}$into above equation.

$\begin{array}{c}{X}_{2p.u.}=0.04\left(\frac{100\text{\hspace{0.17em}MVA}}{16\text{\hspace{0.17em}MVA}}\right)\\ =0.25\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Draw the equivalent impedance diagram.

Figure 1

Determine the equivalent impedance

$\begin{array}{c}{X}_{eq}=(\left(0.375\text{\hspace{0.17em}p.u+0.25\hspace{0.17em}p.u}\right)\parallel 0.375\text{\hspace{0.17em}p.u})+0.25\text{\hspace{0.17em}p.u}\\ =\frac{\left(0.625\right)\left(0.375\right)}{0.625+0.375}+0.25\\ =0.234+0.25\\ =0.4844\text{\hspace{0.17em}p.u}\end{array}$

Redraw the figure 1 of equivalent impedance.

Figure 2

Determine the total impedance of the current.

$\begin{array}{c}{Z}_1=0.5\text{\hspace{0.17em}p.u}\parallel \text{0.4844\hspace{0.17em}p.u}\\ =\frac{\left(0.5\right)\left(0.4844\right)}{0.5+0.4844}\\ =0.246\text{\hspace{0.17em}p.u}\end{array}$

Determine the line current corresponding to $100\mathrm{MVA}$at $\text{13.2\hspace{0.17em}kV}$.

$\begin{array}{c}I=\frac{100\times {10}^6}{\sqrt{3}(13.2\times {10}^3)}\\ =4.37\text{\hspace{0.17em}kA}\end{array}$

Determine the fault current in $\text{kA}$.

Substitute $4.37$for $l$and $0.246$for $Z_{1p.u}$into equation (1).

Therefore the fault current is $17.764\text{\hspace{0.17em}kA}$.

Determine the fault current in MVA.

Substitute$1000\mathrm{MVA}$for$\text{Generator\hspace{0.17em}MVA}$and $0.246$for$Z_{1\text{p.u}}$into equation (2).

$\begin{array}{c}\text{Fault\hspace{0.17em}MVA}=\frac{100\text{\hspace{0.17em}MVA}}{0.246}\\ =406.504\text{\hspace{0.17em}MVA}\end{array}$

Therefore, the fault current in$\hspace{0.17em}\mathrm{MVA}$is $\text{406.504\hspace{0.17em}MVA}$.