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 threephase short-circuit occurs at fault point, 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 as the base$\mathbf{MVA}$ for the system.

Short answer

The fault of a three-phase interconnected generator-reactor system is.$\text{406.504\hspace{0.33em}MVA}$

The fault current of a three-phase interconnected generator-reactor system is.$\text{17.764\hspace{0.33em}kA}$

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.$\text{13.2\hspace{0.33em}kV}$

Base$\text{MVA}$ for the system is.$\text{100\hspace{0.33em}MVA}$

Step 2: Determine the MVA 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 .

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

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

Step 3: Determine the fault current of a three-phase interconnected generator-reactor system

Determine the per-unit impedance for generator. $1$

localid="1656163900426" $G_{1\text{p.u.}}=\left(G_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right){\left(\frac{V_{\text{baseold}}}{V_{\text{basenew}}}\right)}^2$

Here,localid="1656163908008" $G_{\text{p.u.old}}$is per-unit old impedance of generator,localid="1656163913921" $S_{\text{basenew}}$is sequence component of new base quantity, localid="1656163922211" $S_{\text{baseold}}$is sequence component of old base quantity,localid="1656163928874" $V_{\text{baseold}}$is old voltage of base quantity and localid="1656163932524" $V_{\text{basenew}}$is new voltage of base quantity.

Substitute localid="1656163938357" $0.1$for localid="1656163943321" $G_{\text{p.u.old}}$, localid="1656163947823" $100$for localid="1656163953280" $S_{\text{basenew}}$,localid="1656163962159" $20$forlocalid="1656163966640" $S_{\text{baseold}}$,localid="1656163978465" $1$for localid="1656163970254" $V_{\text{baseold}}$and localid="1656163974404" $1$for localid="1656163983113" $V_{\text{basenew}}$into above equation.

localid="1656163991296" $\begin{array}{c}{G}_{\text{1p.u.}}=0.1\left(\frac{100}{20}\right){\left(1\right)}^2\\ =0.5\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Determine the per-unit impedance for generator.localid="1656164000032" $2$

localid="1656163995639" $G_{1\text{p.u.}}=\left(G_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right){\left(\frac{V_{\text{baseold}}}{V_{\text{basenew}}}\right)}^2$

Here,localid="1656164006111" $G_{\text{p.u.old}}$is per-unit old impedance of generator, localid="1656164012136" $S_{\text{basenew}}$is sequence component of new base quantity,localid="1656164019005" $S_{\text{baseold}}$is sequence component of old base quantity, localid="1656164024756" $V_{\text{baseold}}$is old voltage of base quantity andlocalid="1656164035733" $V_{\text{basenew}}$is new voltage of base quantity.

Substitute localid="1656164043424" $0.15$forlocalid="1656164049447" $G_{\text{p.u.old}}$, localid="1656164056096" $100$for,localid="1656164087211" $S_{\text{basenew}}$,localid="1656164092616" $40$forlocalid="1656164098117" $S_{\text{baseold}}$, localid="1656164104842" $1$for localid="1656164111080" $V_{\text{baseold}}$and localid="1656164117045" $1$for localid="1656164122048" $V_{\text{basenew}}$into above equation.

localid="1656164130142" $\begin{array}{c}{G}_{\text{1p.u.}}=0.15\left(\frac{100}{40}\right){\left(1\right)}^2\\ =0.375\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Determine the per-unit reactance for reactor.$1$

localid="1656164136013" $X_{1p.u.}=\left(X_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right)$

Here,localid="1656164143420" $X_{\text{p.u.old}}$is per-unit old reactance,localid="1656164152573" $S_{\text{basenew}}$is sequence component of new base quantity and localid="1656164162133" $S_{\text{baseold}}$ is sequence component of old base quantity.

Substitute localid="1656164173532" $0.05$for,localid="1656164187376" $X_{\text{p.u.old}}$,localid="1656164195401" $100$for localid="1656164202264" $S_{\text{basenew}}$and localid="1656164211758" $20$for localid="1656164219162" $S_{\text{baseold}}$into above equation.

localid="1656164224233" $\begin{array}{c}{X}_{1p.u.}=0.05\left(\frac{100}{20}\right)\\ =0.25\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Determine the per-unit reactance for reactor.$2$

localid="1656164231865" $X_{2p.u.}=\left(X_{\text{p.u.old}}\right)\left(\frac{S_{\text{basenew}}}{S_{\text{baseold}}}\right)$

Here, localid="1656164239903" $X_{\text{p.u.old}}$s per-unit old reactance, localid="1656164246629" $S_{\text{basenew}}$is sequence component of new base quantity and localid="1656164252599" $S_{\text{baseold}}$is sequence component of old base quantity.

Substitutelocalid="1656164266845" $0.04$for,localid="1656164258372" $X_{\text{p.u.old}}$flocalid="1656164273526" $100$Forlocalid="1656164282671" $S_{\text{basenew}}$and$16$for localid="1656164290076" $S_{\text{baseold}}$into above equation.

localid="1656164297649" $\begin{array}{c}{X}_{2p.u.}=0.04\left(\frac{100\text{\hspace{0.33em}MVA}}{16\text{\hspace{0.33em}MVA}}\right)\\ =0.25\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Draw the equivalent impedance diagram.

Determine the equivalent impedance

localid="1656164305718" $\begin{array}{c}{X}_{eq}=\left(\left(0.375\text{\hspace{0.33em}p.u+0.25\hspace{0.33em}p.u}\right)\parallel 0.375\text{\hspace{0.33em}p.u}\right)+0.25\text{\hspace{0.33em}p.u}\\ =\frac{\left(0.625\right)\left(0.375\right)}{0.625+0.375}+0.25\\ =0.234+0.25\\ =0.4844\text{\hspace{0.33em}p.u}\end{array}$

Redraw the figure 1 of equivalent impedance.

Determine the total impedance of the current.

$\begin{array}{c}{Z}_1=0.5\text{\hspace{0.33em}p.u}\parallel \text{0.4844\hspace{0.33em}p.u}\\ =\frac{\left(0.5\right)\left(0.4844\right)}{0.5+0.4844}\\ =0.246\text{\hspace{0.33em}p.u}\end{array}$

Determine the line current corresponding to $\text{100\hspace{0.33em}MVA}$at.$\text{13.2\hspace{0.33em}kV}$

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

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

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

$\begin{array}{c}{I}_{\text{fault}}=\frac{4.37\text{\hspace{0.17em}kA}}{0.246}\\ =17.764\text{\hspace{0.33em}kA}\end{array}$

Therefore the fault current is.$17.764\text{\hspace{0.33em}kA}$

Determine the fault current in MVA.

Substitute$\text{100\hspace{0.33em}MVA}$for$\text{Generator\hspace{0.33em}MVA}$and$0.246$for $Z_{1\text{p.u}}$into equation (2).

$\begin{array}{c}\text{Fault\hspace{0.33em}MVA}=\frac{100\text{\hspace{0.33em}MVA}}{0.246}\\ =406.504\text{\hspace{0.33em}MVA}\end{array}$

Therefore, the fault current in$\text{MVA}$is.$\text{406.504\hspace{0.33em}MVA}$