Question

The main busbars in a generating station are divided into four section, each section being connected to a tie bar through a similar reactor. One 20 MVA, 11 kV generator having substransient reactance 20% is connected to each section bus bar. When a symmetrical fault takes place between the phases of one of the section busbar, the voltage on the remaining section falls to 65% of the normal value. The reactance of each reactors is _____ ohms.

• A. 0.556

Answer 1

The correct option is A 0.556
Let ${\left(MVA\right)}_B=20;$

$\left(k{V}_b\right)=11kV$


Let fault occurs on busbars of $G_4$

Equivalent reactance $=\frac{0.20+X_{p.u.}}{3}=0.066+0.33X_{p.u.}$

Eq. reactance of three groups to fault via tie-bar
$=0.066+0.33X_{p.u.}+X_{p.u.}$

$=0.066+1.33X_{p.u.}$

According to given condition,

$0.066=0.35(0.066+1.33X_{p.u.})$

(or) $X_{p.u.}=0.092p.u.$

$\therefore X=0.092\times \frac{{11}^2}{20}=0.556ohm.$