Question

A generator with constant 1.0 p.u. terminal voltage supplies power through step-up transformer of 0.12 p.u. reactance and a double- circuit line to an infinite bus bar as shown in the figure. The infinite bus voltage is maintained at 1.0 p.u. Neglecting the resistances and susceptances of the system, the steady state stability power limit of the system is 6.25 p.u. If one of the double-circuit is tripped, then resulting steady stability power limit in p.u. will be

• A. 12.5 p.u.
• B. 3.125 p.u.
• C. 10.0 p.u.
• D. 5.0 p.u

Answer 1

The correct option is D 5.0 p.u

Induced emf of the generator$=|E_g|=1p.u.$

terminal voltage $=|V_t|=1p.u.$

Reactance of transformer $=X_t=0.12p.u.$

When double-circuit line is connected
Reactance of line $=X||X=\frac{X}{2}$

Steady state stability power limit

$P_1=\frac{|E_g||V_t|}{X_t+\frac{X}{2}}=\frac{1}{0.12+\frac{X}{2}}$

$6.25=\frac{1}{0.12+\frac{X}{2}}$

$X=0.08p.u.$

When one of the double circuit is tripped, steady state stability power limit

$P_2=\frac{|E_g||V_t|}{X_t+X}=\frac{1}{0.12+0.08}=5p.u.$