Question

A generator with constant 1 pu terminal voltage supplies power through a step-up transformer of 0.10 reactance and a double circuit line to an infinite bus maintained at 1 pu. The steady state stability power limit of the system is 5.25 pu. If one of the double circuit line is tripped, then resulting steady state stability power limit in pu will be

• A. 3.57
• B. 2.32
• C. 1.58
• D. 2.00

Answer 1

The correct option is A 3.57

Given system can be represented by assuming transmission line reactance as X pu.

Initially when both lines are working,
Steady state stability power limit,

$P_{1-max}=\frac{E_f{V}_t}{X_{total}}=\frac{1\times 1}{0.1+\left(X||X\right)}=\frac{1}{0.1+\frac{X}{2}}=5.25$

$\frac{X}{2}=0.190-0.10$

$X=0.180pu$

Now, when one line is tripped, steady state power limit,

$P_{2-max}=\frac{1\times 1}{0.1+X}=\frac{1\times 1}{0.1+0.18}=\frac{1}{0.28}=3.57pu$