Question

A lossless transmission line having operating voltage $V_S=V_R=400kV$ has surge impedance of 400 $\Omega$. It is connected with a 25% of shunt capacitance compensation at the load end. If the generator is delivering a power of 200 MW through these lines, then the load angle $\delta$ of the generator in degree is

• A. ${24.76}^o$
• B. ${28.72}^o$
• C. ${26.56}^o$
• D. ${23.42}^o$

Answer 1

The correct option is C ${26.56}^o$

$SIL=\frac{400\times 400}{Z_{sc}}$

$Z_{sc}=\frac{Z_s}{\sqrt{1+K_{csh}}}=\frac{400}{\sqrt{1+0.25}}=357.77\Omega$

$SIL=\frac{400\times 400}{357.77}\times {10}^6=447.21MW$

$P=P_{max}\sin \delta$

or $200=447.21\sin \delta$

or $\delta =si{n}^{-1}\left(\frac{200}{447.21}\right)={26.56}^o$