GATE 2016 E E – SET 2 Question 15

Given: $$C=10nF/km/ph$$ and $$L=1.6mH/km/ph$$

Shunt compensation: It is the method that is used to improve the power factor of a system. When inductive load is connected to the transmission line, the power factor lags because of lagging load current. In order to compensate , a shunt capacitor is connected which draws current leading the source voltage. This improves the power factor of the system.

Surge impedance $$Z_{s}=\sqrt{\frac{L}{C}}\Rightarrow \sqrt{\frac{1.6\times 10^{-3}}{10\times 10^{-9}}}=400\Omega $$

Load impedance $$Z_{L}=\frac{(kv)^{2}}{P_{L}}=\frac{(400\times 10^{3})^{2}}{300\times 10^{6}}=533.33\Omega $$

Since load impedance is greater than surge impedance $$Z_{L}> Z_{s}$$, the receiving end voltage will rise more than sending end voltage, $$V_{r}> V_{s}$$. So to reduce the voltage to 400kv, we need inductive compensation.