GATE 2023 EE – Question 33
A 50 Hz, 275 kV line of length 400 km has the following parameters:
Resistance, $R=0.035\ \Omega$/km;
Inductance, $L=1$ mH/km;
Capacitance, $C=0.01\ \mu$F/km;
The line is represented by the nominal-π model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by $V_S$ and $V_R$, respectively) maintained at 275 kV, the phase angle difference ($\theta$) between $V_S$ and $V_R$ required for maximum possible active power to be delivered to the receiving end, in degree is _____________ (Round off to 2 decimal places).
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Correct answer: 83.22 to 84.06
Explanation
The series impedance is $Z=400(0.035+j2\pi\times50\times10^{-3})=14+j125.66\ \Omega$, with magnitude 126.4 Ω and angle $\beta=\tan^{-1}\dfrac{125.66}{14}=83.64^\circ$. The shunt capacitors sit at the buses (whose voltages are fixed) and do not change the power through $Z$. The received power is $P_R=\dfrac{V_SV_R}{|Z|}\cos(\beta-\theta)-\dfrac{V_R^2}{|Z|}\cos\beta$, which is maximum when $\theta=\beta=83.64^\circ$.