GATE EE 2024 Set 1 — Question 40

Given:
Line-to-line voltage $V_{LL} = 11$ kV
Apparent power $S = 10$ MVA
Line reactance $X_{line} = 5 \, \Omega$
Synchronous reactance $X_s = 30 \, \Omega$
Load power factor $\cos\phi_L = \sqrt{3}/2$ (lagging), so $\phi_L = 30^\circ$. Phase voltage $V_{ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{11000}{\sqrt{3}}$ V.
Rated line current $I_{L,rated} = \frac{S}{\sqrt{3} V_{LL}} = \frac{10 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = \frac{10000}{\sqrt{3} \times 11} \approx 524.86$ A.
For a star-connected generator, phase current $I_{ph} = I_{L,rated} \approx 524.86$ A. The problem states the generator is producing rated current at rated voltage. This means the voltage at the generator terminal is $V_{ph}$ and the current flowing is $I_{ph}$.
The magnitude of the total impedance from the generator terminal to the load is:
$|Z_{total\_load}| = \frac{V_{ph}}{I_{ph}} = \frac{11000/\sqrt{3}}{524.86} \approx \frac{6350.85}{524.86} \approx 12.10 \, \Omega$. Let the load impedance be $Z_{load} = R_L + jX_L$. We know $\cos\phi_L = \frac{R_L}{|Z_{load}|} = \frac{\sqrt{3}}{2}$ and $\sin\phi_L = \frac{X_L}{|Z_{load}|} = \frac{1}{2}$.
So, $R_L = |Z_{load}| \frac{\sqrt{3}}{2}$ and $X_L = |Z_{load}| \frac{1}{2}$.
The total impedance from the generator terminal to the load is $Z_{total\_load} = R_L + j(X_L + X_{line})$.
$|Z_{total\_load}|^2 = R_L^2 + (X_L + X_{line})^2$.
$(12.10)^2 = \left(|Z_{load}| \frac{\sqrt{3}}{2}\right)^2 + \left(|Z_{load}| \frac{1}{2} + 5\right)^2$
$146.41 = \frac{3}{4} |Z_{load}|^2 + \frac{1}{4} |Z_{load}|^2 + 5|Z_{load}| + 25$
$146.41 = |Z_{load}|^2 + 5|Z_{load}| + 25$
$|Z_{load}|^2 + 5|Z_{load}| - 121.41 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$|Z_{load}| = \frac{-5 \pm \sqrt{5^2 - 4(1)(-121.41)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 485.64}}{2} = \frac{-5 \pm \sqrt{510.64}}{2}$
$|Z_{load}| = \frac{-5 \pm 22.597}{2}$. Since $|Z_{load}|$ must be positive:
$|Z_{load}| = \frac{-5 + 22.597}{2} = \frac{17.597}{2} \approx 8.798 \, \Omega$. $R_L = 8.798 \times \frac{\sqrt{3}}{2} \approx 8.798 \times 0.866 \approx 7.627 \, \Omega$.
$X_L = 8.798 \times \frac{1}{2} \approx 4.399 \, \Omega$. The total impedance from the generator terminal to the load is $Z_{total\_load} = R_L + j(X_L + X_{line})$.
$Z_{total\_load} = 7.627 + j(4.399 + 5) = 7.627 + j9.399 \, \Omega$.
The power factor angle $\phi_{terminal} = \arctan\left(\frac{9.399}{7.627}\right) = \arctan(1.2323) \approx 50.93^\circ$.
The power factor at the terminal is $\cos(\phi_{terminal}) = \cos(50.93^\circ) \approx 0.630$. Since the reactive component is positive, it is lagging.Therefore, the power factor at the terminal of the generator is approximately 0.63 lagging.The final answer is $\boxed{\text{0.63 lagging.}}$