Q3GATE 2024MCQ

A 3-phase. $11 \mathrm{kV} .10$ MVA synchronous generator is connected to an inductive load of power factor $(\sqrt{3} / 2)$ via a lossless line with a per-phase inductive reactance of $5 \Omega$ . The per-phase synchronous reactance of the generator is $30 \Omega$ with negligible armature resistance. If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is

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📖 Explanation

$\mathrm{pF}=\frac{\sqrt{3}}{2}=30^{\circ} \text { lagging }$

\begin{aligned} E & =\sqrt{(V \cos 30+I R)^{2+}\left(V \sin 30^{\circ}+I X\right)^2} \\ E & =\sqrt{(V \cos +I R)^{2+}(V \sin +I X)} \\ \frac{11 \mathrm{~K}}{\sqrt{3}} & =\sqrt{\left(V \cdot \frac{\sqrt{3}}{2}\right)^{2+}\left(\frac{V}{2}+524.86(5)\right)^2} \\ \left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)^2 & =V^2 \cdot \frac{3}{4}+\left(\frac{V}{2}+2624.319\right)^2 \\ \left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)^2 & =\frac{3}{4} V^{2+}\frac{V^2}{4}+2624.319 \mathrm{~V}+2624.319^2 \\ V_t & =4618.105 \mathrm{~V} \\ (I Z)^2 & =E^{2+}V^{2-}2 E V \cos \delta \\ (2624.319)^2 & =\left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)^{2+}(4618.105)^{2-}2\left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)(4618.105) \cos \delta \\ \delta & =20.96 \\ \mathrm{pF} & =\cos (20.96+30)=0.629 \text { lagging } \end{aligned}

Q20GATE 2017NAT

Two parallel connected, three-phase, 50Hz, 11kV, star-connected synchronous machines A and B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactances of machine A and machine B are 1 $\Omega$ and 3 $\Omega$ respectively. Assuming the magnetic circuit to be linear, the ratio of excitation current of machine A to that of machine B is ________.

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📖 Explanation

Synchronous machine A: 11 kV, 50 Hz, Y connected, $X_s=1\Omega$ Synchronous machine B: 11 kV, 50 Hz, Y connected, $X_s=3\Omega$ Given both supplying only reactive power 50MVAR tp 11 kV grid equally

\begin{aligned} \therefore \; Q_A &=\sqrt{3}VI_A \sin \phi =25mVAR \\ Q_B &= \sqrt{3}VI_B \sin \phi =25mVAR\\ \therefore \;\;I_A &=\frac{25 \times 10^6}{\sqrt{3} \times 11 \times 10^3}=1312.2A \\ \therefore \;\; I_B&= 1312.2A\\ E_A &= V-jI_AX_s\\ &=\frac{11 \times 10^3}{\sqrt{3}}-j1312.2\angle 90 \times 1 \\ E_B &=V_jI_BX_s \\ &= =\frac{11 \times 10^3}{\sqrt{3}}-j1312.2\angle 90 \times 3 \end{aligned}

Given magnetic circuit to be linear,

\begin{aligned} \therefore \;\; E &\propto I_f \\ \therefore \;\; \frac{I_{f_A}}{I_{f_B}}&=\frac{E_A}{E_B} \\ E_A&=7663.05V \\ E_B &= 10287.45V \end{aligned}

Therefore, ratio of excitation current of machine A to machine B $\frac{E_A}{E_B}=0.745$

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