Q21GATE 2004MCQ

A 800 kV transmission line is having per phase line inductance of 1.1 mH/km and per phase line capacitance of 11.68 nF/km. Ignoring the length of the line, its ideal power transfer capability in MW is

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📖 Explanation

Parameter of the line Inductance =L = 1.1 mH/km Capacitance =C =11.68 nF/km Surge impedance of the line $Z_s=\sqrt{\frac{L}{C}}=\sqrt{\frac{1.1 \times 10^{-3}}{11.68 \times 10^{-9}}}=306.88\Omega$ Ideal power transfer capability $=\frac{V{l-l}^2}{Z_s}=\frac{(800 \times 10^3)^2}{306.88}\approx 2085 \times 10^6\approx 2085MW$

Q22GATE 2004MCQ

A new generator having $E_{g}$ = 1.4 $\angle 30^{\circ}$ pu [equivalent to (1.212 + j0.70) pu] and synchronous reactance ' $X_{s}$ ' of 1.0 pu on the system base, is to be connected to a bus having voltage $V_{t}$ in the existing power system. This existing power system can be represented by Thevenin's voltage $E_{th} = 0.9\angle 0^{\circ}$ pu in series with Thevenin's impedance $Z_{th} = 0.25\angle 90^{\circ}$ pu. The magnitude of the bus voltage $V_{t}$ of the system in pu will be

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📖 Explanation

Thevenin's equivalent of existing system, wher,

\begin{aligned} Z_{th}&=0.25\angle 90^{\circ}\; p.u. \\ E_{th}&=0.9 \angle 0^{\circ}\; p.u. \end{aligned}

\begin{aligned} I&=\frac{E_g-E_{th}}{jX_s+Z_{th}}\\ &=\frac{1.4\angle 30^{\circ}-0.9\angle 0^{\circ}}{j1+0.25\angle 90^{\circ}}\\ &=0.56-j0.25\;p.u.\\ V_t&=E_g-jIX_s\\ &=1.4\angle 30^{\circ}(0.56-j0.25)\times (j1)\\ &=0.973\angle 8.27^{\circ}\; p.u.\\ |V_t|&=0.973\; p.u. \end{aligned}

Q23GATE 2004MCQ

A 50 Hz, 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated megavoltamperes at 0.8 power factor. Suddenly a fault occurs reducing in electric power output by 40%. Neglect losses and assume constant power input to the shaft. The accelerating torque in the generator in MNm at the time of fault will be

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📖 Explanation

Before fault, mechanical input $(P_m)=$ Electrical output $(P_e)$ $= 500 \times 0.8=400MW$ After fault, mechanical input remains same $=P_m=400MW$ While electrical output reduces by 40 %. So, electrical output at the time of fault $=(P_e)_f= 0.6 \times 400 =240MW$ Accelerating power $=P_a=P_m-(P_e)_f$ $=400-240=160MW$ Synchronous speed (in rpm) $N_s=\frac{120 \times f}{P}=\frac{120 \times 50}{4}=1500\; rpm$ Synchronous speed (in rad/sec) $\omega _s=\frac{2 \pi N_s}{60}=\frac{2 \pi \times 1500}{60}=157 \text{ rad/sec}$ $P_a=T_a\omega$ $\Rightarrow \;T_a\omega$ (accelerating Torque) $=\frac{P_a}{\omega _s}=\frac{160}{157}MN-m=1.018MN-m$

Q24GATE 2003MCQ

A generator delivers power of 1.0 p.u. to an infinite bus through a purely reactive network. The maximum power that could be delivered by the generator is 2.0 p.u. A three-phase fault occurs at the terminals of the generator which reduces the generator output to zero. The fault is cleared after $t_{c}$ second. The original network is then restored. The maximum swing of the rotor angle is found to be $\delta _{max}$ = 110 electrical degree. Then the rotor angle in electrical degrees at $t =t_{c}$ is

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📖 Explanation

Power generated by the generator

\begin{aligned} P_e &=P_{max} \sin \delta \\ \text{At }\delta _0& \\ P_e&=1 pu \\ P_{max}&=2pu \\ \delta _{cr}&=\text{roto angle at }t =t_c\\ P_e&= P_{max} \sin \delta \\ 1&=2 \sin \delta _0 \\ \delta _0 &=30^{\circ} \end{aligned}

During fault, $P_e$ become zero and the fault is cleared at $\delta _{cr}$ , Mechanical input to the generator remains constant, $P_m=1pu$ Applying equal areal criterion,

\begin{aligned} A_1&= A_2\\ A_1&= \int_{\delta _0}^{\delta _{cr}} (P_m-0)d\delta \\ &= P_m(\delta _{cr} -\delta _0)\;\;...(i)\\ A_2&=\int_{\delta _{cr}}^{\delta _{max}}(P_{max} \sin \delta -P_m)d\delta \\ &= P_{max}(\cos \delta _{cr} - \cos \delta _{max})\\ &-P_m(\delta _{max}-\delta _{cr})\;\;...(ii)\\ &\text{Equating eq. (i) and (ii),} \\ P_m(\delta _{cr}-\delta _0)&=P_{max}( \cos \delta _{cr} - \cos \delta _{max}) \\ &-P_m(\delta _{max} -\delta _{cr}) \\ \cos \delta _{cr} &= \frac{P_m}{P_{max}} (\delta _{max} - \delta _{0}) +\cos \delta _{max}\\ \cos \delta _{cr}&= \frac{1}{2} \times (110-30) \times \frac{\pi}{180}+\cos 110\\ \delta _{cr}&=69.14^{\circ} \end{aligned}

Q25GATE 2002MCQ

A transmission line has a total series reactance of 0.2 pu. Reactive power compensation is applied at the midpoint of the line and it is controlled such that the midpoint voltage of the transmission line is always maintained at 0.98 pu. If voltage at both ends of the line are maintained at 1.0 pu, then the steady state power transfer limit of the transmission line is

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📖 Explanation

$P_{ss}=\frac{1 \times 0.98}{0.1}=9.8\;pu$