Q7GATE 2017NAT

A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} S&=250 MVA \\ \cos \phi &=0.8 \\ K.E. &=1000MJ \\ P_e&= 60MW\\ \delta _0&=10^{\circ} \\ t&=10\; cycles\\ &=\frac{10}{50}=0.2\; sec \\ t&=5\; cycle =0.1sec \\ \text{Load}&\text{ is removed,} \\ P_e&=0 \\ P_a&=P_m-P_e \\ P_m &=60MW \\ M&=\frac{HS}{180f}=\frac{K.E.}{180f} \\ &= \frac{1000}{180 \times 50}=0.111\\ \frac{d^2 \delta }{dt^2}&=\frac{P_a}{M}=545.45 \text{ ele.degree}/s^2 \\ \text{Integrating}&\text{ the above eq.} \\ \int \left ( \frac{d^2 \delta }{dt^2} \right )dt&=\int \frac{P_a}{M}dt=545.45 \times t\\ \text{Integrating}&\text{ tit once again with dt} \\ &=545.45 \times t(dt)\\ \delta &=\frac{545.45 \times t^2}{2}=2.7^{\circ}\\ \text{So, new}&\text{ value of power angle}\\ &=10^{\circ}+2.7^{\circ}=12.7^{\circ} \end{aligned}

Q10GATE 2014NAT

The figure shows the single line diagram of a single machine infinite bus system. The inertia constant of the synchronous generator H = 5 MW-s/MVA. Frequency is 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor angle $\delta$ equal to 30 $^{\circ}$ . A three phase short circuit fault occurs at a certain location on one of the circuits of the double circuit transmission line. During fault, electrical power in pu is $P_{max} \sin \delta$ . If the values of $\delta \; and \; d\delta/dt$ at the instant of fault clearing are 45 $^{\circ}$ and 3.762 radian/s respectively, then $P_{max}$ (in pu) is ______.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2}&=(P_m-P_e) \\ \text{Let, } \omega _r&=\frac{d\delta }{dt} \\ \frac{2H}{\omega _s}\frac{d\omega _r }{dt} &=(P_m-P_e) \\ \text{Multiplying with } &\omega _r \text{ \in both side}\\ \frac{H}{\omega _s}\left ( 2\omega _r \frac{d\omega _r }{dt}\right )&= (P_m-P_e)\frac{d\delta }{dt}\\ \Rightarrow \; \frac{H}{\omega _s} 2d(\omega _r)^2&=(P_m-P_e)d\delta \\ \Rightarrow \; \frac{H}{\omega _s}\int_{\omega _{r_1}}^{\omega _{r_2}}d(\omega _r^2) &= \int_{\delta _1}^{\delta _2}(P_m-P_e)d\delta \\ \text{where, } & \\ \omega _{r_1}&=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right|_{\delta =\delta _1} =\omega (\delta =\delta _1)-\omega _s\\ \omega _{r_2}&=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right|_{\delta =\delta _2} =\omega (\delta =\delta _2)-\omega _s\\ \frac{H}{\omega _s}(\omega _{r_2}^2 -\omega _{r_1}^2)&=\int_{\delta _1}^{\delta _2}(P_m-P_e)d\delta \\ \text{In problem, }&\\ \frac{d\delta }{dt}=0, &\frac{d\delta }{dt}=3.762\\ \frac{5}{314}(3.762^{2-}0^2)&=\int_{30^{\circ}}^{45^{\circ}}(1-P_{max} \sin \delta )d\delta \\ \Rightarrow \;\; P_{max}&=0.23p.u. \end{aligned}

Q15GATE 2008MCQ

A loss less single machine infinite bus power system is shown below : The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power, then the critical clearing time of the circuit breaker will

🚀 Solve in Practice Mode

📖 Explanation

For such faults, $\delta _{cr}$ is given by $\delta _{cr}=\cos ^{-1} [(\pi-2\delta _0) \sin \delta _0 -\cos \delta _0]\;...(i)$ and

\begin{aligned} t_{cr}&=\text{critical cleaning time } \\ &= \left ( \frac{2H(\delta _{cr}-\delta _0)}{\pi f P_m} \right )^{1/2}\;...(ii) \end{aligned}

CASE-1: When only once generator is connected Mechanical input to the generator $(P_{m_1})=$ Electrical power delivered by the generator $(P_{e_1})$ $\Rightarrow \; P_{m_1}=P_{e_1}=1p.u.$ CASE-2: When two generator connected in parallel Electrical power delivered by each generator $P_{e_2}=0.5p.u.$ Mechanical inout to each generator $(P_{m_2})=P_{e_2}=0.5p.u.=\frac{P_{m_1}}{2}$ Assuming $\delta _0$ in each case same $\delta _{0_1}=\delta _{0_2}=\delta _{0}$ So, $\delta _{cr}$ will also be same in both cases, (from equation (i)),

\begin{aligned} t_{cr}&\propto \frac{1}{\sqrt{P_m}}\\ &(\delta _{cr}, H \text{ and } \delta _0 \text{ are same}) \\ \therefore \; t_{cr_2} &=\sqrt{\frac{P_{m_1}}{P_{m_2}}}\times t_{cr_1} \\ t_{cr_2}&= \sqrt{\frac{P_{m_1}}{P_{m_1}/2}}\times 0.28\\ &= \sqrt{2} \times 0.28=0.396\;sec \end{aligned}

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