Question 1 |
In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is _________
12.8 | |
28.4 | |
20.5 | |
32.6 |
Question 1 Explanation:
\begin{aligned} X &=0.25+0.2+0.4||0.4 \\ &=0.45+0.2=0.65pu \\ P&=V_{pu} \times I_{pu} \cos \phi \\ 0.8 &=1 \times I_{pu} \times 0.8 \\ I_{pu} &= 1pu\\ E &= V+jI_aX_s\\ &=1+1\angle -36.86^{\circ} \times j0.65 \\ &= 1.484 \angle 20.51^{\circ}pu\\ \delta &= 20.51^{\circ} \end{aligned}
Question 2 |
Consider a lossy transmission line with V_{1} \; and \; V_{2} as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be
greter than |\frac{V_{1}V_{2}}{X}| | |
less than |\frac{V_{1}V_{2}}{X}| | |
equal to |\frac{V_{1}V_{2}}{X}| | |
equal to |\frac{V_{1}V_{2}}{Z}| |
Question 2 Explanation:
With only x:
P_{max}=\left | \frac{V_1V_2}{x} \right |
With Lossy Tr, Line
P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )
Therefore, with Lossy Line P_{max} \lt \left | \frac{V_1V_2}{x} \right |
P_{max}=\left | \frac{V_1V_2}{x} \right |
With Lossy Tr, Line
P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )
Therefore, with Lossy Line P_{max} \lt \left | \frac{V_1V_2}{x} \right |
Question 3 |
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.
1.27 | |
2.7 | |
12.7 | |
127 |
Question 3 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}
Question 4 |
The figure shows the single line diagram of a power system with a double circuit transmission
line. The expression for electrical power is 1.5 sin\delta, where \delta is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of \delta as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is P_{max} sin\delta, the valueof P_{max} , in pu is _________.
0.729 | |
1.22 | |
2.6 | |
4.8 |
Question 4 Explanation:
Using equal area criteria:
A_1=A_2
\int_{\delta _0}^{\delta _c}(P_{m0}-P_{max1} \sin \delta )d\delta =\int_{\delta _c}^{\delta _2}(P_{max1} \sin \delta -P_{m0})d\delta
By solving above integration,
P_{max1}=\frac{P_{m0}(\delta _2-\delta _0)}{\cos \delta _0- \cos \delta _2}
Given data:
\begin{aligned} P_{m0} &=1pu \\ P_e&= 1.5 \sin (\delta _0)=P_{m0}=1pu\\ \delta _0&41.8^{\circ}=0.7295\; rad \\ \delta _2 &=1.221\; rad =96.95^{\circ} \end{aligned}
Substitute above values in above equation,
P_{max1}=\frac{1(1.221-0.7295)}{\cos 41.8- \cos 69.95}=1.220 p.u.
Question 5 |
The synchronous generator shown in the figure is supplying active power to an infinite bus via two short, lossless transmission lines, and is initially in steady state. The mechanical power input to the generator and the voltage magnitude E are constant. If one line is tripped at time t_1 by opening the circuit breakers at the two ends (although there is no fault), then it is seen that the generator undergoes a stable transient. Which one of the following waveforms of the rotor angle \delta shows the transient correctly?
A | |
B | |
C | |
D |
Question 5 Explanation:
Initial value of \delta =\delta _0 at t =t_1 with both lines are in service of one of the line circuit breakes opened the \delta is increased for same time and stabilizes to new value of \delta.
Question 6 |
A 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating in steady state at synchronous speed, and producing 1 pu of real power. The initial value of the rotor angle \delta is 5^{\circ}, when a bolted three phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the value of \delta in degrees, 0.02 second after the fault is _________.
0.5 | |
2.8 | |
4.2 | |
5.9 |
Question 6 Explanation:
Let 3-\phi fault occurs at t=0
\Rightarrow \; P_e=0\text{ for }t\geq 0^+
Now, swing equation for t\geq 0^+
\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2} &=P_m \\ \frac{d^2\delta }{dt^2}&= \frac{P_m \omega _s }{2H}\\ \frac{d\delta }{dt}&= \frac{P_m \omega _s }{2H}t+C\\ C=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right| _{t=0^+}&=\omega (t=0^+)-\omega _s\\ C=\omega _s-\omega _s&=0\;(\omega (t-0^+)=\omega _s) \\ \frac{d\delta }{dt}&=\frac{P_m \omega _s }{2H}t \\ \Rightarrow \; \delta &= \frac{P_m \omega _s }{4H}t^2+C'\\ C'&= \delta |_{t=0^+}=\delta _0\\ \delta &= \frac{P_m \omega _s }{4H}t^2+\delta _0\\ \text{Now in problem,}\\ \delta (t=0.02)&\\ &=\frac{1 \times 2 \times 180 \times 50 \times 0.02^2 +5^{\circ}}{4 \times 2}\\ &=5.9^{\circ} \end{aligned}
\Rightarrow \; P_e=0\text{ for }t\geq 0^+
Now, swing equation for t\geq 0^+
\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2} &=P_m \\ \frac{d^2\delta }{dt^2}&= \frac{P_m \omega _s }{2H}\\ \frac{d\delta }{dt}&= \frac{P_m \omega _s }{2H}t+C\\ C=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right| _{t=0^+}&=\omega (t=0^+)-\omega _s\\ C=\omega _s-\omega _s&=0\;(\omega (t-0^+)=\omega _s) \\ \frac{d\delta }{dt}&=\frac{P_m \omega _s }{2H}t \\ \Rightarrow \; \delta &= \frac{P_m \omega _s }{4H}t^2+C'\\ C'&= \delta |_{t=0^+}=\delta _0\\ \delta &= \frac{P_m \omega _s }{4H}t^2+\delta _0\\ \text{Now in problem,}\\ \delta (t=0.02)&\\ &=\frac{1 \times 2 \times 180 \times 50 \times 0.02^2 +5^{\circ}}{4 \times 2}\\ &=5.9^{\circ} \end{aligned}
Question 7 |
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 ______.
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 ______.
0.11 | |
0.23 | |
0.48 | |
0.64 |
Question 7 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}
Question 8 |
There are two generators in a power system. No-load frequencies of the generators are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of 1Hz/MW. Total load in the system is 2.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in the steady state is ______.
25 | |
50 | |
75 | |
100 |
Question 8 Explanation:
Let no-load frequency of generator-1 be 51.5 Hz and no-load frequency of generator-2 be 51 Hz.
Given total load = 2.5 Mwand drop of both machines =1 Hz/MW
Let machine-1 shares a load of P_1 MW them machine-2 will share a load of (2.5-P_1) MW
Le the steady state frequency of the system be f.
\begin{aligned} \text{Now, }\frac{51.5-f}{P_1}&=1=droop \\ P_1&= (51.5-f)\;\;...(i)\\ \text{Also, } \left ( \frac{51-f}{2.5-P_1} \right )&=1=droop \\ 51-f &=2.5-P_1 \\ P_1 &=f-48.5 \;\;...(ii) \\ \text{equating eq. }&\text{(i) and (ii), we get,} \\ 51.5-f &=f-48.5 \\ 2f&=100 \\ f&=50Hz \end{aligned}
Given total load = 2.5 Mwand drop of both machines =1 Hz/MW
Let machine-1 shares a load of P_1 MW them machine-2 will share a load of (2.5-P_1) MW
Le the steady state frequency of the system be f.
\begin{aligned} \text{Now, }\frac{51.5-f}{P_1}&=1=droop \\ P_1&= (51.5-f)\;\;...(i)\\ \text{Also, } \left ( \frac{51-f}{2.5-P_1} \right )&=1=droop \\ 51-f &=2.5-P_1 \\ P_1 &=f-48.5 \;\;...(ii) \\ \text{equating eq. }&\text{(i) and (ii), we get,} \\ 51.5-f &=f-48.5 \\ 2f&=100 \\ f&=50Hz \end{aligned}
Question 9 |
A synchronous generator is connected to an infinite bus with excitation voltage E_f= 1.3 pu. The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is _____.
0.8 | |
0.11 | |
0.45 | |
1.82 |
Question 9 Explanation:
Given, E_f=1.3pu, X_s=1.1pu, V=1 pu, P=0.6pu
Neglecting stator resistance i.e. \theta _s=90^{\circ}
Output power delivered,
\begin{aligned} P &=\frac{VE_f}{X_s}\sin \delta \\ \sin \delta &=\frac{PX_s}{VE_f} \\ &= \frac{0.6 \times 1.1}{1 \times 1.3}\approx 0.5\\ \therefore \;\; \delta &=30^{\circ}=\text{Power angle} \\ \therefore \;\; \cos \delta &= \cos 30^{\circ}=\frac{\sqrt{3}}{2} \end{aligned}
Reactive power is given by,
\begin{aligned} Q&=\left [ \frac{VE_f}{X_s} \cos \delta -\frac{V^2}{X_s} \right ]\\ &=\left [ \frac{1 \times 1.3}{1.1} \times \frac{\sqrt{3}}{2}-\frac{1^2}{1.1} \right ]\\ Q&=(1.023-0.909)=0.114pu\\ &\approx 0.11pu \end{aligned}
Neglecting stator resistance i.e. \theta _s=90^{\circ}
Output power delivered,
\begin{aligned} P &=\frac{VE_f}{X_s}\sin \delta \\ \sin \delta &=\frac{PX_s}{VE_f} \\ &= \frac{0.6 \times 1.1}{1 \times 1.3}\approx 0.5\\ \therefore \;\; \delta &=30^{\circ}=\text{Power angle} \\ \therefore \;\; \cos \delta &= \cos 30^{\circ}=\frac{\sqrt{3}}{2} \end{aligned}
Reactive power is given by,
\begin{aligned} Q&=\left [ \frac{VE_f}{X_s} \cos \delta -\frac{V^2}{X_s} \right ]\\ &=\left [ \frac{1 \times 1.3}{1.1} \times \frac{\sqrt{3}}{2}-\frac{1^2}{1.1} \right ]\\ Q&=(1.023-0.909)=0.114pu\\ &\approx 0.11pu \end{aligned}
Question 10 |
A cylinder rotor generator delivers 0.5 pu power in the steady-state to an infinite
bus through a transmission line of reactance 0.5 pu. The generator no-load voltage
is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator
is 5MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle,
in degrees, for a three-phase dead short circuit fault at the generator terminal is
53.5 | |
60.2 | |
70.8 | |
79.6 |
Question 10 Explanation:
\begin{aligned} P_e&= \frac{|E||V|}{(X_d+X_T)}\sin \delta _0\\ 0.5 &=\frac{1.5 \times 1}{(1+0.5)}\sin \delta _0 \\ \delta _0 &=\sin ^{-1}(0.5)=30^{\circ}=\frac{\pi}{6} \end{aligned}
Critical clearing angle
\begin{aligned} \delta _{cr}&=\cos ^{-1}[(\pi-2\delta _0) \sin \delta _0 -\cos \delta _0]\\ &=\cos ^{-1}\left [ \left ( \pi-\frac{\pi}{3} \right ) \times \frac{1}{2} -\frac{\sqrt{3}}{2}\right ]\\ &= \cos ^{-1} (0.18)=79.56^{\circ} \end{aligned}
There are 10 questions to complete.