# Power System Stability

 Question 1
A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of 15 MJ/MVA. If the input and output powers of the alternator are 15 MW and 10 MW, respectively, the angular acceleration in mechanical $degree/s^2$ is __________. (round off to nearest integer)
 A 25 B 50 C 75 D 100
GATE EE 2022      Power System Stability
Question 1 Explanation:
We have, swing equation
$\frac{2HS}{\omega _s}\frac{d^2\delta }{dt^2}=P_a$
Put the values,
\begin{aligned} \frac{2 \times 15 \times 20}{360 \times 50}\frac{d^2\delta }{dt^2}&=15-10\\ \frac{d^2\delta }{dt^2}&=\frac{5 \times 360 \times 50}{2 \times 15 \times 20}\\ &=150 \text{ electric degree per} sec^2\\ \frac{d^2\delta }{dt^2}&=\alpha =\frac{2}{4} \times 150 \text{ Mech. degree/}sec^2\\ &=75\text{ Mech. degree/}sec^2 \end{aligned}
 Question 2
Two generating units rated for 250 MW and 400 MW have governor speed regulations of 6% and 6.4%, respectively, from no load to full load. Both the generating units are operating in parallel to share a load of 500 MW. Assuming free governor action, the load shared in MW, by the 250 MW generating unit is _________. (round off to nearest integer)
 A 100 B 150 C 200 D 250
GATE EE 2022      Power System Stability
Question 2 Explanation:
Let no-load frequency is 50 Hz.
Draw the curve :

From the curve,
\begin{aligned} \frac{50-f}{3}&=\frac{P_1}{250} \;\;\;..(1) \\ \frac{50-f}{3.2}&=\frac{P_2}{400} \;\;\;..(2) \end{aligned}
From eq. (1) & (2),
\begin{aligned} \frac{3P_1}{250}&= \frac{3.2P_2}{400} \\ 3P_1&=2P_2\;\;\;...(3) \end{aligned}
Given: $P_1+P_2=500$
From eq. (3),
$P_1+1.5P_1=500$
$P_1=200MW$
 Question 3
In the figure shown, self-impedances of the two transmission lines are $1.5j\:p.u$ each, and $Z_{m}=0.5j \:p.u$ is the mutual impedance. Bus voltages shown in the figure are in p.u. Given that $\delta>0$, the maximum steady-state real power that can be transferred in p.u from Bus-1 to Bus-2 is

 A $\left | E \right |\left | V \right |$ B $\frac{\left | E \right |\left | V \right |}{2}$ C $2\left | E \right |\left | V \right |$ D $\frac{3\left | E \right |\left | V \right |}{2}$
GATE EE 2021      Power System Stability
Question 3 Explanation:

\begin{aligned} L_{e q}&=\frac{L_{1} L_{2}-M^{2}}{\left(L_{1}+L_{2}-2 M\right)} \\ X_{\mathrm{eq}}&=\frac{1.5 \times 1.5-0.5^{2}}{1.5+1.5-2 \times 0.5}=1 \mathrm{p.u.} \end{aligned}

\begin{aligned} P_{\max }&=\frac{|E||V|}{1}\\ P_{\max }&=|E||V| \end{aligned}
 Question 4
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 _________
 A 12.8 B 28.4 C 20.5 D 32.6
GATE EE 2019      Power System Stability
Question 4 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 5
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
 A greter than $|\frac{V_{1}V_{2}}{X}|$ B less than $|\frac{V_{1}V_{2}}{X}|$ C equal to $|\frac{V_{1}V_{2}}{X}|$ D equal to $|\frac{V_{1}V_{2}}{Z}|$
GATE EE 2018      Power System Stability
Question 5 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 $P_{max} \lt \left | \frac{V_1V_2}{x} \right |$
 Question 6
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.
 A 1.27 B 2.7 C 12.7 D 127
GATE EE 2017-SET-2      Power System Stability
Question 6 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 7
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 _________.
 A 0.729 B 1.22 C 2.6 D 4.8
GATE EE 2017-SET-1      Power System Stability
Question 7 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 8
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 A B B C C D D
GATE EE 2015-SET-2      Power System Stability
Question 8 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 9
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 _________.
 A 0.5 B 2.8 C 4.2 D 5.9
GATE EE 2015-SET-1      Power System Stability
Question 9 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}
 Question 10
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 ______.
 A 0.11 B 0.23 C 0.48 D 0.64
GATE EE 2014-SET-3      Power System Stability
Question 10 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}
There are 10 questions to complete.