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 |


There are 5 questions to complete.