Power System Stability Questions and Answers

Answer: A

Both statements are true. A power swing is an oscillation of the machine rotor angles after a disturbance, such as a line switching or fault clearing, and a stable system rides through it and settles down, so a swing does not necessarily mean that the system is unstable. Because a swing can look to a distance relay like a fault (the impedance moves into the relay's zone), the relay must tell the two apart, which is done by power swing blocking, so that it trips for a fault but not for a stable swing.

Hence true and true is the right pair, option A.

Answer: A

Detailed models of the synchronous machine for stability and dynamic studies are developed with Park's transformation (the dq0 transformation), which converts the stator quantities from the a, b, c frame to a frame rotating with the rotor. In that frame the inductances, which vary with rotor position in the abc frame, become constants and the equations become much simpler.

Maxwell's equations and Gauss's equations describe the electromagnetic field in general and are not used to build machine models in this way, and Kirchhoff's laws are only the circuit laws applied after the model is set up. The answer is option A.

Answer: C

Transient stability is improved by fast fault clearing, auto reclosing of breakers (the fault is often temporary), high speed excitation systems, increasing the system voltage (which raises the power that the lines can transfer), reducing the transfer reactance, and so on. These are the correct techniques.

The key gives decreasing the X/R ratio as the one that is not a correct technique. Lowering X/R by adding resistance does not raise the power limit of the line and wastes power, so it is not a way of improving stability; the usual aim is a low reactance. The answer is option C.

Answer: A

The key marks both statements true. The equal area criterion is a graphical method of testing the transient stability of a single machine connected to an infinite bus, so statement 1 is true. The swing equation, \(\dfrac{2H}{\omega_s}\dfrac{d^2\delta}{dt^2} = P_m - P_e\), is the basic equation of rotor motion; it is solved step by step for transient stability, and when it is linearised for a small disturbance it gives the criterion for steady state (small signal) stability, so it is used in both studies and statement 2 is accepted as true.

So the pair true and true, option A, is the key. The statement 2 is only loosely worded, since the swing equation is more often linked with transient stability.

Answer: C

The inertia constant \(H\) is the kinetic energy stored in the rotor at synchronous speed per MVA of machine rating, so the stored energy is the product of \(H\) and the rating.

\[ \begin{aligned} W = H \times S \\ &= 8.0\ \text{MJ/MVA} \times 100\ \text{MVA} \\ &= 800\ \text{MJ} \end{aligned} \]

The frequency, number of poles and voltage are not needed. The values 8 MJ and 80 MJ would be for 1 MVA and 10 MVA, and 8000 MJ for 1000 MVA. The answer is 800 MJ, option C.

Answer: D

The power angle asked for is the angle of the generator terminal voltage \(V_t\) ahead of the infinite bus. The reactance between them is the transformer in series with the two parallel lines of j0.5 each.

\[ X = 0.1 + \frac{0.5}{2} = 0.35\ \text{pu} \]

\[ \begin{aligned} P = \frac{V_tV}{X}\sin\delta \\ &= \frac{1 \times 1}{0.35}\sin\delta \\ &= 1 \Rightarrow \sin\delta = 0.35 \end{aligned} \]

\[ \delta = 20.5^\circ \]

The generator reactance of 0.25 pu would only be used for the angle of the internal voltage behind it, so it is not needed here. The answer is option D.

Answer: A

The power transferred is \(P_e = P_{max}\sin\delta\). Before the fault \(P_{max} = 2\) pu and the load is 1 pu.

\[ \begin{aligned} \sin\delta_0 = \frac{1}{2} \Rightarrow \delta_0 \\ &= 30^\circ = \frac{\pi}{6} = 0.5236\ \text{rad} \end{aligned} \]

The fault reduces the limit to 0.5 pu, which is what sets the swing afterwards, but the initial angle only depends on the pre fault values. The other options are larger angles of 36, 42 and 48 degrees. The answer is 0.523 rad, option A.

Answer: A

The inertia constant \(M\) in seconds squared per electrical degree is related to the stored energy per MVA, \(H\), by

\[ M = \frac{GH}{180 f} \]

Taking the stored energy of 2.52 MJ/MVA as H on a per MVA basis (G = 1):

\[ \begin{aligned} M = \frac{2.52}{180 \times 50} \\ &= 2.8 \times 10^{-4}\ \text{s}^2/\text{elec degree} \end{aligned} \]

This is the key, option A. If the 20 MVA rating were put in as G, the value would be 20 times larger (5.6e-3), which is not offered, so the item is read per unit MVA. The load of 18 MW is not needed.