The following problem appeared in GATE 2002 conducted by IISc Bangalore. GATE 2002 was the last GATE where subjective numericals were posed.The Electrical Engineering paper especially had some really nice problems. One such problem is given below. Anybody who can solve this problem without help, can rest assured that he/she has a really good understanding of power system stability concepts. Unfortunately such problems no longer appear since the test pattern was changed to objective type/ multiple choice since 2003.
Problem
A synchronous generator is to be connected to an infinite bus through a transmission line of reactance X = 0.2 pu as shown in the fig below. The generator data is as follows:
X'= 0.1 pu, E' = 1.0 pu, H = 5MJ/MVA, mechanical power $P_{m}$ = 0.0 pu, $\omega=2\pi 50$ rad/sec. All quantities are expressed on a common base.
The generator is initially running on an open circuit with the frequency of the open circuit voltage slightly higher than that of the infinite bus. If at the instant of closure $\displaystyle\delta=0$ and $\displaystyle\omega=\displaystyle\frac{d\delta}{dt}=\displaystyle\omega_{o}$. compute the maximum value of $\displaystyle\omega_{o}$ so that the generator pulls into synchronism.
Solution:
The swing equation derived in most text books is for the transient operation of the generator due to load fluctuations or faults. However, in this particular numerical, at the instant of closure the load angle $\delta=0$. After switching in as the generator tries to synchronize with the grid, $\delta$ increases and during this period the rotor angle $\theta_{r}=\delta$*.
Therefore, rotor speed during this period will be $\omega_{r}=\displaystyle\frac{d\theta_{r}}{dt}=\displaystyle\frac{d\delta}{dt}$
Using the swing equation for per unit quantities and the above equations we get
$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}\frac{d\omega_{r}}{dt}=P_{m}-\frac{EV}{X}\sin{\delta}}$
$P_{m}=0$ p.u. is given.
$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}\frac{d\omega_{r}}{dt}=-\frac{EV}{X}\sin{\delta}}$
$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}\frac{d\omega_{r}}{d\delta}\frac{d\delta}{dt}=-\frac{EV}{X}\sin{\delta}}$
$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}^2\frac{d\omega_{r}}{d\delta}=-\frac{EV}{X}\sin{\delta}}$
The maximum value of $\omega$ will occur when $\displaystyle\frac{d\omega_{r}}{d\delta}=0$. Substituting in above we get $\sin{\delta}=0\implies\delta=\frac{\pi}{2}$
$\boxed{\displaystyle\int_{\omega_{o}}^{\omega_{B}}\frac{2H}{\omega_{B}^2}\omega_{r}^2{d\omega_{r}}=-\int_{0}^{\delta}\frac{EV}{X}\sin{\delta}{d\delta}}$
$\boxed{\displaystyle\int_{\omega_{o}}^{\omega_{B}}\frac{2H}{\omega_{B}^2}\omega_{r}^2{d\omega_{r}}=-\int_{0}^{\displaystyle\frac{\pi}{2}}\frac{EV}{X}\sin{\delta}{d\delta}}$
Solving the above integral we get the solution as
$\boxed{\displaystyle\omega_{o}=\omega_{B}\left(1+\frac{3EV}{2HX\omega_{B}}\right)^{\displaystyle\frac{1}{3}}}$
$\omega_{o}=2{\pi}$(50.053) rad/sec. Slip frequency limit is seen to be 0.053 Hz which is agreeable with practical situations.
*Note that this is different from the equation for rotor angle in the case when a change in load or fault condition causes transients in a synchronized generator $\theta_{r}=\delta+\displaystyle\omega_{s}t$
