# Transient Stability

Transient stability is concerned with the effect of large disturbances. These are usually due to faults, the most severe of which is the three-phase short circuit which governs the transient stability limits in the UK. Elsewhere, limits are based on other types of fault, notably the single-line-to-earth fault which is by far the most frequent in practice.

When a fault occurs at the terminals of a synchronous generator the real power output of the machine is greatly reduced as the voltage at the point of fault approaches zero and the only load on the machine is that of the inductive circuits of the generator. However, the input power to the generator from the turbine has not time to change during the short period of the fault and the rotor gains speed to store the excess energy. If the fault persists long enough the rotor angle will increase continuously and synchronism will be lost. Hence the time of operation of the protection and circuit breakers is all important.

An aspect of importance is the use of auto-reclosing circuit breakers. These open when the fault is detected and automatically reclose after a prescribed period (usually less than 1 s). If the fault persists the circuit breaker reopens and then recloses as before. This is repeated once more, when, if the fault still persists, the breaker remains open. Owing to the transitory nature of most faults, often the circuit breaker successfully recloses and the rather lengthy process of investigating the fault and restoring the line is avoided. The length of the auto-reclose operation must be considered when assessing transient stability limits; in particular, analysis must include

the movement of the rotor over this period and not just the first swing.

*dd*

If, in equation (8.2), both sides are multiplied by 2 — then

If the machine remains stable during a system disturbance, the rotor swings until its *dd . dd*

angular velocity — is zero; if — does not become zero the rotor will continue to move *dt dt*

and synchronism is lost. The integral of *JDPdd* in Equation (8.3) represents an area on the *P-d* diagram. Hence the criterion for stability is that the area between the P-d curve and the line representing the power input P_{0} must be zero. This is known as the *equal- area criterion.* It should be noted that this is based on the assumption that synchronism is retained or lost on the first swing or oscillation of the rotor, which may not always be the case. Physically, the criterion means that the rotor must be able to return to the system all the energy gained from the turbine during the acceleration period.

A simple example of the equal-area criterion may be seen by an examination of the switching out of one of two parallel lines which connect a generator to an infinite busbar (Figure 8.3). Initially the generator delivers *P** _{0}* at an angle d

_{0}through both lines. When one line is switched out, the reactance of the circuit and hence the angle d increases. If stability is retained, the two shaded areas (A

_{1}and A

_{2}) are equal and the swinging rotor comes initially to rest at angle d

_{2}, after which the damped oscillation converges to d

_{1}. In this particular case the initial operating power and angle could be increased to such values that the shaded area between d

_{0}and d

_{1}(A

_{1}) could be equal to the area between

*d*

*and d*

_{1}_{3}, where d

_{3}= 180 — d

_{:}; this would be the condition for maximum input power. If it swings beyond d

_{3}, the rotor continues to accelerate and instability results.

The power-angle curves for the condition of a fault on one of two parallel lines are shown in Figure 8.4. The fault is cleared when the rotor has swung to *d** _{1}* and the shaded area d

_{0}to d

_{1}(A

_{1}) indicates the energy stored. The rotor continues to swing until it reaches d

_{2}when the two areas A

_{1}and A

_{2}are equal. In this particular case

*P*

*is the maximum operating power for a fault clearance time corresponding to*

_{0 }*d*

*and, conversely,*

_{t}*d*

_{t}*is the critical clearing angle*for

*P*

_{0}*.*If the angle

*d*

*is decreased (for*

_{t}**Figure 8.3 **Power-angle curves for one line and two lines in parallel. Equal-area criterion. Resistance neglected

**Figure 8.4 **Fault on one line of two lines in parallel. Equal-area criterion. Resistance neglected. d_{1} is critical clearing angle for input power P_{0}

example, by faster clearance of the fault) it is possible to increase the value of P_{0 }without loss of synchronism.

The general case where the clearing angle d_{1} is not critical is shown in Figure 8.5. Here, the rotor swings to d_{2}, where the shaded area from d_{0} to d_{1} (A_{1}) is equal to the area d_{1} to d_{2} (A_{2}).

The time corresponding to the critical clearing angle is called the *critical clearing time* for the particular (normally full-load) value of power input. The time is of great importance to protection and switchgear engineers as it is the maximum time allowable for their equipment to operate without instability occurring. The critical clearing angle for a fault on one of two parallel lines may be determined as follows: Applying the equal-area criterion to Figure 8.5:

from which

and the critical clearing angle is obtained

If d_{1} is the critical clearing angle then it may be seen that

**Figure 8.5 **Situation as in Figure 8.4, but 8] not critical

It should be noted that a three-phase short circuit on the generator terminals or on a closely connected busbar absorbs zero power and prevents the generator outputting any real power to the system. Consequently, *P _{2}* = 0 in Figures 8.4 and 8.5.

Example 8.2

A generator operating at 50 Hz delivers 1 p.u. power to an infinite busbar through a network in which resistance may be neglected. A fault occurs which reduces the maximum power transferable to 0.4 p.u., whereas before the fault this power was 1.8 p.u. and after the clearance of the fault it is 1.3 p.u. By the use of the equal-area criterion, determine the critical clearing angle.

**Solution**

The appropriate load-angle curves are shown in Figure 8.4. * P_{0}* — 1p.u., P

_{2}— 0.4 p.u.,

*—1.3 p.u., and*

**P**_{2}*—1.8 p.u.*

**P**_{m}

Applying equation (8.4) (note that electrical degrees must be expressed in radians)

In a large system it is usual to divide the generators and spinning loads into a single equivalent generator connected to an infinite busbar or equivalent motor. The main criterion is that the rotating machines should be electrically close when forming an equivalent generator or motor. If stability with faults in various places is investigated, the position of the fault will decide the division of machines between the equivalent generator and motor. A power system (including generation) at the receiving end of a long line would constitute an equivalent motor if not large enough to be considered an infinite busbar.