...Contd... Equal Area Criterion

Now consider the case when the line is reclosed at δ_c such that the area of acceleration is larger than the area of deceleration, i.e., A₁ > A₂ . The generator load angle will then cross the point δ_m , beyond which the electrical power will be less than the mechanical power forcing the accelerating power to be positive. The generator will therefore start accelerating before is slows down completely and will eventually become unstable. If, on the other hand, A₁ < A₂ , i.e., the decelerating area is larger than the accelerating area, the machine will decelerate completely before accelerating again. The rotor inertia will force the subsequent acceleration and deceleration areas to be smaller than the first ones and the machine will eventually attain the steady state. If the two areas are equal, i.e., A₁ = A₂ , then the accelerating area is equal to decelerating area and this is defines the boundary of the stability limit. The clearing angle δ_c for this mode is called the Critical Clearing Angle and is denoted by δ_cr. We then get from Fig. 9.3 by substituting δ_c = δ_cr

We can calculate the critical clearing angle from the ab move equation. Since the critical clearing angle depends on the equality of the areas, this is called the equal area criterion.

Now a frequently asked question is what does the critical clearing angle mean?

Since we are interested in finding out the maximum time that the circuit breakers may take for opening, we should be more concerned about the critical clearing time rather than clearing angle. Furthermore, notice that the clearing angle is independent of the generalized inertia constant H . Hence we can comment that the critical clearing angle in this case is true for any generator that has a d-axis transient reactance of 0.20 per unit. The critical clearing time, however, is dependent on H and will vary as this parameter varies.

To obtain a description for the critical clearing time, let us consider the period during which the fault occurs. We then have P_e = 0. We can therefore write from

Integrating the above equation with the initial acceleration being zero we get

Further integration will lead to

Replacing δ by δ_cr and t by t_cr in the above equation, we get the critical clearing time as

To illustrate the response of the load angle δ , the swing equation is simulated in MATLAB. The swing equation of (9.14) is then expressed as

where Δω_r is the deviation for the rotor speed from the synchronous speed ω_s . It is to be noted that the swing equation of (9.21) does not contain any damping. Usually a damping term, that is proportional to the machine speed Δω_r, is added with the accelerating power. Without the damping the load angle will exhibit a sustained oscillation even when the system remains stable when the fault cleared within the critical clearing time.

Fig. 9.4 Stable and unstable system response as a function of clearing time.

Fig. 9.4 depicts the response of the load angle δ for two different values of load angle. It is assumed that the fault occurs at 0.5 s when the system is operating in the steady state delivering 0.9 per unit power. The load angle during this time is constant at 23.96° . The load angle remains stable, albeit the sustained oscillation when the clearing time t_cl is 0.253 s. The clearing angle during this time is 88.72° . The system however becomes unstable when the clearing time 0.2531s and the load angle increases asymptotically. The clearing time in this case is 88.77° . This is called the Loss of Synchronism. It is to be noted that such increase in the load angle is not permissible and the protection device will isolate the generator from the system.

The clearing time of (8.20) is derived based on the assumption that the electrical power P_e becomes zero during the fault as in (8.19). This need not be the case always. In that even we have to resort to finding the clearing time using the numerical integration of the swing equation. See example 9.5 to illustrates the point.

Example 9.5