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For one machine system and infinite bus-bars a method known as ‘equal area criterion of stability’ is employed. The use of this method eliminates partially or wholly the calculations of swing curves and thus saves a considerable amount of work. The method is applicable to any two machine system. This method is not applicable to multi-machine system directly.

The principle of this method consists of the basis that when δ oscillates around the equilibrium point with constant amplitude, transient stability will be maintained.

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Starting with swing equation –

Multiplying both sides of the above equation by dδ/dt, we get –

Rearranging, multiplying by dt and integrating, we have –

where δ_{0} is the torque angle at which the machine is operating while running at synchronous speed under normal conditions. Under the above conditions the torque angle was not changing, i.e., before the disturbance dδ/dt = 0. Also if the system has transient stability the machine will again operate at synchronous speed after the disturbance, i.e., dδ/dt = 0.

Hence the condition for the transient state stability is given by the equation –

This means that the area under the curve P_{A} should be zero which is possible only when P_{A} has both accelerating and decelerating powers, i.e., for a part of the curve P_{s} > P_{E} and for the other P_{E} > P_{s} (Fig. 7.7). For a generator action P_{s} > P_{E} for positive area A_{1} and P_{E} > P_{s} for negative area A_{2} for stable operation.

Hence the name equal area criterion. The area A_{1} represents the kinetic energy stored by the rotor during acceleration and the area A_{2} represents the kinetic energy given up by the rotor to the system and when it is all given up, the machine has returned to its original speed. It is to be noted that the kinetic energy involved in our explanation is fictitious as it is being calculated corresponding to relative speed rather than the actual speed.

The equal area criterion is also useful in determining the maximum limit on the load that the system can take without exceeding the stability limit. This can happen as long as area between P_{s} line and the P_{E} curve is equal to the area between the initial torque angle δ_{0} and the P_{s} line. In case the area A_{2} is less than the area A_{1}, the system will become unstable.

**The main cases for the study of transient stability problems in a system are: **

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1. Sudden changes in load.

2. Switching of one of the lines which causes a change in the reactance of the system and hence a change in load conditions.

3. Sudden fault on the system which causes reduction in output, requiring arrangement for clearance of fault rapidly, and study of after-fault condition which may cause part of the system outage.

In each case, the procedure will be to determine the power angle curve for the initial condition of the system, for the condition under fault, and for the after-fault condition and plot the curves in per unit values. Then locate the points for the initial load conditions finding out δ_{0}.

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Then using equal-area criterion, determine the new angle of displacement δ. The maximum angle δ_{max} which may be allowed and the corresponding maximum permissible load can also be determined.

**Effect of Increase in Load**:

**There are two possibilities of increasing load from the point of view of transient disturbance and system stability:**

(i) Total load exceeds the steady-state limit of the system.

(ii) The rate of increase in load is so high as to set up oscillations which cross the critical point.

Consider a synchronous motor connected to an infinite bus-bar and operating at synchronous speed with mechanical output P_{0}. Let the load angle corresponding to output of P_{0} be δ_{0} determined from power angle diagram (Fig. 7.8).

The power angle diagram is drawn using the relationship.

where V_{G} is the voltage of the infinite bus-bars, V_{M} is the voltage behind the transient reactance of the synchronous motor, X is the transient reactance of the system (motor, transmission line, transformers etc.), and δ is the load or torque angle.

Under these conditions the mechanical power output P_{0} and electrical input P_{E} are equal .Now let the load on the motor be suddenly increased to P_{s}. Momentarily there is no change in load angle δ corresponding to electrical input to the motor P_{E} < P_{s}. Therefore, the motor decelerates supplying extra power required, as a result the load angle δ increases and the electrical input power P_{E} starts increasing.

Thus electrical input keeps on increasing till it becomes equal to output power. This operating point is given by the point b on the diagram and the load angle corresponding to P_{s} is δ_{S}. At point b electrical input power P_{E} is equal to mechanical output power, P_{s} and therefore, decelerating force is zero but due to inertia of the rotor, the load angle goes on increasing.

The speed of the motor beyond point b starts increasing. The speed is minimum at operating point b. When the speed of the motor becomes equal to synchronous speed beyond point b, say at point c, the load angle stops increasing.

But between operating points c and b, electrical power input P_{E} > mechanical output power P_{s}; therefore the motor accelerates and the load angle δ starts decreasing till it reaches the point b where this time the speed is maximum and is more than the synchronous speed.

At point b the acceleration ceases and the rotor moves towards point a with a finite deceleration. In this way the cycle keeps on repeating itself between points a and c till the oscillations are damped out. Applying ‘equal area criterion of stability’ and ignoring the losses, the area of the sectioned part above the curved portion ab must be equal to the sectioned part below the curved potion bc.

The first area A_{1} represents the rotational energy lost by the rotor which is due to the retarding torque given by –

T_{R} = T_{S} – T_{E}

Also the energy gained by the rotor is represented by the sectioned area A_{2}, i.e., area under the portion be.

This is due to accelerating torque given by –

T_{A} = T_{E} – T_{S}

Since speeds at points a and c are synchronous, so the two areas are equal.

Let the load angle corresponding to the point c, known as the point of maximum swing of the rotor, be δ_{m}. This angle should be such that the stability of the system is maintained.

It follows from the above discussion that the point c should be so located that A_{1} = A_{2}.

In fact if the point c can lie anywhere on the power angle diagram above the line b b’ the system will have transient stability. The point b’ on the curve will give the load angle δ_{C} beyond which P_{E} will be less than P_{s }and an unstable operation will set in. At this instant the synchronism will be lost and the load angle δ will continue to increase till the motor stops.

The value of load angle beyond which the operation becomes unstable is called the critical angle of load, δ_{critical}.

It is thus obvious that for maintaining transient stability it is necessary that rotor swing must not be beyond the critical load angle δ_{C}. This condition will reach either if the initial operating load angle δ_{0} is too large or the sudden increase of power to be met was too high.

**Effects of Switching Operation**:

Consider a generator supplying power to an infinite bus over two transmission lines operating in parallel, as shown in Fig. 7.9. The power angle diagram for the two lines operating in parallel is represented by curve A in Fig. 7.10.

Curve B represents the power angle diagram after isolating one of the lines. Let the mechanical power input be P_{s} corresponding to point a on power angle curve A (power angle curve drawn for two lines), the load angle being δ_{0}. Now let one of the lines be opened instantaneously and simultaneously at both ends.

As a result the equivalent impedance between the bus-bars is increased and hence curve B lies below curve A. As soon as one of the line is opened, the output of the generator is reduced, say to P_{1} (say operating point b on curve B) and since input P_{s} remains constant which is higher than the output, the rotor accelerates and therefore, the torque angle δ increases and the operating point moves along curve B towards d from b.

As it reaches d, the accelerating power becomes zero and the speed of the generator becomes more than that of infinite bus and the speed continues to increase. From d to e the rotor experiences deceleration but the speed is more than that of the infinite bus till it reaches point e where the relative speed is zero and the load angle ceases to increase. At point e, the output exceeds the input and hence the rotor decelerates and the speed goes down relative to the infinite bus till it reaches again point d where the speed is minimum.

This load angle continues to decrease till it reaches point b where again the speed is equal to the speed of infinite bus. The cycle repeats itself in absence of damping. In practice due to damping present the rotor operates at point d on curve B and the torque angle is δ_{S}.

In order to determine the transient stability limit for this case the input line P_{s }should be so raised that area below the line P_{s} and the curve B and above the line P_{s} and the curve B at the intersection of line P_{s} and curve B are equal. This is illustrated in Fig. 7.11. The value of P_{s} corresponding to this situation is known as the transient stability limit.

**The transient stability in this case depends upon: **

(i) Steady state limit of the remaining system and

(ii) The difference in initial and final load angles.

**Faults with Subsequent Circuit Isolation**:

Curve A in Fig. 7.12 represents the power angle diagram corresponding to healthy condition of the system shown in Fig. 7.9. Curve B represents corresponding to fault on one of the two lines and fault allowed to exist for some time. Curve C corresponds to the situation when the faulted line is removed.

Let P_{s} be the power supplied by the generator to the infinite bus in normal conditions and let δ_{0} be the corresponding load angle. At the time of fault, the output of the generator reduces and becomes as at a’.

Hence the rotor accelerates and rotor moves along the curve B to point b’ when the faulty line is removed and the operating point becomes b on curve C where the output exceeds the input and the rotor decelerates till the speed becomes equal to the speed of the infinite bus and the load angle ceases to increase at point c.

It is thus clear that the transient stability limit not only depends upon the type of disturbance but it also depends upon the clearing time of the breaker. Faster the breaker operation smaller will be the area A_{1} and therefore, larger will be the transient stability limit.

**Generator Connected to Infinite Bus through a Line****: **

Now let us consider another case of a generator connected to the infinite bus through a single line, as shown in Fig. 7.13. Let P_{s} be the power supplied by generator to the infinite bus in normal conditions and let δ_{ 0} be the corresponding load angle. Now say there occurs a 3-phase fault on the line temporarily. The power angle curve (Fig. 7.14) will correspond to horizontal axis because output power becomes zero.

If the breaker recloses after some time corresponding to clearing angle δ_{C} when the fault vanishes, the output will be more than the input and, therefore, the rotor decelerates. Finally, if the clearing angle δ_{C} is such that A_{1} = A_{2}, the system becomes stable.