transient stability of synchronous machines (and power systems). .... In order to answer these questions we now consider the mathematics of the swing equation.
SYSTEM STABILITY TRANSIENT STABILITY (Kundur – Chapter Thirteen) Introduction
Transient stability is the ability of the power system to maintain synchronism when subjected to a severe transient disturbance. It is primarily associated with parallel operation of the machines in the system, and in particular the synchronous generators. A synchronous generator loses stability when it falls out of step, or slips, with respect to the rest of the ac network; what happens in such a case is that the rotor of the synchronous generator advances beyond a certain critical angle at which the magnetic coupling between the rotor (and hence the turbine / prime mover) and the stator fails. The rotor, no longer held in synchronism with the rotating field created by the stator currents, rotates relative to this field and pole slipping occurs. Each time the poles traverse the angular region where stability obtains, synchronising forces attempt to pull the rotor back into step. Large synchronous machines falling out of step in this manner cause violent fluctuations in the neighbouring system voltages as well as violent groaning in the boilers feeding their turbines. A machine or group of machines that has fallen out of step must quickly be brought back into step or disconnected from the system. From the theory of the synchronous machine we have already seen that under steady state conditions the synchronising power between the rotor and stator reaches zero when the rotor angle reaches 90 degrees (ie. the steady-state stability limit). However, as we shall see, under transient conditions this angle may temporarily exceed 90 degrees without loss of stability (synchronism). This is an important difference between steady state stability and transient stability of synchronous machines (and power systems). The transient stability limit refers to the maximum possible flow of power past a point that will allow stability to be maintained following a sudden (generally large) disturbance. As we shall see, the transient stability limit is lower than the steady state stability limit and is therefore more important. Another extremely important aspect of transient stability is that it involves large excursions in the system variables and hence the non-linearities of the system have a considerable influence. The characteristics of a non-linear system are different at each different operating point and for each different disturbance and so transient stability by its very nature is analysed on a contingency basis. For example, even for a given steady state operating point, the transient stability limit varies with the type, duration and location of disturbances occuring in the transmission system. Since it is impossible to examine (and make a stability judgement on) every possible contingency, a good knowledge of the system is vital to ensure stability is assessed for all credible contingencies.
Analysis of Transient Stability Concepts With the advent of modern computers and powerful simulation programs it is now possible to simulate the behaviour of power systems using fairly detailed mathematical models that include the large-signal nonlinearities of the synchronous generators, excitation systems etc, all of which have a significant bearing on transient stability. This then is generally the manner in which transient stability is evaluated nowadays. However a great deal of conceptual insight into the fundamental nature of the transient stability problem, and the factors that influence it, can be obtained using the very simple models and graphical techniques that were employed prior to the advent of computers. For such conceptual analysis, the mathematical model employed is the swing-equation model, in which the synchronous generator is represented by its rms equivalent circuit model with constant field current (and hence constant field flux linkages). In addition, stator & transmission line resistance and mechanical friction in the turbine are neglected such that there are no sources of damping in the model. The only dynamics represented are those of the mechanical system (turbine and rotor inertia). The governor dynamics are neglected and the per unit torque and per unit power are considered equal and are used interchangeably [see Kundur pg 829]. The system model then appears as below.
Kinetic energy stored in rotating mass
E =
Angular momentum
M
ie
E =
1 2
M ω0
=
1 2
J ω0
2
Joules Joules – seconds / radian
J ω0
[ ω 0 synch speed (rad/s) , J polar moment of inertia (kg-m 2 ) ]
It has been found that energy E is proportional to the machine rating G by a certain factor known as the inertia constant H . By definition, for a machine of rating G MVA, the stored kinetic energy E = GH hence
=
H
=
1 Mω0 2
G
1 Mω0 2
=
MW sec MVA
stored energy at synch speed rated power
ie. the unit of H is time. The actual rotor speed
pθ
= ω 0 + pδ
Rotor acceleration
p 2θ
=
p 2δ
= ω
(1) (2)
[ where p =
d ] dt
Assuming the machine is generating (ie. δ is positive) with an input mechanical torque (power) Tm and an opposing electrical torque Te . The equation of motion is then J ( p 2δ ) = Tm − Te
= ∆T
= Ta
(3)
ie a positive accelerating torque Ta causes an increase in the rotor angle as one would expect.
p 2δ
Also from (3)
and since E =
∆T J
=
(∆T ω 0 ) ω 0 2
Jω0 2
2
=
∆Pω 0 2E
Mω0 2 p 2δ
Then
Also,
=
Pe
=
EV sinδ XT
=
∆P M
(4)
= Pmax sinδ
so M p 2δ
= ∆P = Pm − Pmax sinδ (5)
J p 2δ
= ∆T
The Swing Equation
= Tm − Tmax sinδ
The swing equation relates the difference between mechanical input power and electrical output power, with any imbalance resulting in acceleration and (hopefully) subsequent deceleration; ie. the system is an electro-mechanical oscillator. Transient stability is concerned with whether or not there is sufficient restorative power (synchronising power) to overcome an initial acceleration / deceleration following a disturbance. If the system is transiently stable, small signal stability is thereafter concerned with whether the electromechanical oscillations decrease in amplitude and eventually cease (positively damped) or persist (insufficient damping) or even increase (negative damping). Before looking at how the swing equation can be used, we need to convince ourselves of a few inherent background characteristics associated with the machine’s rotor variables under dynamic conditions.
At steady state: ∆ω
= ω − ω0
= 0 , thus δ is constant at steady state.
For δ to change from one operating point to another requires temporary non-zero ∆ω to enable rotor to change position relative to stator mmf moving at ω 0 . The rate at which δ changes is determined by the size of the speed deviation ∆ω ie.
d (δ ) = ∆ω dt
= ω − ω0
ie. we need to recognise that speed deviation is proportional to the rate of change of load angle, as well as an important consequence – speed deviation and angle deviation can never be zero simultaneously during the oscillation (90 degrees out of phase). Hence, during the electromechanical oscillations that follow a disturbance, speed deviation is at its largest when the angle is at its equilibrium point; conversely, the angle deviation is at its largest when ∆ω is zero (ie. when ω = ω 0 ). Now consider some examples ¾Generator experiences a sudden increase in shaft input power from Pm 0 to Pm1 According to steady-state theory the machine moves from operating point ( Pm0 ,δ 0 ) to ( Pm1 ,δ 1 )
But how does the change occur and is the system transiently stable for this change in operating point (disturbance)? We can deduce the answers from the swing equation and graphical techniques: Swing eqn (eqn 3) tells us that immediately following disturbance → (input Pm > output Pe ) and ∴ rotor angle increases.
+ve acceleration
Furthermore, this situation (accel. and ↑ in δ ) continues while Pm > Pe ie. until δ reaches δ 1 when Pe once again matches Pm1 .
At point b: Although Pe = Pm1 , (ie. no accel power) the rotor arrives at this point with nonzero speed, ∴ δ continues to increase [upwards past b towards c] Beyond b: Pe now > Pm1 and rotor decelerates; this continues till excess speed ∆ω decreases to zero (ie. rotor returns to synch. speed) at some (as yet undetermined) angle δ m . At point c: Now, although ∆ω is instantaneously zero, Pe still > Pm1 so decel. continues and ∆ω becomes negative and (as decel. goes on) becomes more and more –ve. Note: as ∆ω goes from +ve to –ve rotor has changed direction, ie. rotor angle stops increasing and starts decreasing at whatever angle δ m the speed decreases to zero.
From c to b: Rotor continues to decel. while Pe > Pm1 , so δ traces trajectory from δ m back towards δ 1 at which point Pe = Pm1 and acc. power = 0. But, rotor arrives back at δ 1 below synch. speed; ∴ δ continues to fall behind [downwards from b to a]. From b to a: At the instant the rotor falls behind correct angle δ 1 , Pm1 > Pe and rotor begins to re-accelerate to get speed back up to synch.; in the finite time this takes, rotor angle decreases back to δ 0 when ∆ω = 0 (ie. ω = ω 0 ) but Pm still > Pe and rotor continues to accel. → whole procedure starts again. Rotor angle oscillates between δ 0 and δ m around new steady-state value δ 1 . If no damping is present, oscillations persist indefinitely. If damping is present angle oscillations decay to δ 1 . Thus far we have assumed that at some angle δ m , ω returns to ω 0 (ie. that ∆ω returns to zero and the rotor ∴ turns around and comes back). How do we tell whether this will happen (ie. whether stability is maintained) and what δ m will be? Equal Area Criterion In order to answer these questions we now consider the mathematics of the swing equation [eqn (5)] in more detail; the equation is non-linear and so cannot be solved directly – thus either we use numerical integration (computer simulation) or graphical techniques. Multiplying dδ we get both sides of eqn (4) by 2 dt dδ 2 dt
d δ 2 dt 2
dδ dt
=
d dδ dt dt
2
2 M
=
2
=
δ
∫ ∆P dδ
2∆P dδ M dt
(6)
δ0
Equation (6) gives a closed-form expression for the speed deviation ∆ω = dδ dt (all squared) of the machine following a disturbance. As we have seen, the speed deviation is initially zero at steady state, but then changes following a disturbance. For stable operation of the system following a disturbance the time response of the rotor angle δ must be bounded, that is to say that the rotor angle δ must reach some maximum value δ m on its first post-disturbance swing. As we have already seen, if this is the case, the speed deviation of the machine is zero when it reaches this maximum rotor angle δ m . Thus from eqn (6) as a criterion for stability we can write 2 M
δm
∫ ( Pm − Pe ) dδ
δ0
= 0
(7)
where δ 0 is the initial rotor angle and δ m is the maximum rotor angle to which the machine swings after the disturbance. In other words, eqn (7) is saying that the system will remain stable (bounded response) provided the system can reach a post-fault situation where the speed deviation reaches zero, and at which the rotor angle will then have stopped increasing (and by extension will subsequently start decreasing from δ m ). Furthermore from (7) we see that we can simplify the criterion further by dividing both sides by 2 M to yield δm
∫ ( Pm − Pe ) dδ
= 0
(8)
δ0
Eqn (8) leads to an insightful graphical and physical understanding of the stability criterion. Firstly, from mathematics we can interpret / equate the integral in eqn (8) to the area under the graph of the function ( Pm − Pe ) plotted against δ , which must then be zero between δ 0 and δ m if the system is to be stable. Splitting the integral in (8) into two parts we have δm
δ1
∫ ( Pm − Pe ) dδ
∫ ( Pm − Pe ) dδ
+
δ0
= 0
(9)
δ1
⇒ (Area A1 ) + (Area A2 ) = 0
(9)
In Fig. 1 this is satisfied because the positive area A1 enclosed between the Pe and Pm curves from δ 0 to δ 1 (positive since Pm > Pe from δ 0 to δ 1 ) is equal to the negative area A2 enclosed between Pe and Pm from δ 1 to δ m (negative since Pm < Pe from δ 1 to δ m ). A further useful insight is gained by recognising that the area A1 from δ 0 to δ 1 is equal to the kinetic energy E1 (positive) gained by the rotor during its acceleration as δ changes from δ 0 to δ 1 . If the system is to have a stable (bounded) response, clearly the energy gained during acceleration must be returned during a later deceleration and hence eqn (9) can be rewritten as E1
+
E2
= 0
(10)
Equations (8) and (9) form the mathematical basis for what is known as the equal-area criterion for assessing the stability of a synchronous machine connected to a power system. The criterion can be used to determine the maximum permissible (step) increase in the mechanical power ∆P = Pm1 − Pm0 in the previous example: in this case the system will remain stable provided an area A2 above the new value of Pm can be found to match the area A1 below the new value of Pm . As already seen in Fig. 1, in finding this area A2 we simultaneously find the value of the maximum rotor angle δ m that occurs in the ensuing transient response (why?). However, what of our earlier question – how big can δ m be before stability is lost?
Firstly, it is important to understand that to cause a bigger excursion in δ m we need to make a bigger disturbance, in this case a bigger increase in Pm (why?). As the new value of Pm increases, the maximum possible positive value of A2 (area above Pm and below Pe ) becomes smaller and the area A1 becomes bigger. Physically this means that as the step increase in Pm gets bigger more energy is being accumulated by the rotor during its initial acceleration but, paradoxically, there is less possibility of returning this excess energy to the transmission system during the subsequent deceleration. This is illustrated in Fig. 2 for three values of ∆P , each progressively larger than the small value of ∆P illustrated in Fig. 1.
The diagram in Fig. 2 (a) shows the situation for a moderate increase in Pm . The diagram in Fig. 2 (a) illustrates firstly that because of the larger step increase in input power, the new equilibrium value of load angle (at which Pm is matched by Pe ) is, as expected, larger than that of Fig. 1 in which the step increase in Pm was small. The diagram also illustrates that, because of this larger increase in Pm the positive area enclosed below the new Pm curve and above the P − δ curve is now larger since a greater kinetic energy is imparted to, and stored in, the rotor by the time it has accelerated from δ 0 to its new equilibrium angle δ 1 . However the diagram shows that despite the increase in area A1 , a matching negative area A2 can still be found above the Pm curve and below the P − δ curve. However, graphically, since the matching area A2 must now be larger, consequently the rotor has to swing to a greater value of δ m in order for the increased energy stored in the rotor during acceleration to be returned to the transmission system during subsequent deceleration (physically speaking, the rotor has to decelerate for longer to return all its excess energy, and in doing so its rotor angle opens out to a larger value of δ m ). Finally, Fig. 2 (a) shows that for this moderate step increase in Pm there is “reserve” (unused but available) decelerating energy corresponding to area A3 : that is the rotor could, if required, open out to a larger angle to continue the deceleration of the rotor and hence there is a stability margin in the system for this magnitude of step increase in Pm .
The diagram in Fig. 2 (b) shows the situation for a large increase in Pm . The diagram illustrates that, even for this large step increase ∆Pm , an area above the Pm and below the P − δ curve (negative accelerating energy) can still be found to match the large area A1 below the Pm curve (positive acc. energy). Again, as expected, this even larger increase in Pm requires an even larger new equilibrium value of load angle δ 1 and hence a larger value of maximum rotor angle δ m as the rotor decelerates for even longer to transfer its excess kinetic energy out into the transmission system. However, in Fig. 2 (b) there is now no unused area A3 above the Pm curve and below the P − δ curve: for this particular disturbance the rotor has transferred as much absorbed kinetic energy to the transmission system as it is able to do, since any further opening out of the rotor angle past δ m = δ L would result in a positive area between the Pm and P − δ curves. Thus there is no unused capacity to decelerate the rotor in the case of this disturbance and the system is at the limit of transient (first-swing) stability.
Finally, the diagram of Fig. 2 (c) shows the situation for an excessive increase in Pm . The diagram illustrates that for this step increase ∆Pm the area A1 below Pm and above the P − δ curve (positive Pm − Pe integral wrt δ ) cannot be matched by an equal area A2 above Pm and below the P − δ curve (negative Pm − Pe integral wrt δ ). That the system is unstable following this increase ∆Pm can be seen both from the graphical and physical interpretations of the diagram in Fig. 2 (c).
Graphically, any area enclosed by Pm and the P − δ curve to the right of the limiting value of load angle δ L is by definition positive and therefore adds to the positive area A1 . Hence if the negative area A2 enclosed by the Pm and P − δ curves between δ 1 and δ L is not equal to or greater than the positive area A1 , the criterion in eqn (9) can never be satisfied – even if δ continues to open out to δ m and beyond (which it will !) – and the system is thus unstable (for this disturbance). Physically, the generator rotor accelerates from its initial steady angle δ 0 to the new angle δ 1 required for the machine’s electrical power conversion Pe to match the new mechanical input power Pm1 , and gains excessive positive kinetic energy E1 = (area A1 ) in doing so. As in all previous cases, the rotor angle continues to open out past the required equilibrium value of δ 1 since it arrives at this value of δ 1 with non-zero speed deviation + ∆ω (excess kinetic energy). Along the subsequent trajectory from b to c the electrical power Pe generated by the machine exceeds the mechanical input power Pm1 and the machine rotor decelerates, its excess kinetic energy being transferred to electrical energy in the transmission system. However by the time the rotor angle opens out to δ L at point c this deceleration process has not succeeded in removing all excess positive kinetic energy from the rotor (because the decelerating energy E 2 = (area A2 ) is less than the positive accelerating energy E1 = (area A1 ) ) and so the rotor arrives at angle δ L still with excess positive kinetic energy (shaft speed ω > synch ω 0 ). With this excess shaft speed the rotor angle δ continues to increase, but for values of δ > δ L the diagram shows that the generated electrical power Pe is less than the input mechanical power
Pm1 , and hence the machine rotor is once again subjected to a net positive accelerating power Pa = Pm − Pe . Thus as soon as the angle opens out past δ L the machine is past the point of no return: the rotor speed is still larger than ω 0 and so the angle δ continues to increase, but as it does so it suffers ever greater positive accelerating power Pa and the machine speed must continue to increase and the rotor angle will run away until the machine starts to pole slip and stability is lost. Note also – and very importantly – that the system in Fig. 2 (c) is steady state stable at the new value of mechanical input power Pm1 : there is a value of load angle δ 1 on the upward slope of the P − δ curve at which the machine is able to match the input mechanical power Pm1 with generated electrical power Pe1 and hence remain at synchronous speed. However the system is not able to move to this stable steady state operating point ( Pe1 , δ 1 ) from the initial steady state operating point ( Pe0 , δ 0 ) via a step change in the input power Pm . The system may however be able to survive a more gradual increase in Pm to take it to point ( Pe1 , δ 1 ). The other interesting counter point [cf. Fig 2 (b)] is that under transient conditions the machine can swing past the steady state stability limit ( δ = 90 $ ) and still remain stable. This illustrates two important points. Firstly, there is a difference between steady-state and transient stability; the fact that the system is steady state stable at a post-disturbance operating point does not necessarily imply that it can stably arrive at that operating point following that disturbance. Secondly, transient stability is highly dependent on the nature of the disturbance – in this case a less severe way of changing the system input from its pre- to post-disturbance values could allow stability to be maintained. As a result, describing a system as transiently stable can strictly only be done with reference to a particular disturbance or contingency (or set of contingencies) and this is where realistic scenario planning and knowledge of the system is important. Examples of such contingencies follow. [ see also: Kundur pp 829 (loss of line) and 833 (short circuit fault) ]
FACTORS INFLUENCING TRANSIENT STABILITY
¾Generator inertia
¾Generator loading
¾Generator output (power transfer) during fault – depends on fault location and fault type
¾Fault clearing time
¾Post-fault transmission system reactance
¾Generator reactance
¾Generator internal voltage magnitude – this depends on field excitation, ie. the power factor of the power sent out at the generator terminals
¾Infinite bus voltage magnitude
EXCITATION CONTROL FOR TRANSIENT STABILITY What do we know about transient stability?
¾Primarily concerned with maintenance of synchronism for large disturbances.
¾Equal area criterion approach to analysis has shown that we need to consider stability on a contingency basis, in each case determining whether the system has sufficient capacity to decelerate (or accelerate) the generator rotor after a fault is cleared.
¾Equal area criterion approach has also shown that the key to ensuring transient stability is maintained is the ability to transfer rapidly any excess energy (either +ve or –ve) out of the rotor and into the transmission system, both during (if possible) and after the fault.
¾Finally, the equal area criterion approach has illustrated (a) An important problem associated with severe faults: the voltage at the terminals of the generator can drop significantly. (Why?)
(b) That the consequence of this temporary drop in terminal voltage is a reduction in the all-important ability to transfer synchronising power out of the generator. (b) That the solution is to get the terminal voltage back up as soon as possible.
The generator’s automatic voltage regulator (AVR) works through the excitation system to maintain (within reason) constant generator terminal voltage. Thus the AVR plays a crucial role with respect to transient stability, by attempting to maintain terminal volts under fault conditions and ensuring a fast terminal voltage recovery profile after the fault is cleared.
Under transient conditions, to get the best (fastest) possible terminal voltage recovery profile, the attributes of an excitation system that are beneficial are high speed and high ceiling voltage so as to overcome the inherent delay of the field circuit. The excitation control solution (for transient stability) is a high-gain, fast-response exciter with high ceiling voltage and an AVR. We will see, however, that this approach is detrimental to the damping of the generator rotor oscillations, but that the problem can be fixed.
How is the AVR working? The AVR works by forcing the field voltage in such a way as to maintain the flux levels in the synchronous machine under fault conditions in opposition to the tendency of the flux to be diminished by the effects of armature reaction. This in turn maintains the terminal voltage and hence the ability of the generator to continue to transfer power onto the busbars.
Note: the effects of armature reaction on flux levels are particularly pronounced during faults because the fault currents are large and predominantly (lagging) reactive in nature (ie. the reactive output power of the generator is large under fault conditions).
The longer the fault duration and the closer the fault is to the generator terminals, the worse this effect gets... consequently, the more difficult it becomes for the AVR to maintain constant flux linkages.
INCREASED RELIANCE ON EXCITATION CONTROL Modern trends have contributed to lower stability margins: ¾Increased generator ratings, lower inertias and higher reactances ¾Increasing interconnection and reliance on the transmission system to carry generator loading This has led to increased reliance on excitation control. The use of fast exciters allows operation with higher system series reactances, but there is a view that these gains cannot go on forever. Indeed, modern exciters are now virtually instantaneous, so any further increases in system performance in this regard will require alternative methods of compensation. METHODS OF IMPROVING TRANSIENT STABILITY These can broadly be categorised as (a) Reduction of the disturbance (fault severity and duration) (b) Increase restorative synchronising forces (c) Reduce accelerating power (torque) mechanically or by artificial load ¾High-speed fault clearing ¾Reduction of transmission system reactance •
Static (series capacitors)
•
Dynamic (FACTS)
¾Regulated shunt compensation (traditional / FACTS) ¾Dynamic braking ¾Single pole switching ¾Fast valving ¾Generator tripping ¾High-speed excitation systems ¾Flexible AC Transmission Systems (FACTS)
