Now multiply the left-hand-side by ω and the right-hand side by dδ/dt (recall ω= dδ/dt) to get:

Transcription

1 Equal Area Criterion.0 Developent of equal area criterion As in previous notes, all powers are in per-unit. I want to show you the equal area criterion a little differently than the book does it. Let s start fro Eq. (.43) in the book. H d e a Re dt () Note in () that the book calls ω Re as ω R ; this needs to be 377 rad/sec (for a 60 Hz syste). We can also write () as H Re d dt e a () Now ultiply the left-hand-side by ω and the right-hand side by dδ/dt (recall ω= dδ/dt) to get: H Re d dt d e dt (3)

2 Note: d ( t) d( t) ( t) dt dt Substitution of (4) into the left-hand-side of (3) yields: (4) H Re d dt d e dt (5) Multiply by dt to obtain: H Re d d e (6) Now consider a change in the state such that the angle goes fro δ to δ while the speed goes fro ω to ω. Integrate (6) to obtain: H Re d e d (7) Note the variable of integration on the left is ω. This results in

3 3 Re d H e (8) The left-hand-side of (8) is proportional to the change in kinetic energy between the two states, which can be shown ore explicitly by substituting H=W k /S B =(/)Jω R /S B into (8), for H: Re d S J e B R (8a) Re d J J S e B R (8b) Returning to (8), let ω be the speed at the initial oent of the fault (t=0 +, δ=δ ), and ω be the speed at the axiu angle (δ=δ r ), as shown in Fig. below. Note that the fact that we identify a axiu angle δ=δ r indicates an iplicit assuption that the perforance is stable. Therefore the following developent assues stable perforance.

4 e 3 e e3 δ δ δ 3 δ c 90 δ r δ 80 Fig. Since speed is zero at t=0, it reains zero at t=0 +. Also, since δ r is the axiu angle, the speed is zero at this point as well. Therefore, the angle and speed for the two points of interest to us are (note the dual eaning of δ : it is lower variable of integration; it is initial angle): e δ=δ δ=δ r ω =0 ω =0 4

5 Therefore, (8) becoes: H Re r 0 e d (9a) We have developed a criterion under the assuption of stable perforance, and that criterion is: r d 0 e (9b) Recalling that a = - e, we see that (9b) says that for stable perforance, the integration of the accelerating power fro initial angle to axiu angle ust be zero. Recalling again (8b), which indicated the left-hand-side was proportional to the change in the kinetic energy between the two states, we can say that (9b) indicates that the accelerating energy ust exactly counterbalance the decelerating energy. Inspection of Fig. indicates that the integration of (9b) includes a discontinuity at the oent when the fault is cleared, at angle δ=δ c. Therefore we need to break up the integration of (9b) as follows: 5

6 6 0 3 c r c d d e e (0) Taking the second ter to the right-hand-side: c r c d d e e 3 () Carrying the negative inside the right integral: c r c d d e e 3 () Observing that these two ters each represent areas on the power-angle curve, we see that we have developed the so-called equal-area criterion for stability. This criterion says that stable perforance requires that the accelerating area be equal to the decelerating area, i.e., A A (3) where c d A e (3a)

7 r A e 3 d (3b) c Figure illustrates. e 3 A A e e3 e δ δ δ 3 δ c 90 δ r δ 80 Fig. Figure indicates a way to identify the axiu swing angle, δ r. Given a particular clearing angle δ c, which in turn fixes A, the achine angle will continue to increase until it reaches an angle δ r such that A =A. 7

8 .0 Stability perforance In the notes called owerangletiedoain.pdf, on pp. 9-0, we considered stability perforance in ters of what causes increased acceleration, or, decreased deceleration. We can consider siilarly here, in ters of A and A. Stability perforance becoe ore severe, or oves closer to instability, when A increases, or if available A decreases. We consider A as being bounded on the right by δ, because, as we have seen in previous notes, δ cannot exceed δ because δ>δ results in ore accelerating energy, not ore decelerating energy. Thus we speak of the available A as being the area within e3 - bounded on the left by δ c and on the right by δ. Contributing factors to increasing A, and/or decreasing available A, are suarized in the following four bullets and corresponding illustrations. 8

9 . increases: A increases, available A decreases e 3 A A e e3 e δ δ 3 δ c 90 δ r δ δ 80 Fig. 3 9

10 . e decreases: A increases. e 3 A A e e3 e δ δ 3 δ c 90 δ r δ δ 80 Fig. 4 0

11 3. t c increases: A increases, available A decreases e 3 A A e e3 e δ δ 3 δ c 90 δ r δ δ 80 Fig. 5

12 4. e3 decreases: available A decreases. e 3 A A e e3 e δ δ δ 3 δ c 90 δ r δ 80 Fig Instability and critical clearing angle/tie Instability occurs when available A <A. This situation is illustrated in Fig. 7.

13 e 3 A A e e3 e δ δ 80 δ 3 δ c Fig. 7 Consideration of Fig. 7 raises the following question: Can we express the axiu clearing angle for arginal stability, δ cr, as a function of and attributes of the three power angle curves, e, e, and e3? The answer is yes, by applying the equal area criterion and letting δ c =δ cr and δ r = δ. The situation is illustrated in Fig. 8. δ 3

14 e 3 A A e e3 e Fig. 8 Applying A =A, we have that δ δ 80 δ 3 δ cr δ cr d e e3 cr d (4) 4

15 The approach to solve this is as follows (this is #7 in your hoework #3):. Substitute e = M sinδ, e3 = M3 sinδ. Do soe calculus and then soe algebra. 3. Define r = M / M, r = M3 / M, which is the sae as r =X /X, r =X /X Then you obtain: cos cr M r r cos r r cos (5) And this is equation (.5) in your text. Your text, section.8., illustrates application of (5) for the exaples.4 and.5 (we also worked these exaples in the notes called ClassicalModel ). We will do a slightly different exaple here but using the sae syste. 5

16 Exaple: Consider the syste of exaples.3-.5 in your text, but assue that the fault is At the achine terinalsr = M / M =0. Teporary (no line outage)r = M3 / M =. The pre-fault swing equation, given by equation () of the notes called ClassicalModel, is H ( t) 0.8.3sin (6) Re with H=5. Since the fault is teporary, the post-fault equation is also given by (6) above. Since the fault is at the achine terinals, then the faulton swing equation has e =0, resulting in: H ( t) 0.8 (7) Re With r =0 and r =, the equation for critical clearing angle (5) becoes: coscr M cos (8) 6

17 Recall δ =π-δ ; substituting into (8) results in cos cr M cos (9) Recall the trig identity that cos(π-x)=-cos(x). Then (9) becoes: cos cr M cos (0) We can solve for δ fro the pre-fault swing equation, (with 0 accelerating power) according to sin 0.368rad.095 () In this case, because the pre-fault and post-fault power angle curves are the sae, δ is deterined fro δ according to () This is illustrated in Fig. 9. 7

18 M e e δ δ 80 Fig. 9 Fro (6), we see that =0.8 and M =.3, and (0) can be evaluated as cos cr cos M δ cr 0.8 (0.368) cos(0.368) δ

19 Therefore δ cr =.638rad= It is interesting to note that in this particular case, we can also express the clearing tie corresponding to any clearing angle δ c by perforing two integrations of the swing equation. H d e dt (3) Re For a fault at the achine terinals, e =0, so H d d Re Re dt dt H (4a) Thus we see that for the condition of fault at the achine terinals, the acceleration is a constant. This is what allows us to successfully obtain t in closed for, as follows. To solve (4a), we recall that ω=dδ/dt, so that (4a) ay be rewritten as dω dt = ω Re H dω = ω Re H dt (4b) (4c) 9

20 Then we ay integrate (4c) fro t=0 to t=t (on the right) and, correspondingly, ω=0 to ω (on the left), resulting in ω 0 dω Integration of (4d) results in t = ω Re o H dt (4d) ω = ω Re H t Again, recalling that ω=dδ/dt, we can express (4e) as (4e) which can be written as dδ dt = ω Re H t (4f) dδ = ω Re H tdt (5) Then we ay again integrate fro t=0 to t=t (on the right) and, correspondingly, δ=δ to δ (on the left), resulting in δ δ dδ t = ω Re o H tdt (6) Now integrate the right-hand side of (6) fro t=0 to t=t and the left-hand-side fro corresponding angles δ to δ, resulting in 0

21 Re ( t) H t (7) Solving for t yields: t 4H Re ( t) (8) So we obtain the tie t corresponding to any clearing angle δ c, when fault is teporary (no loss of a coponent) and fault is at achine terinals, using (8), by setting δ(t)=δ c. Returning to our exaple, where we had =0.8, H=5sec, δ =0.368rad, and δ cr =.638rad=93.86, we can copute critical clearing tie t cr according to t cr 4(5) (377)(0.8) The units should be seconds, and we can check this fro (8) according to the following: sec ( rad / sec)( pu) rad sec

22 I have used y Matlab nuerical integration tool to test the above calculation. I have run three cases: t c =0.8 seconds (6.8 cycles) t c =0.90 seconds (7.4 cycles) t c =0.903 seconds (7.48 cycles) Results for angles are shown in Fig. 0, and results for speeds are shown in Fig.. Fig.

23 5 0 tclear=0.903 seconds 5 tclear=0.90 seconds Speed (rad/sec) tclear=0.8 seconds Tie (seconds) Fig. Soe interesting observations can be ade for the two plots in Figs. and. In the plots of angle: The plot of asterisks has clearing tie seconds which exceeds the critical clearing tie of 0.90 seconds by just a little. But it is enough; exceeding it by any aount at all will cause instability, where the rotor angle increases without bound. 3

24 The plot with clearing tie 0.8 seconds looks alost sinusoidal, with relatively sharp peaks. In contrast, notice how the plot with clearing tie 0.90 seconds (the critical clearing tie) has very rounded peaks. This is typical: as a case is driven ore closely to the arginal stability point, the peaks becoe ore rounded. In the plots of speed: The speed increases linearly during the first ~ seconds of each plot. This is because the accelerating power is constant during this tie period, i.e., a =, since the fault is at the achine terinals (and therefore e =0). In the solid plot (clearing tie 0.8 seconds), the speed passes straight through the zero speed axis with a constant deceleration; in this case, the turn-around point on the power-angle curve (where speed goes to zero) is a point having angle less than δ. But in the dashed plot (clearing tie 0.90 seconds), the speed passes through the zero speed axis with decreasing deceleration; in this case, the turn-around point on the power angle curve (where speed goes to zero) is a 4

25 point having angle equal to δ. This point, where angle equals δ, is the unstable equilibriu point. You can perhaps best understand what is happening here if you think about a pendulu. If it is at rest (at its stable equilibriu point), and you give it a push, it will swing upwards. The harder you push it, the closer it gets to its unstable equilibriu point, and the ore slowly it decelerates as it turns around. If you push it just right, then it will swing right up to the unstable equilibriu point, hover there for a bit, and then turn around and coe back. In the speed plot of asterisks, corresponding to clearing tie of 0.90 seconds, the speed increases, and then decreases to zero, where it hovers for a bit, and then goes back positive, i.e., it does not turn-around at all. This is equivalent to the situation where you have pushed the pendulu just a little harder so that it reaches the unstable equilibriu point, hovers for a bit, and then falls the other way. It is interesting that the speed plot of asterisks (corresponding to clearing tie of 0.90 seconds) 5

26 increases to about 4 rad/sec at about.4 second and then sees to turn around. What is going on here? To get a better look at this, I have plotted this to 5 seconds, as shown in Fig. 3. Fig. 3 In Fig. 3, we observe that the oscillatory behavior continues forever, but that oscillatory behavior occurs about a linearly increasing speed. This oscillatory 6

27 behavior ay be understood in ters of the power angle curve, as shown in Fig. 4. ower e Decelerating energy δ Accelerating energy Fig. 4 We see that Fig. 4 indicates that the achine does in fact cycle between a sall aount of decelerating energy and a uch larger aount of accelerating energy, and this causes the oscillatory behavior. The fact that, each cycle, the accelerating energy is uch larger than the decelerating energy is the reason why the speed is increasing with tie. 7

28 You can think about this in ters of the pendulu: if you give it a push so that it goes over the top, if there are no losses, then it will continue to go round and round. In this case, however, the average velocity would not increase but would be constant. This is because our analogy of a one-push differs fro the generator case, where the generator is being pushed continuously by the echanical power into the achine. You should realize that Fig. 4 fairly reflects what is happening in our plot of Fig. 3, i.e., it appropriately represents our odel. However, it differs fro what would actually happen in a synchronous achine. In reality, once the angle reaches 80 degrees, the rotor agnetic field would be reconfigured with respect to the stator agnetic field. This is called slipping a pole, and without out-of-step relaying (OOR), the unit will experience ultiple pole slips in rapid succession thereafter. Most generators have out-of-step protection that is able to deterine when this happens and would then trip the achine. We will study OOR at the end of the course. 8

29 4.0 A few additional coents 4. Critical clearing tie Critical clearing tie, or critical clearing angle, was very iportant any years ago when protective relaying was very slow, and there was great otivation for increasing relaying speed. art of that otivation cae fro the desire to lower the critical clearing tie. Today, however, we use protection with the fastest clearing ties and so there is typically no option to increase relaying ties significantly. erhaps of ost iportance, however, is to recognize that critical clearing tie has never been a good operational perforance indicator because clearing tie is not adjustable once a protective syste is in place. 4. Sall systes What we have done applies to a one-achine-infinite bus syste. It also applies to a -generator syste (see proble.4 in the book which is your assigned #8 on HW3). It does not apply to ultiachine systes, except in a conceptual sense. 9

30 4. Multiachine systes We will see that nuerical integration is the ain way we have of analyzing ultiachine systes. We will take a brief look at this in the next set of notes. 30