In the Previous Notes (Stability 3), We Developed the Equal Area Criterion, Which Says That
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Stability 4 1.0 Introduction In the previous notes (Stability 3), we developed the equal area criterion, which says that For stability, A1=A2, which means the decelerating energy (A2) must equal the accelerating energy (A1) in order for the system response to be stable. Analytically, we have that
clear max P0 P d P P0 d M fault fault M (1) 0 clear A1 A2 Figure 1 below illustrates a stable case. 2.2 2.0 P P e 1.8 pre
1.6 (f) (g) Ppost
1.4 (e) A 1.2 2 1.0 ● ▲ (d) ▲ ● 0.8 A1 0.6 (a) 0.4 (c) Pfault 0.2 (b) 0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax
Fig. 1
1 Figure 2 below illustrates an unstable case. 2.2 2.0 P P e 1.8 pre
1.6 Ppost 1.4
1.2 A2 1.0 ● ▲ ▲ ● 0.8 A1 0.6 0.4 (c) Pfault 0.2 (b) 0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax
Fig. 2
Everything about Figs. 1 and 2 are the same with one exception, the clearing angle (δclear) in Fig. 2 is greater than the clearing angle in Fig. 1. In other words, Fig. 2 assumes that the speed of the protection system is slower than the speed of the protection system in Fig. 1.
In these notes, we want to develop expressions for computing critical clearing angle.
2 2.0 Critical clearing angle
The critical clearing angle will occur when the equal-area criterion is satisfied and the maximum angle is δmax=180-δ0. Such a case is illustrated in Fig. 3. 2.2 2.0 P P e 1.8 pre
1.6 Ppost 1.4 A 1.2 2 1.0 ● ▲ ▲ ● 0.8 A1 0.6 0.4 Pfault 0.2 0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax
Fig. 3
We want to compute the critical clearing time. We will denote it as δclear=δcr.
3 To do this, let’s define the following:
Ppre Ppre max sin (2)
Pfault Pfault max sin r1Ppremax sin (3)
Ppost Ppost max sin r2 Ppremax sin (4) where Ppremax, Pfaultmax, and Ppostmax are the amplitudes of the power-angle curves for the pre-fault, fault- on, and post-fault networks, respectively; 0 Let’s first compute A1. 4 clear A P0 P d 1 M fault 0 clear (P0 P sin )d M fault max (5) 0 0 clear PM Pfault max cos 0 0 PM ( clear 0 ) Pfault max (cos clear cos 0 ) Now let’s compute A2. max A P P 0 d 2 post M clear max P sin P 0 d post max M (6) clear 0 max Ppost max cos PM clear 0 Ppost max (cos clear cos max ) PM ( clear max ) If the system is stable, then A1=A2. So let’s equate the expressions in eq. (5) and (6), below. 5 0 A1 PM (clear 0 ) Pfault max (cosclear cos0 ) 0 (7) Ppost max (cosclear cosmax ) PM (clear max ) A2 Expand: 0 0 PMclear PM0 Pfault max cosclear Pfault max cos0 0 0 (8) Ppost max cosclear Ppost max cosmax PMclear PMmax 0 Notice there is a PM cr on both sides, and so: 0 PM0 Pfault max cosclear Pfault max cos0 0 (9) Ppost max cosclear Ppost max cosmax PM max Let’s put all terms with cosδclear on the left side and everything else on the right: Pfault max cosclear Ppost max cosclear 0 0 (10) Pfault max cos0 Ppost max cosmax PMmax PM0 Factor out the cosδclear term on the left and the 0 PM term on the right: cosclear (Pfault max Ppost max ) 0 (11) Pfault max cos0 Ppost max cosmax PM (0 max ) Divide by the term in parentheses on the left: cos clear 0 Pfault max cos 0 Ppost max cos max PM (0 max ) (12) (Pfault max Ppost max ) 6 Now eq. (12) is true as long as the system response is stable (if it is not stable, then the equal-area criterion is not satisfied and therefore eq. (7) is invalid). But if the system response is marginally stable, then the clearing angle will be the maximum possible angle for which we can clear and still retain stability, i.e., it is the critical clearing angle, and so in this case, δclear=δcr. In addition, the maximum angle must be the unstable equilibrium, which is δmax=180-δ0. Making these substitutions into eq. (12) results in coscr P cos P cos( ) P0 ( ) fault max 0 post max 0 M 0 0 (Pfault max Ppost max ) (13) Recalling that cos(π-x)=-cos(x), and noting on the right-hand-side that we can combine the two δ0 terms inside the brackets, we get: 7 coscr 0 Pfault max cos0 Ppost max cos0 PM (20 ) (14) (Pfault max Ppost max ) Now recall eqs. (3) and (4), which imply that: Pfault max r1Ppremax (15) Ppost max r2 Ppremax (16) Substituting eqs. (15) and (16) into (14), we get: coscr 0 r1Ppre max cos0 r2Ppre max cos0 PM (20 ) (17) (r1Ppre max r2Ppre max ) Factoring out the Ppremax from the bottom and dividing it through all terms in the top, and rearranging, results in 0 PM r1 cos0 r2 cos0 (20 ) Ppre max (18) coscr (r1 r2 ) Let’s consider a few cases: 8 Case 1, Temporary fault at machine terminals: The fact that it is a temporary fault means that the post-disturbance network is the same as the pre-disturbance network, therefore r2=1. The fact that it is a three-phase fault at the machine terminals means that the ability to transmit power to the infinite bus, during the fault-on period, is zero. Therefore r1=0. Applying these values to eq. (18) results in: P0 cos M (2 ) 0 P 0 cos pre max cr 1 (19) 0 PM ( 20 ) cos0 Ppre max But recall that 0 PM Ppre max sin0 (20) Substitution of (20) into (19) results in Ppre max sin0 coscr ( 20 ) (21) Ppre max 9 Or, coscr sin0 ( 20 ) cos0 (22) The above is a closed form solution for the critical clearing angle for the condition of a temporary three phase fault at the machine terminals. Recall the example introduced in the notes called “Stability 2” for the below system: j0.1 j0.4 X’d=j0.2 V= 1.0<0° j0.4 Bus 1 Bus 3 Bus 2 |Vt|= |V1|=1.0 Fig. 4 In those notes, we determined that the angle between the generator internal voltage and the infinite bus is δa=28.44°. This is δ0, and it is for the same system that is characterized in these notes by Fig. 3. Using δ0=28.44°=0.4964 rad in eq. (22) results in 10 coscr sin0 ( 20 ) cos0 sin(0.4964)( 2(0.4964)) cos(0.4964) 0.4763( 0.998) 0.8793 1.021 0.8793 0.1417 Therefore we have that 1 coscr 0.1417 cr cos 0.1417 1.4286rad In degrees, this is 81.85°. Reference to Fig. 5, which is a “hand-approximation” for this case, suggests this angle is quite reasonable. 2.2 2.0 P P e 1.8 pre 1.6 Ppost 1.4 A2 1.2 1.0 ● ● 0.8 0.6 A1 0.4 0.2 0 Pfault 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax Fig. 5 11 3.0 Critical clearing time Let’s consider our case of a temporary three- phase fault at the machine terminals. To obtain information on clearing time, we need to look at the differential equation characterizing this system, which is: 2H ˙˙ 0 (t) Pa, pu PM Pe (23) e0 The right hand-side is of course 0 before the fault (no acceleration), but just after the fault, in this case, Pe goes instantly to 0. We therefore have that 2H ˙˙ 0 (t) PM (24) e0 And so we see that just after the fault, there is non-zero acceleration, but that acceleration is constant since the right-hand-side is constant! Equation (22) may be rewritten as 12 ˙˙(t) e0 P0 (25) 2H M Rewrite the left-hand-side of eq. (25) as d 2 d ˙˙(t) e0 P0 (26) dt 2 dt 2H M Multiply both sides by dt: d e0 P0 dt (27) 2H M Now integrate on the left from ω(0)=0 (initial state is zero velocity) to ω(t) and on the right from t=0+ to t: (t) t d(t) e0 P0 dt M (28) (0)0 2H 0 (t) e0 P0 t (29) 2H M Now express the left-hand-side as the derivative of δ(t) 13 d e0 P0 t (30) dt 2H M Multiply both sides by dt: d e0 P0 tdt (31) 2H M Now integrate the left-hand-side from δ0 to δ(t), and the right-hand-side from t=0+ to t: (t) t d e0 P0 tdt M (32) 2H 0 0 (t) e0 P0 t 2 (33) 0 4H M Now recall we have the critical clearing angle δ(t)=δcr, and we are attempting to find the time for which we reach this angle. So solve eq. (33) for time to obtain: 4H t (t) 0 0 (34) PMe0 14 When the angle is the critical clearing angle, we obtain: 4H tcr cr (t) 0 0 (35) PMe0 So, let’s compute the critical clearing time for our machine. The only other thing we need to know is the inertia constant H. We can assume that it is H=3.0 sec on the machine base. With 0 PM 1.0 , ωe0=377, δ0=28.44°=0.4964 rad, and cr 1.4286rad , we have 4*3 t 1.4286 0.4964 cr 1*377 0.1723sec What if the inertia constant was 5? In this case, we would obtain: 4*5 t 1.4286 0.4964 cr 1*377 0.2224sec The larger the machine, the longer it takes to accelerate it. 15