<p> 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</p><p>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 </p><p>1.6 (f) (g) Ppost </p><p>1.4 (e) A 1.2 2 1.0 ● ▲ (d) ▲ ● 0.8 A1 0.6 (a) 0.4 (c) Pfault 0.2 (b) 0 </p><p>0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax </p><p>Fig. 1</p><p>1 Figure 2 below illustrates an unstable case. 2.2 2.0 P P e 1.8 pre </p><p>1.6 Ppost 1.4 </p><p>1.2 A2 1.0 ● ▲ ▲ ● 0.8 A1 0.6 0.4 (c) Pfault 0.2 (b) 0 </p><p>0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax </p><p>Fig. 2</p><p>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.</p><p>In these notes, we want to develop expressions for computing critical clearing angle.</p><p>2 2.0 Critical clearing angle</p><p>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 </p><p>1.6 Ppost 1.4 A 1.2 2 1.0 ● ▲ ▲ ● 0.8 A1 0.6 0.4 Pfault 0.2 0 </p><p>0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax </p><p>Fig. 3</p><p>We want to compute the critical clearing time. We will denote it as δclear=δcr. </p><p>3 To do this, let’s define the following:</p><p>Ppre  Ppre max sin  (2)</p><p>Pfault  Pfault max sin  r1Ppremax sin (3)</p><p>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<r1<1 where r1=0 corresponds to a three-phase fault at the machine terminals, and r1=1 corresponds to no-fault at all.  0<r2<1 where r2=0 corresponds to a three-phase fault at the machine terminals that is not cleared, and r2=1 corresponds to a temporary fault (fault is removed without protective relay action to also remove a circuit and weaken the transmission)</p><p>Let’s first compute A1.</p><p>4  clear A  P0  P d 1  M fault  0</p><p> clear  (P0  P sin )d  M fault max (5)  0</p><p>0  clear  PM   Pfault max cos  0 0  PM ( clear  0 )  Pfault max (cos clear  cos 0 )</p><p>Now let’s compute A2.</p><p> max A  P  P 0 d 2  post M  clear</p><p> max  P sin  P 0 d  post max M (6)  clear</p><p>0  max   Ppost max cos  PM   clear 0  Ppost max (cos clear  cos max )  PM ( clear  max )</p><p>If the system is stable, then A1=A2. So let’s equate the expressions in eq. (5) and (6), below.</p><p>5 0 A1  PM (clear 0 )  Pfault max (cosclear  cos0 ) 0 (7)  Ppost max (cosclear  cosmax )  PM (clear max )  A2 Expand: 0 0 PMclear  PM0  Pfault max cosclear  Pfault max cos0 0 0 (8)  Ppost max cosclear  Ppost max cosmax  PMclear  PMmax 0 Notice there is a PM  cr on both sides, and so: 0  PM0  Pfault max cosclear  Pfault max cos0 0 (9)  Ppost max cosclear  Ppost max cosmax  PM max</p><p>Let’s put all terms with cosδclear on the left side and everything else on the right:</p><p>Pfault max cosclear  Ppost max cosclear 0 0 (10)  Pfault max cos0  Ppost max cosmax  PMmax  PM0</p><p>Factor out the cosδclear term on the left and the 0</p><p>PM term on the right: cosclear (Pfault max  Ppost max ) 0 (11)  Pfault max cos0  Ppost max cosmax  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 )</p><p>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). </p><p>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</p><p> coscr 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:</p><p>7 coscr 0 Pfault max cos0  Ppost max cos0  PM (20   ) (14)  (Pfault max  Ppost max ) Now recall eqs. (3) and (4), which imply that:</p><p>Pfault max  r1Ppremax (15)</p><p>Ppost max  r2 Ppremax (16) Substituting eqs. (15) and (16) into (14), we get:</p><p> coscr 0 r1Ppre max cos0  r2Ppre max cos0  PM (20   ) (17)  (r1Ppre max  r2Ppre max )</p><p>Factoring out the Ppremax from the bottom and dividing it through all terms in the top, and rearranging, results in 0 PM r1 cos0  r2 cos0  (20   ) Ppre max (18) coscr  (r1  r2 )</p><p>Let’s consider a few cases:</p><p>8 Case 1, Temporary fault at machine terminals:</p><p>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.</p><p>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.</p><p>Applying these values to eq. (18) results in: P0 cos  M (2   ) 0 P 0 cos  pre max cr 1 (19) 0 PM  (  20 )  cos0 Ppre max But recall that 0 PM  Ppre max sin0 (20) Substitution of (20) into (19) results in</p><p>Ppre max sin0 coscr  (  20 ) (21) Ppre max</p><p>9 Or,</p><p> coscr  sin0 (  20 )  cos0 (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.</p><p>Recall the example introduced in the notes called “Stability 2” for the below system:</p><p> j0.1 j0.4 X’d=j0.2 V= 1.0<0° </p><p> j0.4 </p><p>Bus 1 Bus 3 Bus 2 </p><p>|Vt|= |V1|=1.0 </p><p>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</p><p>10 coscr  sin0 (  20 )  cos0  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 coscr  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 </p><p>1.6 Ppost 1.4 A2 1.2 1.0 ● ● 0.8 0.6 A1 0.4 0.2 0 Pfault </p><p>0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 δa δ0 δclear δmax </p><p>Fig. 5</p><p>11 3.0 Critical clearing time</p><p>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!</p><p>Equation (22) may be rewritten as</p><p>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)</p><p>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) PMe0</p><p>14 When the angle is the critical clearing angle, we obtain: 4H tcr  cr (t) 0  0 (35) PMe0 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</p><p>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.</p><p>15</p>