max clear Figure 1 below illustrates a stable case. P post (f) (g) (e) A 2 (d) (c)

Transcription

1 Stability 4. Introduction In the previous notes (Stability 3, we developed the equal area iterion, which says that For stability, A =A, which means the decelerating energy (A must equal the accelerating energy (A in order for the system response to be stable. Analytically, we have that A fault d fault A d Figure below illustrates a stable case. ( e (a (b pre post A (c (e (d (f (g A fault δ δ δ Fig. δ a

2 Figure below illustrates an unstable case. e pre post A (c (b A fault δ δ δ Fig. δ a Everything about Figs. and are the same with one exception, the ing angle (δ in Fig. is greater than the ing angle in Fig.. In other words, Fig. assumes that the speed of the protection system is slower than the speed of the protection system in Fig.. In these notes, we want to develop expressions for computing itical ing angle.

3 . Critical ing angle The itical ing angle will occur when the equal-area iterion is satisfied and the imum angle is δ =8-δ. Such a case is illustrated in Fig. 3. e pre post A A fault δ δ δ Fig. 3 δ a We want to compute the itical ing time. We will denote it as δ =δ. 3

4 To do this, let s define the following: pre pre sin ( sin r sin fault post where fault pre (3 post sin r pre sin (4 pre, fault, and post are the amplitudes of the power-angle curves for the pre-fault, fault-on, and post-fault networks, respectively; <r < where r = corresponds to a threephase fault at the machine terminals, and r = corresponds to no-fault at all. <r < where r = corresponds to a threephase fault at the machine terminals that is not ed, and r = corresponds to a temporary fault (fault is removed without protective relay action to also remove a circuit and weaken the transmission Let s first compute A. 4

5 A fault d ( fault sin d ( fault cos fault (cos cos (5 Now let s compute A. A post post post post sin cos (cos d d cos ( (6 If the system is stable, then A =A. So let s equate the expressions in eq. (5 and (6, below. 5

6 A ( post Expand: post (cos cos Notice there is a post fault cos fault cos fault post cos cos (cos cos ( cos fault cos A on both sides, and so: post fault cos cos (9 (7 (8 Let s put all terms with cosδ on the left side and everything else on the right: fault fault cos cos post post cos cos ( Factor out the cosδ term on the left and the term on the right: cos fault ( fault cos post post cos ( Divide by the term in parentheses on the left: cos fault cos ( post fault cos post ( ( ( 6

7 Now eq. ( is true as long as the system response is stable (if it is not stable, then the equal-area iterion is not satisfied and therefore eq. (7 is invalid. But if the system response is marginally stable, then the ing angle will be the imum possible angle for which we can and still retain stability, i.e., it is the itical ing angle, and so in this case, δ =δ. In addition, the imum angle must be the unstable equilibrium, which is δ =8-δ. aking these substitutions into eq. ( results in cos fault cos post ( fault cos( post ( (3 Recalling that cos(π-x=-cos(x, and noting on the right-hand-side that we can combine the two δ terms inside the brackets, we get: 7

8 cos fault cos ( post fault cos post ( (4 Now recall eqs. (3 and (4, which imply that: fault r pre (5 post r pre (6 Substituting eqs. (5 and (6 into (4, we get: cos r pre cos r ( r pre pre r cos pre ( (7 Factoring out the pre from the bottom and dividing it through all terms in the top, and rearranging, results in cos r cos r cos ( r r Let s consider a few cases: pre ( (8 8

9 Case, 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 r =. 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 r =. Applying these values to eq. (8 results in: cos pre cos pre ( cos ( (9 But recall that pre sin ( Substitution of ( into (9 results in pre sin cos ( pre ( 9

10 Or, cos sin ( cos ( The above is a closed form solution for the itical ing angle for the condition of a temporary three phase fault at the machine terminals. Recall the example introduced in the notes called Stability for the below system: X d =j. j. j.4 V =.< j.4 Bus Bus 3 Bus V t = V =. Fig. 4 In those notes, we determined that the angle between the generator internal voltage and the infinite bus is δ a =8.44. This is δ, and it is for the same system that is characterized in these notes by Fig. 3. Using δ =8.44 =.4964 rad in eq. ( results in

11 cos sin ( cos sin(.4964( (.4964 cos( ( Therefore we have that cos.47 cos rad In degrees, this is Reference to Fig. 5, which is a hand-approximation for this case, suggests this angle is quite reasonable. e pre post A A fault δ δ δ Fig. 5 δ a

12 3. Critical ing time Let s consider our case of a temporary threephase fault at the machine terminals. To obtain information on ing time, we need to look at the differential equation characterizing this system, which is: H e ( t a, pu e (3 The right hand-side is of course before the fault (no acceleration, but just after the fault, in this case, e goes instantly to. We therefore have that H e ( t (4 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 ( may be rewritten as

13 H e ( t Rewrite the left-hand-side of eq. (5 as d dt d dt H e ( t ultiply both sides by dt: d H e dt (5 (6 (7 Now integrate on the left from ω(= (initial state is zero velocity to ω(t and on the right from t= + to t: ( t e d( t H ( (8 ( t H e t t dt (9 Now express the left-hand-side as the derivative of δ(t 3

14 d e t dt H (3 ultiply both sides by dt: d H e tdt (3 Now integrate the left-hand-side from δ to δ(t, and the right-hand-side from t= + to t: ( t d t e H tdt (3 ( t e t 4H (33 Now recall we have the itical ing angle δ(t=δ, and we are attempting to find the time for which we reach this angle. So solve eq. (33 for time to obtain: t ( t 4H e (34 4

15 When the angle is the itical ing angle, we obtain: t ( t 4H e (35 So, let s compute the itical ing 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. sec on the machine base. With, ω e =377, δ =8.44 =.4964 rad, and, we have.. 486rad 4*3 t *377.73sec What if the inertia constant was 5? In this case, we would obtain: 4*5 t *377.4sec The larger the machine, the longer it takes to accelerate it. 5