SCHOOL OF ELECTRICAL, MECHANICAL AND MECHATRONIC SYSTEMS. Transient Stability LECTURE NOTES SPRING SEMESTER, 2008

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1 SCHOOL OF ELECTRICAL, MECHANICAL AND MECHATRONIC SYSTEMS LECTURE NOTES Transient Stability SPRING SEMESTER, 008 October 7, 008

2 Transient Stability Transient stability refers to the ability of a synchronous generator connected to the grid to maintain synchronism when subjected to severe disturbances due to heavy load changes, loss of generation in other units or faults in the system. When there is a sudden disturbance, the machine rotor starts to oscillate about its equilibrium position. These oscillations are modeled using swing equation. 1 Swing equation When there is an unbalance between mechanical torque (T m ) and electrical torque (T e ), it will cause the rotor to accelerate or decelerate. The difference in the torque (T a ) is given by, J dω m T a = T m T e = T m T e (1) where J is the moment of inertia of the rotor in kg.m and ω m is the angular velocity of the rotor in rad/s. Equation (1) can be normalised as follows: Defining inertia constant H as where T base = V A base ω 0m H = 1 Jω 0m V A base J = HV A base ω 0m = H ω 0m T base (). Combining (1) and () we get, H d ( ) ωm ω 0m H d ω r = T m T e T base = T m T e (3) = ωr ω 0 where the variables superscripted with indicates per unit quantities and ωm ω 0m ω r is the velocity of the rotor in electrical rad/s and ω 0 is its rated value. If δ is the angular position of the rotor with respect to the rotating reference frame, = ω r and δ = ω r t ω 0 t + δ 0 (4) where δ 0 = δ at t = 0. Differentiating (4) twice with respect to time, dδ = ω r ω 0 = ω r (5) = d ( ω r) = d (ω 0 ω r ω 0 ) (6) = ω 0 d ω r 1 (7)

3 Substituting in (3), H ω 0 = T m T e (8) To include the rotor damping effects, we include rotor damping constant K D and proportional to change in speed and we have, H ω 0 = T m T e K D ω r (9) Equation (9) is known as swing equation. In pu representation, torque and power are equal, so we can represent (9) as, H ω 0 = P m P e K D ω r (10) Single machine on infinite bus (SMIB) model Infinite bus is the one with constant frequency and voltage. Consider a generator G connected to infinite bus through a transformer and a transmission line as shown in Figure 1(a). The reactance of the line X line, transformer X trans and the transient reactance X d are indicated in Figure 1(b). E is the generator voltage behind X d and E bus is the infinite bus voltage. Generator voltage E leads the E bus by an angle δ. The given network can be reduced as shown in Figure 1(c) where X T = X d + X trans + X line. Let E be the reference vector, the current I t is given by, Apparent power behind X d is given by, I t = E 0 E bus δ jx T = E 0 E bus (cos δ j sin δ) jx T (11) S = P + jq = E I t = E E bus sin δ X T + j E (E E bus cos δ) X T (1) If we neglect the stator resistance loss, the airgap power will be the terminal power P e. P e = E E bus sin δ X T = P max sin δ (13) where P max = E E bus X T. Equation (13) is diagrammatically represented in Figure. Under steady

4 Figure 1: Single machine on infinite bus system state P m = P e, and let the corresponding rotor angle be δ. From (13), P max is inversely proportional to X T. With X T increased to X T 1, P max reduces to P max1 as shown in Figure and also the steady state angle δ increases to δ 1. Combining (8) and (13) we get, H ω 0 = P m P max sin δ (14).1 Rotor angle response to sudden change in P m When the mechanical power P m is increased to P m1 as shown in Figure 5, since P m1 > P e, rotor accelerates increasing rotor angle towards δ 1. At y, P m1 = P e, but the rotor speed is higher than the synchronous speed ω 0 and so the rotor angle continues to increase. This will increase P e, now since P m1 < P e, rotor will start decelerate. At z with δ, rotor speed reaches syn.speed but still P m1 < P e rotor continues to decelerate with speed falling below ω 0. The rotor retraces the path, after reaching x the rotor continues to oscillate about the new equilibrium point δ 1 3

5 Power Angle Figure : Power angle curve. Equivalent single machine system We can represent a system consisting of many generators operating in parallel by an equivalent single machine system. Find the equivalent single machine system of the following multimachine systems. 1. Each machine 30 kva, H = 6 s, xd = 0.5 pu. (Refer Figure 3).. G 1 = 30 kva, H 1 = 6 s, xd 1 = 0.5 pu and G = 60 kva, H = 4 s, xd = 0.3 pu (Refer Figure 4). 3 Stability based on equal area criterion Using SMIB model, the stability of rotor angle oscillations can be determined graphically without solving the swing equations. Even though this method cannot be applied to multimachine systems, it can provide basic understanding and information about critical clearing angle and stability limits. From equation (14) we have = ω 0 H (P m P e ) (15) 4

6 Figure 3: 3 M/C system Figure 4: M/C system Multiplying by dδ, Integrating (16) we get, dδ = ω 0 H (P m P e ) dδ ( ) d dδ = ω 0 H (P m P e ) dδ (16) ( ) dδ ω0 = H (P m P e ) dδ (17) Under steady state conditions dδ = 0, during disturbance δ will be changed, but after the removal of disturbance the system should regain steady state conditions without δ deviating unboundedly. For stability equation (17) should be equal to zero i.e, δ ω 0 δ 0 H (P m P e ) dδ = 0 (18) 5

7 Power Angle Figure 5: Rotor angle oscillations where δ 0 is the steady state rotor angle, δ is the maximum rotor angle (see Figure 5). For the equation (18) to be zero, the area described by the function P m P e must be zero. From Figure 5, the kinetic energy gained is given by area A 1 and is given by, A 1 = δ1 δ 0 (P m P e ) dδ (19) The energy lost during deceleration is given by the area A and is given by, for stability, A = δ δ 1 (P m P e ) dδ (0) Area A 1 = Area A (1) i.e, kinetic energy gained = energy lost during deceleration Equation (1) represents the principle of stability through equal area criterion. 3.1 Stability during sudden input power change For sudden change in input power stability can be found by using the stability criteria given by (1). Stability can be maintained only if we can find area A in the power angle curve such that (1) is satisfied. The maximum rotor angle δ max is given by the rotor angle corresponding to P m for which π < δ > π as shown in Figure 6. 6

8 Power Angle Figure 6: Power angle curve for sudden change in P m Using (13) and (1) we can write, P m (δ 1 δ 0 ) On simplification, δ1 δ 0 P max sin δ dδ = δmax δ 1 P max sin δ dδ P m (δ max δ 1 ) () ] P m (δ max δ 0 ) = P max [( cos δ) δ 1 δ 0 + ( cos δ) δmax δ 1 At stability limit P m = P max sin δ max, substituting in (3), = P max (cos δ 0 cos δ max ) (3) sin δ max (δ max δ 0 ) + cos δ max = cos δ 0 (4) Equation (4) is a nonlinear equation which can be solved by iterative technique to get δ max. With δ max known we can find δ 1 and P m as follows: 3. Stability during 3 phase fault δ 1 = π δ max and P m = P max sin δ 1 (5) Fault at sending end. When fault occurs near the sending end as in Figure 7, no power is send to the infinite bus. If we neglect the resistance of transformer and generator P e is zero during fault. Since P e is zero, P m is used to accelerate rotor and this increases rotor angle to δ 1. At 7

9 Figure 7: Fault at the sending end δ 1, the fault is cleared and P e takes the value corresponding to y as shown in Figure 8(a). At y since P e < P m, rotor decelerates to z and then returns back to x and starts to oscillate about δ 0. Critical clearing angle δ c is reached when area A becomes less than A 1 [see Figure 8(b)] and at z the rotor angle becomes δ max. Using (1) solving for δ c, P m (δ c δ 0 ) = δmax δ c (P max sin δ P m ) dδ = P max (cos δ c cos δ max ) P m (δ max δ c ) P m (δ max δ 0 ) = P max cos δ c P max cos δ max (6) [ δ c = cos 1 cos δ max + P ] m (δ max δ 0 ) P max To find the critical time t c, substitute P e = 0 in (15), Integrating both sides from time 0 to t we get, (7) = ω 0 H P m (8) dδ = ω 0 H P mt δ = ω 0 4H P mt + δ 0 (9) at δ = δ max, t = t c, substituting in (9), t c = 4H (δ max δ 0 ) ω 0 P m (30) 8

10 Power Power m m Angle 0 c max Angle Figure 8: Power angle curve for fault at sending end 4 Numerical Example Let us consider a 50 Hz single machine on infinite bus system. The machine is connected to infinite bus through a transformer and a transmission line. The system data is included in the Figure 9. A temporary 3 phase fault occurs at sending end and the system is restored back to normal after fault clearance. Find the critical clearing angle and critical clearing time above which the system will become unstable. 9

11 Figure 9: Power system for numerical example. 0.9 j0.08 Generator Current I G = = 0.9 j0.08 pu 1 0 Total reactance X T 1 = = pu Generator voltage behind x d, E q = V bus I G X T 1 = j (0.9 j0.08) = pu ( ) Initial power angle δ 0 = sin 1 Pe X T 1 = E q V bus Maximum power angle δ max = 180 δ 0 = [ Critical clearing angle δ c = cos 1 cos δ max + P ] m (δ max δ 0 ) = P max 4H (δ max δ 0 ) Critical clearing time t c = = 0.71 s ω 0 P m 10

12 Figure 10: Power angle curve for the numerical example. 11