Part A: From (14.19) in your text (see notes Stability1 ), we have

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1 Prob Consider a round-rotor generator delivering a steady-state power P G =.5 to an infinite bus through a transmission line with reactance X L =.4. Assume that E a =1.8, V =1/_, H=5sec, and X d =X q =1.. Neglect resistance. A.Find the two possible steady-state (equilibrium values of power angle δ=/_e a -/_V that lie in the interval [,π]. B. Which of these two equilibrium values are we likely to observe in practice? Hint: Linearize around the equilibrium value and consider the natural frequencies found from the linear differential equation in Δδ. Part A: From (14.19 in your text (see notes Stability1, we have M ( t D ( t P G ( PM The problem implies that P M =.5, therefore we have M D PG (.5 Here: Ea V 1.8*1. PG ( sin( sin( 1.86 sin( X X 1..4 d L 1.86 sin( sin

2 Part B: What does it mean to linearize around the equilibrium value? It means to find a linear approximation to the function at that value. If the function is a nonlinear function, then the behavior will only be valid for small distances away from the value. The figure below illustrates: If our function is in terms of δ, then our linearized function is in terms of Δδ. Wherever there is a linear terms in δ of our original function, we may replace it directly with Δδ. But if there is a nonlinear term in our original function, then we must use a Taylor series expansion. Our function is M ( t D ( t P G ( PM M D 1.86 sin. 5 (1

3 Now let the angle δ change by a small amount about an equilibrium δ. Then d d dt dt, d d dt dt 3 ( Recall the Taylor series expansion of a function is: f ( x ( x ( f ( x ( f' ( x ( x ( 1 f'' ( x ( ( x and recognizing that for small Δx, the higher-order terms are extremely small and may be neglected, we obtain, ( ( ( ( d sin sin sin d ( f ( x x f ( x f'( x x And so the sin function can be expanded to (3 sin sin (cos Substituting (3 and ( into (1, we obtain: M D 1.86(sin (cos Distributing through the 1.86, we get ( M D 1.86 sin 1.86 cos....5 And bringing the 1.86 sinδ to the right-hand-side, we obtain: M D 1.86 cos sin (4 Now, consider that δ is either.89 or Either way, 1.86 sin Therefore,.5 M D 1.86 cos.5

4 This is the result of Linearization of swing equation given on pg The LaPlace Transform of this equation is: M ( s s D ( s s 1.86 cos ( s Factoring out the Δδ(s, we have ( s Ms Ds 1.86 cos (5 Now what if the right-hand-side of (4 is not zero but rather, what if we perturb this system ever so slightly so that 1.86 sin.5 In this case, u( t Ms Ds 1.86 cos U( s ( s and we have ( s Ms Ds U( s cos We observe that the denominator of the previous equation is the characteristic polynomial of this system. When the system is in equilibrium, we may divide both sides of (5 by Δδ(s to obtain the characteristic equation of the system:

5 o Ms Ds 1.86 cos( H 5 M.65 f o 6 THEN.65s Ds 1.86 cos( o If we plug in δ o =.89 degrees into the above equation, we obtain:.65s.65s This has roots of D D D Ds 1.86 cos(.89 Ds.5 4(.65 (.5 (.65 D D.53 D Consider D is very small, e.g., D=.1. Then the above is j3.915 In this case, both roots are in the Left-Hand-Plane. It should be clear that any smaller value of D will not change this situation at all (since the first term will still be real and negative and the second term will still be imaginary. Now consider that D is large, e.g., D=1. Then the above is 5

6 , In this case, both roots are in the Left-Hand-Plane. It should be clear that any larger value of D will not change this situation at all (since the first term will still be large and negative, and the second term will never be greater than the first term in absolute value. A system having all of the roots of the characteristic equation in the Left-Half-Plane is a stable system. This corresponds to our understanding of the stable equilibrium point that we discussed in class. If on the other hand we make δ o = degrees one coefficient is negative, meaning that the natural frequencies are in the right hand plane. If we plug in δ o =.89 degrees into the above equation, we obtain:.65s.65s This has roots of D D D Ds 1.86 cos( Ds (.65 ( (.65 D D.53 D It should be clear that the second term will be real and it will be larger in magnitude than the first term, 6

7 independent of what D is. For example, if D=1, then we have , Thus, there will be one root of the characteristic equation in the Right-Half-Plane. A system having one or more roots in the Right-Half-Plane in an unstable system. This corresponds to our understanding of the unstable equilibrium point that we discussed in class. To answer the question, we will never observe the unstable equilibrium in practice because small perturbations are always happening in the power system and would perturb the system off of this point, and once perturbed, it would either move to the stable equilibrium or it would go unstable. The stable equilibrium is the only one we will ever observe in practice. 7