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A Power System Example Starrett Mini-Lecture #3. Power System Equations Start with Newton again.... T = I  We want to describe the motion of the rotating.

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Presentation on theme: "A Power System Example Starrett Mini-Lecture #3. Power System Equations Start with Newton again.... T = I  We want to describe the motion of the rotating."— Presentation transcript:

1 A Power System Example Starrett Mini-Lecture #3

2 Power System Equations Start with Newton again.... T = I  We want to describe the motion of the rotating masses of the generators in the system

3 The swing equation 2H d 2  = P acc  o dt 2 P = T   = d 2  /dt 2, acceleration is the second derivative of angular displacement w.r.t. time  = d  /dt, speed is the first derivative

4 l Accelerating Power, P acc l P acc = P mech - P elec l Steady State => No acceleration l P acc = 0 => P mech = P elec

5 Classical Generator Model l Generator connected to Infinite bus through 2 lossless transmission lines l E’ and x d ’ are constants  is governed by the swing equation

6 Simplifying the system... l Combine x d ’ & X L1 & X L2 l jX T = jx d ’ + jX L1 || jX L2 l The simplified system...

7 Recall the power-angle curve P elec = E’ |V R | sin(  ) X T

8 Use power-angle curve l Determine steady state (SEP)

9 Fault study l Pre-fault => system as given l Fault => Short circuit at infinite bus  P elec = [E’(0)/ jX T ]sin(  ) = 0 l Post-Fault => Open one transmission line yX T2 = x d ’ + X L2 > X T

10 Power angle curves

11 Graphical illustration of the fault study

12 Equal Area Criterion 2H d 2  = P acc  o dt 2 rearrange & multiply both sides by 2d  /dt 2 d  d 2  =  o P acc d  dt dt 2 H dt => d {d  } 2 =  o P acc d  dt {dt } H dt

13 Integrating, {d  } 2 =  o P acc d  {dt} H dt For the system to be stable,  must go through a maximum => d  /dt must go through zero. Thus...  m  o P acc d  = 0 = { d  2 l H { dt }  o

14 The equal area criterion... l For the total area to be zero, the positive part must equal the negative part. (A1 = A2) P acc d  = A1 <= “Positive” Area P acc d  = A2 <= “Negative” Area  cl oo mm

15 For the system to be stable for a given clearing angle , there must be sufficient area under the curve for A2 to “cover” A1.

16 In-class Exercise... Draw a P-  curve l For a clearing angle of 80 degrees yis the system stable? ywhat is the maximum angle? l For a clearing angle of 120 degrees yis the system stable? ywhat is the maximum angle?

17 Clearing at 80 degrees

18 Clearing at 120 degrees

19 What would plots of  vs. t look like for these 2 cases?


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