Equal Area Criterion for Transient Stability Assessment Prof
Equal Area Criterion for Transient Stability Assessment Prof. V. B. Pandya Asst. Prof. (EE)
Roadmap § Concept of equal area criterion § Transient stability assessment using equal area criterion for sudden increase in mechanical input to turbine § First swing stability and instability for sudden increase in mechanical input § Fault at generator terminals – cleared quickly
Concept of Equal Area Criterion • Single machine connected to infinite bus (SMIB) system can be used to assess transient stability by means of simple criterion known as Equal Area Criterion without going for numerical solution of swing equation • For a single machine connected to infinite bus system Pm j. Xd’ + Pe E’/δ j. XL Fixed (Infinite Bus) V/0
Concept of Equal Area Criterion • If system is unstable, continues to increase indefinitely with time and machine loses synchronism. • On the other hand, performs oscillations whose amplitude decreases in actual practice because of damping. These two situations are shown in below fig. • It can be visualized that for a stable system, indication of stability will be given by observation of the first swing, where will go to maximum and will start decreasing. This fact can be stated as a stability criterion.
Concept of Equal Area Criterion • Thus criterion for stability mathematically • This leads to concept of equal area criterion as follows:
Concept of Equal Area Criterion • Here δ 0 is initial rotor angle before it begins to swing due to perturbation • The condition of stability can be stated as: Ø “The system is stable if the area under Pa (accelerating power)-curve reduces to zero at some value of ”. Ø In other words, the positive (accelerating) area under Pa- curve must equal the negative (decelerating) area and hence the name “equal area” criterion of stability.
Concept of Equal Area Criterion • What does dδ/dt signify? • Ans: - Synchronous machine relative speed – The equation gives relative speed of machine with respect to the synchronous revolving reference frame – If stability of system needs to be maintained, the relative speed equation must be zero sometimes after the disturbance
Sudden Change in Mechanical Input • Consider an SMIB system with machine operating at A (δ 0, Pe 0) • Increase Pm 0 suddenly to Pm 1. • Pm 1> Pe 0 =>Pa increases i. e dω/dt = (Pm-Pe)/M =>ω increases from ωs and since ω> ωs and dδ/dt = ω- ωs => δ 0 increases to δ 1 and hence Pe = E’ V sin δ /(Xd’+XL)=>Pe increases to point B from A. • At B Pm 1 = Pe now so Pa=0 and rotor starts decelerating such that ω stops increasing but still ω> ωs so δ 1 further increases till δ 2 at C where ω = ωs i. e. dδ/dt = 0. • As the generator continues to slow, dδ/dt Pe becomes negative due to ω < ωs. Since ω< ωs, δ (δ 2) decreases towards B being first swing stable! Here Area A 1=A 2 i. e. C A 2 A 1 B A δ 0 δ 1 δ 2 Π State point will overshoot point B (where Pa=0) also due to rotor inertia towards A & δ oscillates around B
First swing stability- Qualitative Behavior Pe Stable Equilibrium C B A Pm 0=Pe 0 δ δ 0 δ 1 δ 2 First swing Stable at δ 2 0 δ 1 δ 0 0 δ Time
First swing stability- Qualitative Behavior Unstable Equilibrium Pe C A 2 A 1 ω-ωsyn B δ A δ 0 δ 1 δ 2 = δmax Π Time • Any further increase in Pm means the area available for A 2 (decelerating) is less than A 1 (accelerating) so that the excess kinetic energy causes rotor angel δ to increase beyond point C and decelerating power becomes accelerating power resulting into instability of system.
Fault at generator terminals – cleared quickly Pm P Pe E D Pmax C A 1 A δ 0 B Pe - pre and post fault Pm=constant A 2 δ 1 ∞ δ 2 Pe during fault, from A to B δ A, AB : Fault occurs Pm>Pe ω ↑ δ↑ (δ 0 → δ 1) B, CD : Fault Cleared Pm < Pe, ω > ωsyn, dδ/dt ↓ but δ↑(δ 1 → δ 2) D Pm<Pe ω =0 ↓ δ mom. constant DE : Pm<Pe ω < ωsyn ↓ δ ↓ Swings back!
Fault at generator terminals – Critical clearing angle(δcr) and critical clearing time(tcr) P Pmax D Pe - pre and post fault C A 2 δ 0 A E Pm=constant Pe during fault, from A to B A 1 B δcr δmax δ If clearing is delayed such that δ 2 becomes δmax as shown where A 1=A 2. Any further delay in clearing fault results into A 1>A 2 making system unstable.
Derivation of δcr and tcr when Pe=0 during fault
Derivation of δcr and tcr when Pe=0 during fault
Sudden short circuit of one of parallel lines Case -1 : Fault at generator end of line
A 2 A 1
Case -2 : Fault away from generator end on line (middle)
δmax
Derivation of δcr
Case -3 : Reclosure
Case -3 : Reclosure If CB of line 2 are successfully reclosed, power transfer once again becomes Since reclosure restores power, changes of stable operation improves.
Case -3 : Reclosure For critical clearing angle,
Example • A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a double circuit transmission line in which resistance is neglected. A fault takes place reducing the maximum power transfer to 0. 5 pu whereas before fault, this power was 2. 0 pu. and after fault clearance, it is 1. 5 pu. By equal area criterion determine the critical clearing angle.
Example
Example
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