1© 2021 Kai Sun

Spring 2021Instructor: Kai Sun

ECE 522 ‐ Power Systems Analysis IISpring 2021

Transient Stability

2© 2021 Kai Sun

Outline

• Transient stability analysis using direct methods– Transient stability of an SMIB system– Direct methods for multi-machine systems

• Time-domain transient stability simulation– Explicit and implicit numerical integration techniques– Simulation of a multi-machine system

• References:– Kundur’s Chapter 13– Saadat’s Chapters 11.5-11.10

3© 2021 Kai Sun

Transient Stability• The ability of the power system to maintain synchronism when subjected to a severe disturbance such as a

fault on transmission facilities, loss of generation or loss of a large load.– The system response to such disturbances involves large excursions of generator rotor angles, power flows, bus

voltages, and other system variables.– Stability is influenced by the nonlinear characteristics of the system– If the resulting angular separation between the machines in the system remains within certain bounds, the system

maintains synchronism.– If loss of synchronism due to transient instability occurs, it will usually be evident within 2-3 seconds after the

initial disturbances

4© 2021 Kai Sun

Two Approaches for Transient Stability Analysis

To analyze the stability of a nonlinear system modeled by DAEs (differential-algebraicequations) following a disturbance:

1. Direct methods (based on Lyapunov’s second method for stability): Determine stability without explicitly solving the DAEs:

1. Define a Transient Energy Function, which is an exact or approximate Lyapunov function. 2. Compare the value of the function to a critical energy to judge whether the system state can stay

inside a stability region of the equilibrium.2. Time-domain simulation (indirect method): Solve an Initial Value Problem of the nonlinear

Differential-Algebraic Equations (DAEs) with a given initial state x=x0 at t=t0 by using step-by-step numerical integration (explicit or implicit).

, E

DE, Attx = f(x,y )

0 = g(x,y )

5© 2021 Kai Sun

Transient Stability Analysis Using Direct Methods

6© 2021 Kai Sun

Stability on a General Dynamical System

Assume origin x=0 is an equilibrium, i.e.

In other words, the system variable will stay in any given small region (<) around the equilibrium point once becoming close enough (<) to that point.

The equilibrium point x=0 is stable in the sense of Lyapunov

such that

x

• Lyapunov Stability: Consider a nonlinear dynamical system

• In mathematics, stability theory addresses the stability of solutions of a set of differential equations, or in other words, stability of trajectories of a dynamical system under small perturbations of an initial condition.

7© 2021 Kai Sun

Lyapunov Stability Criterion (Lyapunov’s second method)

Its equilibrium xs, which satisfies f(xs)=0, is stable if there exists a positive definite function V(x), called a Lyapunov function, such that its time-derivative is not positive, i.e.

• V(x)0 “=0” if and only if x= xs (xS has the minimum V=0)

• (V does not increase with time)

Equilibrium xs is asymptotically stable if the time-derivative of V is negative definite, i.e.

• “=0” if and only if x= xs

, E

DE, Attx = f(x,y )

0 = g(x,y )

( )tx = f(x )

( ) 0dVdt

x

( ) 0dVdt

x

8© 2021 Kai Sun

Transient Energy Function (TEF) Method

• It is a special case of Lyapunov’s second method that relaxes V(x) to an approximate Lyapunov function, i.e. called a Transient Energy function (TEF)

• Consider “a rolling ball” analogy. two quantities are required to determine whether “ball” (state x) can leave “bowl” (stability region of xs)– Initial kinetic energy of the ball– The height of the bowl rim at the crossing point

(depending on the direction of the initial motion)

Application to a Power System• Initially the system operates at stable equilibrium point (SEP) xs. • If a fault occurs, the system gains kinetic energy VKE and

potential energy VPE during the fault-on period and then moves away from xs.

• After fault clearing, VKE is converted into VPE. • To avoid instability, the system must be capable of absorbing

VKE while moving towards the old SEP xs or a new SEP. • For a given post-disturbance network configuration, there is a

critical, maximum amount of transient energy Vcr that the system can absorb without loss of synchronism

• Therefore, assessment of transient stability for the system state at fault clearing (xcl) requires:1. Define a TEF V(xcl) that adequately describes the transient

energy responsible for separating one or more generators from the rest of the system

2. Estimate the critical energy Vcr and calculate transient stability margin, i.e. Vcr V(xcl).

VKE+VPE

VPE=Vcr

9© 2021 Kai Sun

Classic Model of a Single‐Machine‐Infinite‐Bus System

0

00 0

2 ( )

r

r Dm e r

dd t

d KH P Pd t

dw w

ww w

w w

ìïï = -ïïïíïï = - - -ïïïî

max sin sinBe

T

E EP PX

d d¢

= = Pe+jQe

Pt+jQtPB+jQB

With all resistances neglected:

0=260377 rad/s

H in s

r in rad/s

in rad

KD in p.u

max0 0

2 sin 0Dm

KH P P d d dw w

+ + - =

Equilibria satisfy Pmax sin Pm=0• s = arcsin(Pm/Pmax)

stable equilibrium point (SEP) • u =180o-s

unstable equilibrium point (UEP)

Pendulum subject to a constant torque

𝑇m

s

2 sin 0mml D mgl T

singT mgl

10© 2021 Kai Sun

Phase‐Space Trajectories

No damping (KD=0)

Small damping

Large damping

max0 0

2 sin 0Dm

KH P P d d dw w

+ + - =

11© 2021 Kai Sun

Pa r

a >0 =0,

b 0 =r, max

c <0 =0 , =max

Response to a step change in Pm

Consider a sudden increase in Pm: Pm0 Pm1.

• New SEP b satisfying Pe(1)=Pm1

• a b: Due to the rotor’s inertia, cannot jump from 0 to 1, so Pa=Pm1-Pe(0)>0 and r increases from 0. When b is reached, Pa=0 but r >0, and continues to increase.

• b c: >1 and Pa<0, so r decreases while increases until c. At c, r=0 and reaches its maxmum max.

• c: Starting from c, the rotor starts to decelerate (since Pa<0), so r<0 and decreases.

• With all resistances (damping) neglected, and r oscillate around new SEP b with a constant amplitude.

Question 1: Does have a sinusoidal waveform in time?

Question 2: When does r reach the minimum speed?

2

max20

2 sin (accelerating power)m e m aH d P P P P P

dtd

dw

= - = - =

Ignore all losses (KD=0):

UEP

12© 2021 Kai Sun

Nonlinear Oscillation of an SMIB System

Let = -s, =KD/(2H)=0Pmax/(2H)

0sinc cos2 2s

max0 0

2 sin 0Dm

KH P P d d dw w

+ + - =(sin sin ) 0s

2H/0=0.1, D=0, Pmax=1, Pm=0.5 =0, =10, s=30o

Small disturbance c s12

o sf

0cos s

Approximately sinusoidalt1

max

Marginally stable

t2

min

1 m ax 2 m in

1( ) ( )

ft t

13© 2021 Kai Sun

Define oscillation amplitude as δmax=δmax − δs or δmin= δs− δmin.

max min1 max 2 min

1( , )( ) ( )

ft t

Frequency‐Amplitude (F‐A) Curve

s

00

max s s s

2( )cos( ) cos ( )sin

x

iH dtP

xx x

,2 ,2 10( ) ( )

i j i jjt x t x

Oscillation amplitude (max -s)

Freq

uenc

y SEPStability margin

(UEP)

max min( , )f

[1] B. Wang, K. Sun, “Formulation and Characterization of Power System Electromechanical Oscillations,” IEEE Trans. Power Systems, Nov. 2016

[2] B. Wang, X. Su, K. Sun, “Properties of the Frequency-Amplitude Curve,” IEEE Trans. Power Systems, Jan. 2017

14© 2021 Kai Sun

Equal‐Area Criterion (EAC)

At max exists, r=d/dt=0:

• Ignore all losses (KD=0):

Moment of inertia in p.u.

0

2 0 ( )r m eP P dH

d

d

ww dD = -ò

( )0

2

00

1 2 ( )2

rm eH P P d

d

d

ww d

w

æ öD ÷ç ÷⋅ = -ç ÷ç ÷çè ø ò

m

0

ax1

1

( )( ) m em eP P d P P dd

d

d

ddd - ---= ò ò

1 2= 0| | | |A A- =

max

0

0 ( )m eP P dd

dd= -ò

Vke=

2 20

22 ( )m ed d d d d P Pdt dt dt dt dt H

wd d d dé ùê ú = ⋅ = -ê úë û

20

2 ( )2 m e

d P Pdt H

wd= -

0

20 ( )m eP Pd d

dt Hd

d

wdd

é ù -ê ú =ê úë û

ò

Question: What if KD>0?

max0 0

2 sin 0Dm

KH P P d d dw w

+ + - =

2 212ke mlV d= ⋅

Pendulum with a constant torque

𝑇m

s

2 sin 0mml D mgl T

Kinetic energy:

singT mgl

• |A1|=Vke gained from a to b

• |A2|

=Vpe at c relative to b

Potential energy (ref: s) max 1 max max 1(cos cos ) ( )mP Pd d d d= - - -

max s(cos cos ) ( )s mpeV P Pd d d d= - - -s(cos cos ) ( )s mpe mgl TV d d d d= - - -

15© 2021 Kai Sun

Equal-Area Criterion (EAC): The stability is maintained only if a decelerating area |A2| accelerating area |A1| can be located above Pm.

UEP SEP • When reaches UEP, if |A2|<|A1|, will continues to increase (since r>0) and accelerate to lose stability.

• Transient stability depends on the size of the step change in Pm.

16© 2021 Kai Sun

• Solve max.

• Thus, transient stability limit for a step change of Pm

• The new SEP:

max

1

1

01 0 m am x xa max 1sin ( )( ) sinm mP P d P d P

d

d

d

ddd d d d dd d- - = - -ò ò

Following a step change Pm0Pm, solve the transient stability limit of Pm:

• Assume | A1 |=| A2 | in order to solve the limit of Pm

max 0 max 0 max( ) (cos cos )mP Pd d d d- = -

max 0 max max 0( )sin cos cosd d d d d- + =

Transient stability limit for a step change of Pm

• Since at the UEP , there ismax maxsinmP P d=

1 m axd p d= -

UEP SEP

max max 0sinm mP P PdD £ -

Question: From Pm0, is there anyway to increase Pm beyond Pmaxsinmax?

0 max 1 0 max max 1 max(cos cos ) (cos cos )m mP P P Pd d d d d d- + - =- - -

17© 2021 Kai Sun

Solve max by the Newton‐Raphson method

• Define nonlinear function:

• Taylor expansion at an estimate:

• Select an initial estimate:

• Calculate iterative solutions by the N-R algorithm:

• Give a solution when a specific accuracy is reached.

max 0 max max 0( )sin cos cosd d d d d- + =

max max 0 max max 0( ) ( )sin cos cosf d d d d d d= - + -

(0)max/ 2p d p< <

( )max

( )max

( ) ( )(

ma

)

x 0 m x

( )max

maxa

max

(w ( ) )here ( ) cos

k

kk

k

k k

fd

d

ff

d

dd

d

dd d d

--= =

-D( 1) ( ) ( )

max max maxk k kd d d+ = +D

( 1) ( )max max

k kd d e+ - £

)( )max

(max

2( ) 2m

( ) ( )max max

m2 axmaxax

1 0) ( )2!

(k k

k kkdf d fd

fd

UEP SEP

18© 2021 Kai Sun

• In general, Pe,during fault ___ Pe, post-fault for any fault

• For a permanent fault cleared by tripping the fault circuit, Pe,post-fault ___ Pe,pre-fault

• For a temporary fault without losing any circuit, Pe, post-fault ___ Pe,pre-fault

<<

<

=

Response to a three‐phase fault

max sin sinBe

T

E EP PX

d d¢

= =

StableUnstable

19© 2021 Kai Sun

Multi‐Swing Stability

• For an SMIB or two-machine system (a 2-body problem), first-swing stability is usually sufficient to judge its transient stability over multiple swings.

• However, this is not enough to determine transient stability of a multi-machine system (an N-body problem in which chaos often arises).

20© 2021 Kai Sun

Critical Clearing Angle (CCA) and Critical Clearing Time (CCT) 

• Consider a simple case: a three-phase fault at the sending end with Pe, during fault=0 (all resistances are neglected)

• Solve Critical Clearing Angle c

|A1| |A2|Integrating both sides:

• Solve the CCT from the CCA:During the fault (Pe, during fault=0):

max

0m xm a m( sin )

c

c

P d P P dd

d

d

dd dd -=ò ò

( ) max max a0 m x(cos cos ) ( )c m cm cP P Pd d d d d d- - -=-

0 0 max 0max

cos ( 2 ) cos (note: )mc

PP

d p d d d p d= - - = -

2

20

2 0mH d P

dtd

w= - 0 0

02 2t

m md P dt P tdt H Hd w w= =ò

2004 mP t

Hw

d d= + 0

0

4 ( )cc

m

HtP

d dw

-=

21© 2021 Kai Sun

Vpe(u1)=|A2|+|A3|

Vpe(u2)=?

Pe, pre-fault

Pe, post-fault =P3maxsin

Pe, during fault= P2maxsin

P3max

P2max A1

A2A3 m max 0 3max max 2max 0

3max 2max

( ) cos coscos cP P P

P Pd d d d

d- + -

=-

s (u)

• |A1| is the kinetic energy Vke at c

• |A1|+|A3| is the total energy V=Vke+Vpegained at c after the fault is cleared

• |A2|+|A3| =Vpe(u)=Vcr, i.e. the maximum potential energy, or in other words, the critical energy

• The generator is stable if and only if V(c)Vcr (i.e. |A1|+|A3| |A2|+|A3| or equivalently, |A1| |A2| ),

m( )( )s

epeV P P dd

dd d--= ò

( )2

00

( ) 1 22ke r

rV H ww

ww

æ öD ÷ç ÷= ⋅ç ÷çD÷çè ø

A more general case with Pe, during fault>0

( )0

max

3ma0 2max x m maxsin (s n )ic

c

m ccP P PP ddd

d

d

dd d dd d d d d= -- -- òò

|A1| = |A2|If the fault is cleared at the CCA c

22© 2021 Kai Sun

Multiple Faults

d1

(0)c1d2 d3c2 c3

Pe,0Pe,1

Pe,3=Pe,2

max

Pe

Pm

23© 2021 Kai Sun

•How heavily the generator is loaded. •The fault‐clearing time•The post‐fault transmission system reactance•The generator reactance: A lower reactance increases peak power and reduces initial rotor angle.

•The generator output during the fault: This depends on the fault location and type

•The generator inertia: The higher the inertia, the slower the rate of change in angle. This reduces the kinetic energy gained during fault; i.e. the accelerating area A1 is reduced.

•The generator internal voltage magnitude (E’): This depends on the field excitation

•The voltage magnitude (EB) of the bus receiving power from the generator

Some factors that influence transient stability of a generator

•How heavily the generator is loaded. •The fault‐clearing time•The post‐fault transmission system reactance•The generator reactance: A lower reactance increases peak power and reduces initial rotor angle.

•The generator output during the fault: This depends on the fault location and type

•The generator inertia: The higher the inertia, the slower the rate of change in angle. This reduces the kinetic energy gained during fault; i.e. the accelerating area A1 is reduced.

•The generator internal voltage magnitude (E’): This depends on the field excitation

•The voltage magnitude (EB) of the bus receiving power from the generator

24© 2021 Kai Sun

Applying EAC to a Two‐Machine System

• Reduce two interconnected machines to an equivalent SMIB system.

21 0 0

1 1 121 1

( )2 2m e a

d P P Pdt H Hd w w

= - =

22 0 0

2 2 222 2

( )2 2m e a

d P P Pdt H Hd w w

= - =

2 2 212 1 2 0 1 12 2 2

1 2

( )2

a ad d d P Pdt dt dt H Hd d d w

= - = -

22 1 1 212

20 1 2

22 1 1 2

1 2

1 1

2 2

1 2

11

22 =a ma m e eH PH PH HH H

H Pd HH PH

P H Pdt H HH HHd

w-

= -+

-+ +-

+

12,max

12,0

0m,12 ,12

12

( ) 0eP P dH

d

d

wd- =ò

2122 ,

12

211

0, 2

2m eP Pd

dtH dw

= -

H1H2

H2 H12 Pm,12 Pe,12

H2= H1 Pm,1 Pe,1

H2=10H1 0.909H1 0.91Pm,10.09Pm,2 0.91Pe,10.09Pe,2

H2=H1 H1/2 (Pm,1Pm,2)/2 (Pe,1Pe,2)/2

25© 2021 Kai Sun

SMIB Equivalent Based Method (EEAC or SIME) for Multi‐Machine Systems [1]‐[3]

[1] Y. Xue, et al, "A Simple Direct Method for Fast Transient Stability Assessment of Large Power Systems". IEEE Trans. PWRS, PWRS3: 400–412, 1988.

[2] Y. Xue, et al, "Extended Equal-Area Criterion Revisited". IEEE Trans. PWRS, PWRS7: 10101022,1992.

[3] M. Pavella, et al, “Transient Stability of Power Systems: a Unified Approach to Assessment and Control”, Kluwer, 2000.

Main Idea:• According to rotor angle curves over a time window

(e.g. being obtained from simulation), partition mmachines into 2 groups– Critical machines (CMs)– Non-critical machines (NMs)

• Only m-1 ways of partitioning need to be studied.

• For each way of partitioning, construct a 2-machine equivalent and consequently an SMIB (i.e. “OMIB” in the papers) equivalent, such that conclusions of the EAC can be applied.

26© 2021 Kai Sun

Main Steps [3]

max

27© 2021 Kai Sun

Applications to real systems [3]

28© 2021 Kai Sun

Application in Commercialized Software

From TSAT v.14 User Manual by Powertech Labs

29© 2021 Kai Sun

Multi‐Machine System in a Simplified Model• Assumptions:

– Each group of coherent machines, which swing together, are represented by one equivalent machine with damping neglected:

– Then, each machine is represented by a voltage source E’ behind X’d(neglecting armature resistance Ra, the effect of saliency and the changes in flux linkages); The mechanical rotor angle of each machine coincides with the angle of E’.

– Pmi is assumed to remain constant during the entire period of simulation (neglecting governor control).

– Using the pre-fault bus voltages, all loads are converted to equivalent admittances to ground. These admittances are assumed to remain constant, i.e. constant impedance load models.

2

20

2 1, 2, ,i imi ei

H d P P i mdt

,= 1, 2, ,

i i d i iE V jX Ii m

Ii

Vi

30© 2021 Kai Sun

• Add m internal generator nodes (i.e. E’i behind X’di) to the n-bus network to form an (n+m)-bus network.• Then node voltage equations with ground as the reference

Ibus=Ybus Vbus

Ibus is the vector of the injected bus currentsVbus is the vector of bus voltages measured from the reference nodeYbus is the bus admittance matrix of (n+m)(n+m):

Yii (diagonal element) is the sum of admittances connected to bus iYij (off-diagonal element) equals the negative of the admittance between buses i and jCompared to the Ybus for power flow analysis, additional m internal generator nodes are added and Yii(in) is modified to include the load admittance at node i

X’d1

X’d2

X’dm

E’1

E’2

E’m Ybus (n+m)x(n+m)

reducebus m mY

31© 2021 Kai Sun

• To simplify the analysis, all nodes other than the generator internal nodes are eliminated as follows

• The electrical power of each machine:

• The power system model:

I1

I2

Im

* * *

1

Re( ) Re( )m

redei i i i ij j

j

P E I E Y E

2

1

2

1

| | cos

sin cos

nred

i ii i j ij ij ijjj i

n

i ii i j ij ij ij ijjj i

E G E E Y

E G E E B G

ij i j

2

1 1

2

0

2

sin cos sin cos

n

i

n

i j ij ij ij ij ii mi i ii mi

i

i i j ij ij ijj jj j i

ei

i

mi

E E B G C D

P

P E GH P E G

P

ij is the angle of Yijred

(a)

(b)

From (a): (a) (b):red

ij ij ijY G jB

32© 2021 Kai Sun

• Consider the simplified system model (classical-model generators without damping + constant impedance loads).

• Define the center of inertia (COI) and the motion of the COI about all generators:

TEF Method for a Multi‐machine Power System 

2

0 1

2 sin cos n

ij ij ijmi

i ij ei ij

i

m

j

i i i iP E G PH C D P

def1 1

1

m m

i i i ii i

COI n

ii

H H

HH

def

0

COTCOI

• Consider the motion of each generator w.r.t. the COI

0

2

ii

Ci I

i

ii

i

e OmH PH

PH P

1

1

0

0

m

i iim

i ii

H

H

Easy to prove:

00

1 1 10

i

i i COICO

m

I COI

m mi

i iii i

H H PP P HH dt H H

d

def

1 0

2( ) 2 m

COI COT COIi

m eiiHP P HP

33© 2021 Kai Sun

+0

Defining the post‐disturbance TEF

, ( )ke i ii

V , ( )pe i i

i

V ,,

( )magnetic ij iji j

V ,,

)trajectory of( dis isipated iji j

jV =Vke Vpe

• Procedure of the TEF method:1. Run time-domain simulation up to the instant of fault clearing (tcl) to obtain angles and speeds of all

generators, which are used to calculate V(xcl)2. Calculate the critical energy Vcr for the post-disturbance system (this is the most difficult step for a

large-scale systems; Vcr may be defined as the maximum Vpe at the closest UEP or controlling UEP)3. Check Vcr-V(xcl)

def2

1 1

1 ( )2

i

Si

m m

i i mi ei iT

Oi

iI

iCV J P P H P

Hd

Assume a linear integration path [4]

((

) ( ))

cl cl

j

S Si j i

ci j ij

jj

l Sij

ii

d d dk

[4] J.N. Qiang, Clarifications on the Integration Path of Transient Energy Function, IEEE Trans Power Systems, May 2005

1

sin co s n

ei ij ij ij ijjj i

P C D

34© 2021 Kai Sun

Time‐domain transient stability simulation

35© 2021 Kai Sun

Simulating a Multi‐Machine System in a Simplified Model• Solve the initial power flow and determine the initial bus

voltage phasors Vi.

• Terminal currents Ii of m generators prior to disturbance are calculated by their terminal voltages Vi and power outputs Si, and then used to calculate E’i

• All loads are converted to equivalent admittances:

• To include voltages behind X’di, add m internal generator buses to the n-bus power system network to form an n+mbus network (ground as the reference for voltages):

2

10

2 sin cosn

ii mi i ii i j ij ij ij ij

jj i

H P E G E E B G

X’d1

X’d2

X’dm

E’1

E’2

E’mYbus nxn

reducebus m mY ij ijG jB

36© 2021 Kai Sun

Today’s Bulk Power System Simulation

• For transient stability simulation of a bulk power system, its phasor model is usually adopted, which includes a set of nonlinear differential-algebraic equations validated by industry using event data.

• Simulation solves its initial value problems ( IVPs) on contingencies using numerical solvers (explicit or implicit methods)

• Commercial software (PSS/E or PSLF) takes 1-5 min to simulate the 70K-node Eastern Interconnection model (e.g. NERC MMWG model) for each wall-clock second (it can be speeded up to 4-5 sec/sec on a supercomputer with 512 processors by a parallel-in-time algorithm [5])

• Best industry practices in North America simulate 1-3K critical contingencies every 10-15 min on a reduced model (~10K nodes & 2K generators), while most of power companies run off-line simulations.

Eastern Interconnection• 70,000 nodes, • 5,000-10,000 generators, • 100,000+ state variables,• 100,000+ other variables.

[5] D. Osipov, N. Duan, S. Allu, S. Simunovic, A. Dimitrovski, K. Sun, “Distributed Parareal in Time with Adaptive Coarse Solver for Large Scale Power System Simulations,” 2019 IEEE PES General Meeting in Atlanta, GA

A

DEE

NI(x, V) = Y Vx = f(x,V)

37© 2021 Kai Sun

Numerical Integration Methods

• The differential equations to be solved for transient stability analysis are nonlinear ordinary differential equations with known initial values x=x0 and t=t0

where x is the state vector of n dependent variables and t is the independent variable (time). The objective is to solve x as a function of t

• Explicit Methods– computing the value of x at any time t using the values of x from only the previous

time steps, e.g. Euler and R-K methods

• Implicit Methods– Using interpolation functions involving future time steps for the expression under the

integral, e.g. the Backward Euler and Trapezoidal Rule methods

( , )x xd f t

dt

1 ( , )i i ii f tt x xx

1 1 1( , )i i ii f tt xx x

(Forward) Euler method:

Backward Euler method:

( , )f t t x x

38© 2021 Kai Sun

(Forward) Euler Method• The Euler method is equivalent to using the first two terms of the Taylor

series about x around the point (x0, t0), referred to as a first-order method (whose error is on the order of t2)– Approximate the curve at x=x0 and t=t0 by its tangent

– At step i+1:

• It is explicit compared to the Backward Euler Method (implicit)

• The forward Euler method results in inaccuracy and a propagation of the truncation error because it uses the derivative only from the beginning of each interval throughout the entire interval.

0

0 0( , )x

dx f x tdt

( , )dx f x tdt

0

0 0 0 0 01 ( , )x

x dxx x x f x td

x t tt

0x

dxd

xt

t

1 ( , )i i ii f txx x t

t0 t1

x0t

x1

x(t1)

1 1 1( , )i i ii f tx xx t

39© 2021 Kai Sun

Modified Euler (ME) Method• Modified Euler method consists of two steps:

(a) Predictor step:

The derivative at the end of the t is estimated using x1p

(b) Corrector step:

• It is a second-order method (error is on the order of t3)

0

1 0 0 0 0( , )x

p tdxx x f x tdt

x t

1

1

1( , )px

pf xdxt

td

0 1

1 0 00 0 1 1( , ) ( , )

2 2

p px xc

dx dxdt

x tdt f x t f tx x x t

1

1 1( , ) ( , )2i

pc i i i i

if x t f x tx x t

t0 t1

x0t

x1p

x1c

x (t1)

( , )dx f x tdt

Slope at the beginning of t

Estimated slope at the end of t

• Step size t must be small enough to obtain a reasonably accurate solution, but at the same time, large enough to avoid the numerical instability with the computer.

40© 2021 Kai Sun

Runge‐Kutta (R‐K) Methods

• General formula of the 2nd order R-K method (RK2): (error is on the order of t3)

• General formula of the 4th order R-K method:(error is on the order of t5)

1 1 1 2 2( )i i kx kx a a t

1 ( , )i ik f x t

2 1( , )i ik f x k t t

1 1 1 20 2( )k kx x a a t

1 0 0( , )k f x t

2 0 1 0( , )k f x k t t

When a1=a2=1/2, ==1, the RK2 method becomes the ME method (i.e. a special case of RK2)

t0 t1

x0t

x0+ k1

( , )dx f x tdt

x(t1)

At Step i+1:

1 2 3 41

2 26i i

k k k kx x t

1 ( , )i ik f x t1

2 ( , )2 2i ik tk f x t

23 ( , )

2 2i ik tk f x t

4 3( , )i ik f x k t t

~O(t2)

~O(t)

~O(t2)

~O(t3)

~O(t4)

~O(t)

41© 2021 Kai Sun

Numerical Stability of Explicit Integration Methods

• Numerical stability is related to the stiffness of the set of differential equations representing the system.

• The stiffness is measured by the ratio of the largest to smallest time constant, or more precisely by |max/min| of the linearized system.

• Stiffness in a transient stability simulation increases with more details (more smaller time constants) being modeled.

• Explicit integration methods have weak stability numerically; with stiff systems, the solution “blows up” unless a small step size is used. Even after the fast modes die out, small time steps continue to be required to maintain numerical stability.

42© 2021 Kai Sun

Implicit Methods• Implicit methods use interpolation functions for the expression under the

integral. “Interpolation” implies the function must pass through the yet unknown points at t1.

• A widely used implicit integration method is the Trapezoidal Rule method. It uses linear interpolation.

• The stiffness of the system being analyzed affects accuracy but not numerical stability. With larger time steps, high frequency modes and fast transients are filtered out, and the solutions for the slower modes is still accurate. For example, for the Trapezoidal rule, only dynamic modes faster than f(xn,tn) and f(xn+1,tn+1) are neglected.

1 0

1

0 0 1 1

1 1

Δ2Δ2 n n nn + n+n+

x = x +

x =

t f x ,t + f x ,t

t f x ,t +x f x+ ,t

1 0 0 10 12

pt f x ,t + f x ,x tx

Compared to ME method:

t0 t1

f(x0,t0) f(x1,t1)

t

A

Bf

t0 t1

x0t

( , )dx f x tdt

x(t1)

1

0

1 0 0 0| | | || |t

t

x x f x,t d t x AxA B

43© 2021 Kai Sun

Backward Euler Method (implicit)

t can be arbitrarily large as long as max has a negative real part

(this method has A-Stability)

Euler Method (explicit)

The method is numerically stabile if

max has a negative real part and

Comparison of Explicit and Implicit Methods

1 1 1( , )i i i ix x f x t t

( , )x f x t

1 ( , )i i i ix x f x t t

max x

1 max 1

1 max(1 )i i

i

x x tx t

1 maxi ix x t

1max

11i ix x

t

0 max(1 )iix x t

0max

11

i

ix xt

max|1 | 1 t

max

2| |

t

44© 2021 Kai Sun

Simulating a General Multi‐Machine System

• Overall system equations are expressed in the general form comprising a set of 1st-order DEs (dynamic devices) and a set of algebraic equations (devices and network)

wherex state vector of the systemV bus voltage vectorI current injection vectorYN node admittance matrix. It is constant

except for changes introduced by network-switching operations; it is symmetrical except for the dissymmetry introduced by phase-shifting transformers.

A

DEE

NI(x, V) = Y Vx = f(x,V)

• Schemes for the solution of DE and AE are characterized by these factors– The integration method used to solve the DE,

either an implicit method or an explicit method.– The method used to solve the AE (power flow

analysis), e.g. the Newton-Raphson method.– The interfacing between the DE and AE: either a

partitioned approach or a simultaneous approach may be used

• Most commercialized power system simulation programs provide the Modified Euler, 2nd order R-K, 4th order R-K and Trapezoidal Rule methods

45© 2021 Kai Sun

Partitioned (alternating) scheme• Partitioned approach with explicit integration:

1. At t=0, the system is in the steady state, initial values of x, V and I are known, and f(x0, V0)=0.

2. Following a disturbance (e.g. a fault), x cannot change instantly. Solve AE for V and I, the corresponding power flows and other non-state variables of interest at t=0+. Then, compute f(x, V), i.e. dx/dt.

3. Perform explicit integration by, e.g., the RK2 method to solve DE for each time step of t: (say step tn+1)

i. Compute k1=f(xn, Vn)t and k2=f(xn+k1, Vn)t at tn. ii. Compute xn+1=xn+(k1+k2)/2

4. Using xn+1, solve AE by, e.g. N-R method, to give Vn+1 and In+1. Thus all values for tn+1 can be obtained. If the system has a switching operation, the network variables change instantly but not the state variables

• Advantage: Programming flexibility, simplicity, reliability and robustness. Since the solution of DE requires values only from the previous step, the DE associated with each device may be solved independently

• Disadvantage: Susceptibility to numerical instability. For a stiff system, a small time step is required throughout the solution period.

• To solve V, x may not be known accurately.• Similarly, to solve x, V may be only approximately

available.• This can lead to interface errors, the elimination of

which would require extrapolation (i.e. Vnk Vn

k-1,

Vnk-2, …) or iteration of the solution process at each

time step until

( , ) 0n n n NI V Y Vx

( , )n n nx f x V

• At each time step tn:

Alternating

1| |k kn n V V

nVnx

A

DEE

NI(x, V) = Y Vx = f(x,V)

46© 2021 Kai Sun

Partitioned scheme with current‐injection models for all devices

0 0

2

r

r rm e D

ddt

dH P P Kdt

( / 2 )

*

( / 2 )

( )

R e( )

jt

t a d t

e t t

jt

i

i

E V

I

eE E R jX I

P E I

e I

G

M…

Dynamic device

iiV I

*1

ji

i ii

N

ijj

YP QI VV

j

Constant current/power

load

Generator

Network AEs:

jjV I

kkV I

Ii

Vi

Generator DEs:

Generator AEs:

( ) 0,n n n NI Vx V Y

( , )nn nx x Vf

AlternatingnVnx

47© 2021 Kai Sun

Simultaneous Solution with Implicit Integration

( , )fx x V

0 ( , ) NI x V Y V

Solved by N-R method:

Trapezoidal rule