Spring Transient Stability

Spring 2021

Instructor: Kai Sun

ECE 522 ‐ Power Systems Analysis II Spring 2021

Transient Stability

© 2021 Kai Sun 2

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

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

© 2021 Kai Sun 4

Two Approaches for Transient Stability Analysis

To analyze the stability of a nonlinear system modeled by DAEs (differential-algebraic equations) 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=x

at t=t

by using step-by- step numerical integration (explicit or implicit).

,

E DE , A t

t x = f(x, y ) 0 = g(x, y )



Transient Stability Analysis Using 

Direct Methods

© 2021 Kai Sun 6

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.

Lyapunov Stability Criterion (Lyapunov’s second method)

Its equilibrium x

, which satisfies f(x

)=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= x

(x

has the minimum V=0)

• (V does not increase with time)

Equilibrium x

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

• “=0” if and only if x= x

,

E DE , A t

t x = f(x, y ) 0 = g(x, y )

( ) t 

x = f(x ) 

( ) 0 dV

dt x  ( ) 0 dV

dt x 

© 2021 Kai Sun 8

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 x_s)

– 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) x_s.

• If a fault occurs, the system gains kinetic energy V_KE and

potential energy V_PE during the fault-on period and then moves away from x_s.

• After fault clearing, V_KE is converted into V_PE.

• To avoid instability, the system must be capable of absorbing V_KE while moving towards the old SEP x_s or a new SEP.

• For a given post-disturbance network configuration, there is a critical, maximum amount of transient energy V_cr that the system can absorb without loss of synchronism

• Therefore, assessment of transient stability for the system state at fault clearing (x_cl) requires:

1. Define a TEF V(x_cl) that adequately describes the transient energy responsible for separating one or more generators from the rest of the system

2. Estimate the critical energy V_cr and calculate transient stability margin, i.e. V_cr V(x_cl).

V_KE+V_PE

V_PE=V_cr

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

0

0

0 0

2 ( )

r

r D

m e r

d d t

d K

H P P

d t

d w w

w w w

w w

ìïï = -

ïïïíï

ï = - - -

ïïïî

max sin ^B sin

e

T

P P E E

d X¢ d

= = P_e+jQ_e

P_t+jQ_t

P_B+jQ_B With all resistances neglected:

₀=260377 rad/s H in s

_r in rad/s

 in rad K_D in p.u

m ax

0 0

2H K ^D sin _m 0

P P

 

d d d

w + w + - =

Equilibria satisfy P_maxsin P_m=0

• _s= arcsin(P_m/P_max)

stable equilibrium point (SEP)

• _u =180^o-_s

unstable equilibrium point (UEP)

Pendulum subject to a constant torque

 𝑇_m

_s

  

       

2 sin _m 0

ml D mgl T



 sin Tg mgl

© 2021 Kai Sun 10

Phase‐Space Trajectories

No damping (K

=0)

Small damping

Large damping

m ax

0 0

2 H K

sin

0

P P

 

d d d

w + w + - =

P_a _r  a >0 =₀,   b 0 =_{r, max}  c <0 =₀ ,  =_max

Response to a step change in P

Consider a sudden increase in P_m: P_m0 P_m1.

• New SEP b satisfying P_e(₁)=P_m1

• a b: Due to the rotor’s inertia,  cannot jump from ₀ to ₁, so P_a=P_m1-P_e(₀)>0 and _r increases from ₀. When b is reached, P_a=0 but _r >₀, and  continues to increase.

• b c: >₁ and P_a<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 P_a<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 

reach the minimum speed?

2 2 max 0

2H d _{m e m} sin _a (accelerating power)

P P P P P

dt

d d

w = - = - =

Ignore all losses (K_D=0):

UEP

© 2021 Kai Sun 12

Nonlinear Oscillation of an SMIB System

Let  = -_s, =K_D/(2H)

⁼₀^P_max^/(2H)

       

               

    

  

  sinc cos 0

2

2

max

0 0

2 H K

sin

0

P P

 

d d d

w + w + - =           (sin   sin ) 0 



2H/ 

=0.1, D=0, P

=1, P

=0.5   =0,  =10, 

=30

Small disturbance 1 c s

2 o

f    

0 cos



  

 

         

Approximately sinusoidal

t₁

_max

Marginally stable

t₂

_min

1 m ax 2 m in

1

( ) ( )

f ^ t





Define oscillation amplitude as δ

=δ

− δ

or δ

= δ

− δ

.

max min

1 max 2 min

( , ) 1

( ) ( )

  ^   

f t t

Frequency‐Amplitude (F‐A) Curve

 

    



 





0 0

max s s s

( ) 2

cos( ) cos ( )sin

x

i H d

t x P

x x

,2 ,2 1

0 ( ) ( )



   





Oscillation amplitude (_max -_s)

Frequency

SEP

Stability 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

© 2021 Kai Sun 14

Equal‐Area Criterion (EAC)

At _max exists, _r=d/dt=0:

• Ignore all losses (K

=0):

Moment of inertia in p.u.

0

2 0 ( )

r Pm P de

H

d d

w w d

D =

ò

( )

0

2

0

0

1 2 ( )

2

r

m e

H ^d P P d

d

w w d

w æD ö÷

ç ÷

⋅çççè ÷÷ø =

ò

m 0

ax 1

1

( )

(P_m P d_e) ^d P_m P d_e

d d

d - d- - - d

=

ò ò

1 2

= | A | - | | A = 0

max 0

0 ^d (P_m P d_e)

d d

=

ò

V

=

2 2

0

2 2 ( _{m e})

d d d d d

P P dt dt dt dt dt H

w

d d d d

é ùê ú = ⋅ = -

ê úë û

2

0

2 ( )

2 ^{m e}

d P P

dt H

w

d = -

0

2

0(P_m P_e)

d d

dt H

d d

d w d

é ù -

ê ú = ê ú

ë û

ò

Question: What if K_D>0?

max

0 0

2 H K

sin

0

P P

 

d d d

w + w + - =

2 2

1

ke

2 ml

V = ⋅ d

Pendulum with a constant torque

 𝑇_m

_s

  

       

2 sin _m 0

ml D mgl T

Kinetic energy:

 sin Tg mgl

• |A₁|=V_ke gained from a to b

• |A₂|

=V_pe at c relative to b

Potential energy (ref: _s) = P_max(cosd₁-cosd_max)-P_m(d_max -d₁)

max(cos _s cos ) _m( s) Vpe =P d - d -P d d- (cos _s cos ) _m( s)

pe mgl T

V = d - d - d d-

Equal-Area Criterion (EAC):

The stability is maintained only if a decelerating area |A

|  accelerating area |A

| can be located above P

.

UEP

• When  reaches UEP, if |A

|<|A

|,  will continues to SEP

increase (since 

> 

) and accelerate to lose stability.

• Transient stability depends on the size of the step

change in P

.

© 2021 Kai Sun 16

• Solve _max.

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

• The new SEP:

max 1 1

0

1 0 max m xa sin ( max 1)

( ) sin

m m

P P d ^d P d P

d d

d d

d d d d d

d - -

ò

ò

Following a step change P_m0P_m, solve the transient stability limit of P_m:

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

max 0 max 0 max

(d -d )P_m = P (cosd -cosd )

max 0 max max 0

( d - d )sin d + cos d = cos d

Transient stability limit for a step change of P

• Since at the UEP , there isP_m = P_max sind_max

1 m ax

d = p - d

SEP UEP

max sin max 0

m m

P P d P

D £ -

Question: From P_m0, is there anyway to increase P_m beyond P_maxsin_max?

0 max(cos 1 cos )0 max(cos max cos )1 max

m m

P d P d d P d d P d

- + - = - - -

Solve 

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

( d - d )sin d + cos d = cos d

max max 0 max max 0

( ) ( )sin cos cos

f d = d - d d + d - d

(0)

/ 2 max

p <d <p

max( )

( ) max

( ) ( )

(

ma )

x 0 m x

( ) max max

a max

w ( ) ( )

here

( ) cos

k

k

k k

k k

f d d

f f

d

d d

d

d

d d d

-

= - =

D -

( 1) ( ) ( )

max max max

k k k

d

= d + D d

( 1) ( )

max max

k k

d

- d £ e

)

max( ) (

max

2

( ) 2 m

( ) ( )

max max

m

2 ax

max ax

1 0

) ( )

( 2!

k k

k df k d f k

f d

d _{ }

 

 

  

  

SEP UEP

© 2021 Kai Sun 18

• 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

sin

e

T

P P E E

d X ¢ d

= =

Stable

Unstable

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).

© 2021 Kai Sun 20

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 0

m x

m

(

sin

)

c

c

P d P P d

d d

d

d

d d

d = -

ò ò

(

)

m c

P d -d = P d - d -P d -d

0 0 max 0

max

cos _c P^m ( 2 ) cos (note: ) d = P p- d - d d = -p d

2 2 0

2H d _m 0

dt P d

w = - ^{0 0}

2 0 2

t

m m

d P dt P t

dt H H

d w w

=

ò

0 2

4 P t^m 0

H

d = w + d ⁰

0

4 ( _c )

c

m

t H

P

d d

w

= -

V_pe(_u1)=|A₂|+|A₃| V_pe(_u2)=?

Pe, pre-fault

Pe, post-fault =P_3maxsin

Pe, during fault= P_2maxsin

P_3max

P_2max A₁

A₂

A₃ ^{m max 0 3max max 2max 0}

3max 2max

( ) cos cos

cos _c P P P

P P

d d d d

d - + -

= -

_s (_u)

• |A₁| is the kinetic energy V_ke at _c

• |A₁|+|A₃|is the total energy V=V_ke+V_pe gained at _c after the fault is cleared

• |A₂|+|A₃|=V_pe(_u)=V_cr, i.e. the maximum potential energy, or in other words, the critical energy

• The generator is stable if and only if V(_c)V_cr (i.e. |A₁|+|A₃| |A₂|+|A₃|or equivalently, |A₁| |A₂|),

( m )

( )

s

e

Vpe ^d P P d d =

ò

( )

2

0

0

( ) 1

2 2

ke r r

V H w

w w

w = ⋅çæççD ö÷÷÷

D çè ÷ø

A more general case with P

>0

( )

0

max

3ma

0 2maxs ni xsin m( max )

c c

m c c

P P d ^d P d P

d d

d d d d

d -d -

ò

ò

|A

| = |A

|

If the fault is cleared at the CCA _c

© 2021 Kai Sun 22

Multiple Faults

_d1 (₀)

_c1

_d2 _c2 _d3 _c3

P_e,0 P_e,1

P_e,3=P_e,2

_max

 P

P

•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 A

is reduced.

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

•The voltage magnitude (E

) 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 A

is reduced.

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

•The voltage magnitude (E

) of the bus receiving power from the generator

© 2021 Kai Sun 24

Applying EAC to a Two‐Machine System

• Reduce two interconnected machines to an equivalent SMIB system.

2

1 0 0

1 1 1

2

1 1

( )

2 ^{m e} 2 ^a

d P P P

dt H H

d w w

= - =

2

2 0 0

2 2 2

2

2 2

( )

2 ^{m e} 2 ^a

d P P P

dt H H

d w w

= - =

2 2 2

12 1 2 0 1 1

2 2 2

1 2

( )

2

a a

d d d P P

dt dt dt H H

d d d w

= - = -

2

2 1 1 2

12 2

0 1 2

2

2 1 1 2

1 2

1 1

2 2

1 2

1 1

2 2

=

a a H Pm m e e

H P H H

H H

H P

d H P H

H

P H P

dt H H H H H

d w

= - -

+

-

+ +

- +

12,max 12,0

0 m,12 ,12

12

( P P

) d 0 H

d d

w - d =

ò

2 12

2 ,

1

2 2

1 1

0

, 2

2

m e

P P

d dt

H d

w = -

H₁ H₂

H₂ H₁₂ P_m,12 P_e,12 H₂= H₁ P_m,1 P_e,1

H₂=10H₁ 0.909H₁ 0.91P_m,10.09P_m,2 0.91P_e,10.09P_e,2 H₂=H₁ H₁/2 (P_m,1P_m,2)/2 (P_e,1P_e,2)/2

SMIB Equivalent Based Method (EEAC or SIME) for Multi‐Machine Systems 

[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 m machines 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.

© 2021 Kai Sun 26

Main Steps 

max

Applications to real systems 

© 2021 Kai Sun 28

Application in Commercialized Software

From TSAT v.14 User Manual by Powertech Labs

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’

(neglecting armature resistance R

, the effect of saliency and the changes in flux linkages); The mechanical rotor angle of each machine coincides with the angle of E’.

– P

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 2 0

2H d^{i i} _{mi ei} 1, 2, ,

P P i m

dt





= ,

1, 2, ,

i i d i i

E V jX I

i m

  



  

 I_i

V_i

© 2021 Kai Sun 30

• Add m internal generator nodes (i.e. E’

behind X’

) to the n-bus network to form an (n+m)-bus network.

• Then node voltage equations with ground as the reference I

=Y

V

I

is the vector of the injected bus currents

V

is the vector of bus voltages measured from the reference node Y

is the bus admittance matrix of (n+m)(n+m):

Y

(diagonal element) is the sum of admittances connected to bus i

Y

(off-diagonal element) equals the negative of the admittance between buses i and j

Compared to the Y

for power flow analysis, additional m internal generator nodes are added and Y

(i  n) is modified to include the load admittance at node i

X’_d1

X’_d2

X’_dm E’₁

E’₂

E’_m

Y

reduce bus m m

Y

• 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

I_m

* * *

1

Re( ) Re( ^{m red} )

ei i i i ij j

j

P E I E Y E



  

 ^{ }  ^



 

 

2

1

2

1

| | cos

sin cos

n red

i ii i j ij ij ij

jj i

n

i ii i j ij ij ij ij

jj i

E G E E Y

E G E E B G

 

 





  

  

  

  





ij i j

   

 

 

1 1

2 0

2

sin cos sin cos

n

i

n

i j ij ij ij ij i

i mi i ii mi

i

i i j ij ij ij

j j

j j i

ei i

mi

E E B G C D

P P E G

H P E G

P

     

 

 



 

       











 



_ij is the angle of Y_ij^red

(a) (b)

From (a): (a)  (b):

red

ij ij ij

Y G jB

   

      

© 2021 Kai Sun 32

• 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 ⁱ _{m i} ⁿ _ij sin _{ij ij} cos _{ij i ei}

j i

m j

i P i E Gi i P

H  C  D P

 



 

   





def 1 1

1

m m

i i i i

i i

COI n

i i

H H

H H

 















def

0 COT COI

 

  



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

0

2

i i

C

i I

i

i i

i

e O

m

H P P H

H  P

  



  



 





1

1

0 0

m

i i i

m

i i i

H H

















Easy to prove:

0 0

1 1 1

0

i

i i COI

CO

m

I COI

m m

i

i i i

i i

H H P

P P H

H dt H H

d     

     

  

  

def

1 0

( ) 2 2

m

COI COT COI

i

mi ei

P P P H













+0

Defining the post‐disturbance TEF

, ( )

ke i i i

V







i

V







,

magnetic ij( )ij i j

V







,

) trajectory of

dissipated ij( i

i j

V

 





=V

V

• Procedure of the TEF method:

1. Run time-domain simulation up to the instant of fault clearing (t_cl) to obtain angles and speeds of all generators, which are used to calculate V(x_cl)

2. Calculate the critical energy V_cr for the post-disturbance system (this is the most difficult step for a large-scale systems; V_cr may be defined as the maximum V_pe at the closest UEP or controlling UEP) 3. Check V_cr-V(x_cl)

def 2

1 1

1 ( )

2

i

iS

m m

i i mi ei i

T

O i

i

I i

V J P P H

H P d





 

 

 

       

Assume a linear integration path ^[4]

( ( ) ( )

)

cl cl

j

S S

i j i

i j i c

j j

j

l S

ij

i i

d k d    d

    

 

      



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

 

1

sin cos

n

ei ij ij ij ij

jj i

P C  D 



 



© 2021 Kai Sun 34

Time‐domain transient stability simulation

Simulating a Multi‐Machine System in a Simplified Model

• Solve the initial power flow and determine the initial bus voltage phasors V_i.

• Terminal currents I_i of m generators prior to disturbance are calculated by their terminal voltages V_i and power outputs S_i, 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+m bus network (ground as the reference for voltages):

 

2 0 1

2 ⁱ _{i mi i ii} ⁿ _{i j ij} sin _{ij ij} cos _ij

jj i

H  P E G E E B  G 

 _



  

  





X’_d1

X’_d2

X’_dm E’₁

E’₂

E’_m

Y

Y

^{ } _ ^G

^{ jB}

^ _

© 2021 Kai Sun 36

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

)

• 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

D E

E

N

I(x, V) = Y V

x = f(x, V) 