Power System Operations Planning 7 Transient Stability

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Spring 2015Instructor: Kai Sun

ECE 422 Power System Operations & Planning

7 ‐ Transient Stability

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References

•Saadat’s Chapter 11.5 ~11.10•EPRI Tutorial’s Chapter 7•Kundur’s Chapter 13

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Transient Stability

•The ability of the power system to maintain synchronism when subjected to a severe transient 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 of the initial disturbances

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Pe= =Pmaxsin

Pmax= E’EB/XT

• Swing Equation:2

max20

2 sinm e mH d P P P P

dtd

dw

= - = -

Pm= Pe

• The rotor will accelerate if Pm increases, or Pe decreases

Pa  is called Accelerating Power

Single‐Machine Infinite Bus (SMIB) System

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Power Angle Relationship

Question: What if both circuits are out of service?

Pe= =Pmaxsin

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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 equilibrium point 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 rincreases from 0. When b is reached, Pa=0 but r >0, so continues to increase.

• b c: >1 and Pa<0, so r decreases until c. At c, r=0 and reaches the peak value max.

• c: At c, the rotor starts to decelerate (since Pa<0) with r<0 and decreases.

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

2

max20

2 sina m e mH d P P P P P

dtd

dw

= = - = -

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Equal‐Area Criterion (EAC)

If max exists where d/dt=0:

(Note: all losses are neglected)

Moment of inertia in p.u.

2 0 ( )r m eP P dHw

w dD = -ò20

20

2 ( )2 r m eH P P dw

w dw

D = -ò

max

0

max

2

0

1 ( )2

rm eJ P P d

d

dd d

wd

w=

æ öD ÷ç ÷ = -ç ÷ç ÷çè ø ò

0

max

1

1

(( ))m e m eP PP P d dd

dd

ddd --= +òò

21 |area | |area | = 0AA =-

At max, r=0 and the integral=0

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• Equal‐Area Criterion (EAC): The stability is maintained only if a decelerating area |A2|  the accelerating area |A1| can be located above Pm (from b to d, i.e. the Unstable Equilibrium Point or UEP).

• If |A2|<|A1|,  will continue increasing at UEP (since r>0), so it will lose stability. • For the case with a step change in Pm,  the new Pm does matter for transient stability. 

UEP SEP

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1 max

0 11 0 max max max 1( ) sin sin ( )m mP P d P d P

d d

d dd d d d d d d 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

• At the UEP, max maxsinmP P d=

1 maxd p d= -

UEP SEP

max maxsinmP P d=

• Solve max to calculate the transient stability limit for a step change of Pm:

Questions: Can we increase Pm beyond that transient limit? If yes, how much further can we increase Pm?

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

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Solve max by the Newton‐Raphson method

• The nonlinear function form:

• Select an initial estimate:

• Calculate iterative solutions by the N‐R algorithm:

• Give a solution when a specific accuracy  is reached, i.e.

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

max 0( ) cosf cd d= =

( )max/ 2 kp d p< <

( )max

( ) ( )( ) max maxmax ( ) ( )

max 0 max

max

( ) ( )where ( ) cos

k

k kk

k k

c f c fdf

dd

d dd

d d dd

- -D = =

-( 1) ( ) ( )max max max

k k kd d d+ = +D

( 1) ( )max max

k kd d e+ - £

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Response to a three‐phase fault

• Pe,during fault << Pe, post-fault

• Pe,post-fault <Pe,pre-fault for a permanent fault (cleared by tripping the fault circuit) or Pe, post-fault =Pe,pre-fault for a temporary fault

Pe= =Pmaxsin

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Stable Unstable

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Critical Clearing Angle (CCA)

• Saadat’s Sec.11.6 (Example 11.5)• Consider a simple case

– A three‐phase fault at the sending end – Pe, during fault=0 if all resistances are neglected

– Critical Clearing Angle c

|A1|                         |A2|Integrating both sides:

max

0m max m( sin )

c

c

P d P P dd d

d dd d d= -ò ò

( )0 max max max(cos cos ) ( )m c c m cP P Pd d d d d d- = - - -

max maxmax

cos ( ) cosmc c

PP

d d d d= - +

0 0max

( ) cosmc

PP

p d d d= - - -

= -0

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Critical Clearing Time (CCT)

• Solve the CCT from the CCA:Since Pe, during fault=0 for this case, during the fault:

2

20

2m

H d Pdtd

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

-=

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(pre-fault)(post-fault) P3max

(fault-on) P2max

• For a more general case: Pe (during fault)>0

( )0

max

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

c

m ccP P PP ddd

d

d

dd d dd d d d d= -- -- òò

m max 3max max 2 max 0

3max 2 max

( ) cos coscos cc

P P PP P

d d d dd

- + -=

-

P3max

P2maxA1

A2A3

s (u)

• |A1| = Vke(c), the kinetic energy at c

• |A1|+|A3| = V(c)=Vke(c)+Vpe(c), total energy at c

• |A2|+|A3| = Vpe(u)=Vcr, i.e. the largest potential energy

• If and only if V(c)Vcr (i.e. |A1||A2|), the generator is stable

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Factors influencing transient stability

• How heavily the generator is loaded. • The generator output during the fault. This depends on the fault location and type

• 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 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 infinite bus voltage magnitude EB

• How heavily the generator is loaded. • The generator output during the fault. This depends on the fault location and type

• 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 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 infinite bus voltage magnitude EB

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EAC for a Two‐Machine System• Two interconnected machines respectively with H1 and H2

– The system can be reduced 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

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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

= -

Pe12

Pm12

1212,max12,0

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Methods for Transient Stability Analysis

Analyzing a system’s transient stability following a given contingency•Time‐domain simulation: 

– At present, the most practical available method of transient stability analysis is time‐domain simulation in which the nonlinear differential equations are solved by using step‐by‐step numerical integration techniques.

•Direct methods: – Those methods determine stability without explicitly solving the system differential equations. The methods are based on Lyapunov’s second method, define a Transient Energy Function (TEF) as a possible Lyapunov function, and compare the TEF to a critical energy, denoted by Vcr, to judge stability

– EAC is a direct method for a SMIB or two‐machine system

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Numerical Integration Methods

• The differential equations to be solved in power system stability analysis are nonlinear ODEs (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). Our objective is to solve x as a function of t

• Explicit Methods– In these methods, the value of x at any value of t is computed from the

knowledge of the values of x from only the previous time steps, e.g. Euler method and R-K methods

• Implicit Methods– These methods use interpolation functions (involving future time steps)

for the expression under the integral, e.g. the Trapezoidal Rule

( , )x xd f t

dt

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Euler Method• The Euler method is equivalent to using the first two

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

– At step i+1

• The standard Euler method results in inaccuracies because it uses the derivative only at the beginning of the interval as though it applied throughout the interval

0

0 0( , )x

fd xxt

td

( , )dx f x tdt

0

0 01x

x dxx x tt

xd

0x

dxd

xt

t

1 i

i ix

dxx x tdt

t0 t1

x0t

x1

x(t1)

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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 0p

x

dxxdt

tx

1

1

1( , )px

pf xdxt

td

0 1

1 0 2

px xc

dx dxdt dt

x x t

1

1 2

pi i

i

x xci

dx dxdt dt

x 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, e.g. increasing round-off errors.

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Runge‐Kutta (R‐K) Methods

• General formula of the 2nd order R-K method: (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 2i ix x a k a k

1( , )i ik f x t t

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

1 0 1 1 2 2x x a k a k

01 0( , )xk f t t

0 12 0( , )f x k t ttk

The ME method is a special case with a1=a2=1/2, ==1

t0 t1

x0t

x0+ k1

( , )dx f x tdt

x(t1)

At Step i+1:

1 1 2 3 41 ( 2 2 )6i ix x k k k k

1 ( , )i ik f x t t

12 ( , )

2 2i ik tk f x t t

23 ( , )

2 2i ik tk f x t t

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

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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 modeling more details (more small time constants are concerned).

• 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

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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

• The simplest 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.

t0 t1

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

t

1 0

1

0 0 1 1

1 1Δ2

Δ2

n+ n n n n+ n+

x = x +

tx = x + f x ,t +

t f x ,t

f x ,

+ f x

t

,t

1 1 n+12n

pn n n n

tx x f x ,t + f x ,t

Compared to ME method:A

B

x(t1)=x(t0)+|A|+|B|

f

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Overall System Equations• The overall system equations are expressed in the general form

comprising a set of first-order differential equations (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; symmetrical except for dissymmetry introduced by phase-shifting transformers

A

EE

D

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

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Solution of the Equations

• Schemes for the solution of equations DE and AE are characterized by the following factors– The manner of interface between the DE and AE. Either a

partitioned approach or a simultaneous approach may be used– The integration method, i.e. an implicit method or explicit method,

used to solve the DE.– The method used to solve the AE (power flow analysis), e.g. the

Newton-Raphson method.

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

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A Simplified Model for Multi‐Machine Systems

• Consider these classic simplifying assumptions:– Each synchronous machine is represented by a voltage source E’ with

constant magnitude |E’| behind X’d (neglecting armature resistances, the effect of saliency and the changes in flux linkages)

– The mechanical rotor angle of each machine coincides with the angle of E’– The governor’s actions are neglected and the input powers Pmi are assumed to

remain constant during the entire period of simulation– Using the pre-fault bus voltages, all loads are converted to equivalent

admittances to ground. Those admittances are assumed to remain constant (constant impedance load models)

– Damping or asynchronous powers are ignored.– Machines belonging to the same station swing together and are said to be

coherent. A group of coherent machines is represented by one equivalent machine

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• 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 a n+m bus network (ground as the reference for voltages): X’d1

X’d2

X’dm

E’1

E’2

E’mYbus nxn

reducebus m mY

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• Node voltage equation with ground as reference

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 :

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

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

I1

I2

Im

where

where ij is the angle of Yij

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needs to be updated whenever the network is changed.