70 numerical integration techniques

8/4/2019 70 Numerical Integration Techniques

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NUMERICAL INTEGRATIONNUMERICAL INTEGRATION

TECHNIQUESTECHNIQUES

Applications to power system

reated ByNIRMAL PATEL

uided ByS.G.


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OutlineOutline

Introduction One Step Methods

Forward Eulers Method

Backward Eulers Method Trapezoidal Method Taylor series based Methods Runge-Kutte Method

2nd order

4th

order Accuracy & Error Analysis Stability Analysis Stiff systems Power System Applications

Transient stability Analysis


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IntroductionIntroduction

In power systems,1. Some times analytical formulae for

integration i.e. is notavailable .2. Some times we dont know the

functional form and have only a

set of tabulated data points.n both cases it is necessaryo use numerical methods to.valuate the integrals


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Mathematics 101Mathematics 101

Mean Value Theorem:-Mean Value Theorem:- If a function is continuous & differentiable between two

points, say (x1 , y1) & (x2 , y2), then the slope of the linejoining these points is equal to the derivative of the function at

least at one other point, say (a , b), between these two points,i.e...

&If th e n w e ca n d e riv e th e..E u le r's E q u a tio n


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One Step MethodOne Step Method

These methods allow us to vary the stepsize.

Use only one initial value After each step is completed the past step

is forgotten We do not use thisinformation.

:T h e te ch n iq u e s a re d e fin e d a s


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F o r w a r do r w a r d u l e r s M e t h o du l e r s M e t h o dConsider the First order ODE

With an initial condition

Then the (i+1)th point of the solution

is obtained from the ith point usingthe following formula


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GeometricalGeometricalInterpretationInterpretation

X

Y

C

B

A

..


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ExampleExample

Let f=y y(0)=1.

For i=1 y1=1+(0.01*1)

y1=1.01NEW INITIAL POINT:-

(x1,y1)=(0.01,1.01)Similarly we will go until desired x as we can

see the step 1 will be forgotten & newinitial point will be (x1,y1).

The complete solution is given in next slide.


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O N T IN U E D O N T IN U E Di h=0.01 h=0.005

xi yi xi yi1 0.01 1.01 0.005 1.0050

2 0.02 1.0201 0.010 1.01003 0.03 1.030301 0.015 1.051

4 0.04 1.040606 0.020 1.0202

5 0.05 1.05101206 0.025 1.0253

6 - - 0.030 1.0304

7 - - 0.035 1.03568 - - 0.040 1.0408

9 - - 0.045 1.046004

10 - - 0.05 1.05123402


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CONTINUEDCONTINUED

Since the exact solution of this equation is y= ex, the correct value at x = 0.05 is1.05127109. It is clear that, to obtain more accuracywith Eulers forward method, we must take aconsiderably smaller value for h. These results are correct to four decimalplaces after the decimal point. Because of therelatively small step size required, Eulersmethod is not commonly used for integratingdifferential equations. We could, of course, apply Taylorsalgorithm of higher order to obtain betteraccuracy, and in general, we would expectthat the higher the order of the algorithm, the


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Backward EulersBackward EulersMethodMethod

Consider the First order ODE





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ExampleExample

Let f=y y(0)=1.

where 0


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CONTINUEDCONTINUED

Solution table is given herei h=0.005

xi yi

1 0 12 0.005 1.0050253 0.01 1.0100754 0.015 1.01515125 0.02 1.02025252

6 0.025 1.025379417 0.03 1.030532088 0.035 1.0357106339 0.04 1.04091520910 0.045 1.04614593811 0.05 .1 0 5 1 4 0 2 9 5 3


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CONTINUEDCONTINUED

Since the exact solution of thisequation isy = ex, the correct value at x= 0.05 is 1.05127109. It is clear that, we can obtain moreaccuracy with Eulers Backward methodw.r.t. Forward Eulers method. These results are correct to four

decimal places after the decimal point.Because it is relatively Implicit, Eulersbackward method is not commonly usedfor integrating differential equations.


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ra p e z o id a l M e th o dra p e z o id a l M e th o dConsider the First order ODE





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ExampleExample

Let f=y y(0)=1.

where 0


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CONTINUEDCONTINUED

For i=1 First we predict the value of y1 by Eulersforward method.

y1=1+(0.005*1) y1=1.005

Also X1=0.005 So the corrected value of y1is given by:-

1=1.005038


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CONTINUEDCONTINUED

i xi yi1 0 1

2 0.005 1.005013

3 0.01 1.010050

4 0.015 1.015113

5 0.020 1.020201

6 0.025 1.025315

7 0.030 1.030454

8 0.035 1.0356199 0.040 1.040810

10 0.045 1.046028

11 0.050 1.051271


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SummarySummary

Trap Rule, Forward-Euler, Backward-Euler

All are one-step methods as xk+1 is

computed using only xk

, not xk-1

, xk-2

, xk-

3...Forward-Euler is the simplest

As No equation solutionexplicit method.

Backward-Euler is more expensive Equation solution each step

implicit method most stable.

Trapezoidal Rule might be moreaccurate


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TaylorTaylorseriesseriesbasedbasedMethodsMethods

Numerical methods we saw earlier have anunderlying derivation from Taylors Theorem.

For IVP the solution by Taylor series is given by

where


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CONTINUEDCONTINUED

The Taylor series method is astraight forward adaptation ofclassic calculus to develop thesolution as an infinite series. Thecatch is that a computer usuallycan not be programmed to

construct the terms and one doesnot know how many terms shouldbe used.

The method is not strictly a

numerical method but is use in


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CONTINUEDCONTINUED

From the Taylor series expansionThe step size is defined as :Using the initial condition, the

higher order derivatives of theequation can be obtained.

( ) ( ) ( ) ( ) ( ) ( )+++++=

0

IV4

0

3

0

2

00 !4!3!2 xy

h

xy

h

xy

h

xyhxyxy

0xxh =


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ExampleExample

Let f=y=y

With initial condition y(0)=1.

The analytical solution is

( )x

exy =


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CONTINUEDCONTINUED

The higher order derivatives can befound with the initial condition,

y(0) = 1, and the equation( ) ( )

( ) ( )

( ) ( )

( ) ( )xyxy

xyxy

xyxy

xyxy

=

=

=

=

IV

( ) ( )

( ) ( )

( ) ( )

( ) ( ) 100

100

100

100

IV==

==

==

==

yy

yy

yy

yy


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CONTINUEDCONTINUED

From the Taylor series expansion

Plug in the initial conditions:

Resulting in the equation:

( ) ( ) ( ) ( ) ( ) ( ) +++++= 0IV

4

0

3

0

2

00 !4

!3

!2

xyh

xyh

xyh

xyhxyxy

( ) ( ) ( ) ( ) ( ) Error1!4

1!3

1!2

11432

+++++=hhh

hhy

( ) Error

24

6

2

1432

+++++=hhh

hhy


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CONTINUEDCONTINUED

h Secondy(h ) Thirdy(h ) Fourthy(h ) ExactSolution

0 1 1 1 1

0.01 1.01005 1.010050167 1.010050167 1.010050167

0.02 1.0202 1.020201333 1.02020134 1.02020134

0.03 1.03045 1.0304545 1.030454534 1.030454534

0.04 1.0408 1.040810667 1.040810773 1.040810774

0.05 1.05125 1.051270833 1.051271094 1.051271096

1 2.5 2.666666667 2.708333333 2.718281828

1.2 2.92 3.208 3.2944 3.320116923

1.4 3.38 3.837333333 3.9974 4.055199967

1.5 3.625 4.1875 4.3984375 4.48168907

2 5 6.333333333 7 7.389056099

The results


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-u n ge K ut tau n g e K u t t a M e t h o de t h o dRunge-Kutta methods are very popular

because of their good efficiency; and areused in most computer programs fordifferential equations. They are single-

step methods.Eulers method is not very useful in practicalproblems because it requires a very smallstep size for reasonable accuracy. Taylorsalgorithm of higher order is unacceptable

as a general-purpose procedure becauseof the need to obtain higher totalderivatives of y(x). The Runge-Kuttamethods attempt to obtain greateraccuracy, and at the same time avoid the

need for higher derivatives, by evaluatingthe function f(x,y) at selected points on


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22ndnd order RK methodorder RK method

To convey some idea of how theRunge-Kutta is developed, lets

look at the derivation of the2ndorder. Two estimates

( )

( )1nn2nn1

21n1n

,

,

kyhxhfk

yxhfk

bkakyy

++==

++=+


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CONTINUEDCONTINUED

( /T h e Ta y lo r se rie s co e fficie n ts 3 e q u a tio n s 4)u n k n o w n s

If you select a as

If you select a as

* : -Note These co efficient would result in a modified Euler or Midpoint Method

# : -Note These Co efficient are derived by mathematician Heun so this method is also

.known as Heun s method

[ ]2

1,

2

1,1 ===+ bbba

2

3,

2

3,

3

1,

3

2#

==== ba

1,2

1

2

1 *==== ba


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ExampleExample

Let f=y=y


The analytical solutionis

The step size & co-efficient

( ) xexy =

1.0=h

2

3,

2

3,

3

1,

3

2==== ba

( )

( )[ ]

21i1i

1ii2

ii1

2

1,

,

kkyy

kyhxhfk

yxhfk

++=

++=

=

+


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CONTINUEDCONTINUED

x y exact k k1 k20 1 1 1.05 1 1.1

0.1 1.105 1.105171 1.16025 1.105 1.2155

0.2 1.221025 1.221403 1.282076 1.221025 1.3431275

0.3 1.349232625 1.349859 1.416694 1.349233 1.484155888

0.4 1.490902051 1.491825 1.565447 1.490902 1.639992256

0.5 1.647446766 1.648721 1.729819 1.647447 1.812191443

0.6 1.820428676 1.822119 1.91145 1.820429 2.0024715440.7 2.011573687 2.013753 2.112152 2.011574 2.212731056

0.8 2.222788925 2.225541 2.333928 2.222789 2.445067817

0.9 2.456181762 2.459603 2.578991 2.456182 2.701799938

1 2.714080847 2.718282 2.849785 2.714081 2.985488931

1.1 2.999059336 3.004166 3.149012 2.999059 3.298965269

1.2 3.313960566 3.320117 3.479659 3.313961 3.645356622

1.3 3.661926425 3.669297 3.845023 3.661926 4.028119068

1.4 4.0464287 4.0552 4.24875 4.046429 4.45107157

1.5 4.471303713 4.481689 4.694869 4.471304 4.918434085

1.6 4.940790603 4.953032 5.18783 4.940791 5.434869663

1.7 5.459573616 5.473947 5.732552 5.459574 6.005530978

1.8 6.032828846 6.049647 6.33447 6.032829 6.636111731

1.9 6.666275875 6.685894 6.99959 6.666276 7.332903463

2 7.366234842 7.389056 7.734547 7.366235 8.102858326


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44thth order RK methodorder RK method

( + )The i 1 thpoint of the solution is obtained fromthe ith point using the following formula

Where

( )This will result in a local error of O h5 and a( )global error of O h4

kyy ii +=+ 1

[ ]4321 226

1kkkkk +++=


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CONTINUEDCONTINUED

:The general form of the equations

[ ]

( )[ ]

( )[ ]34

23

12

1

4321

,

2

1,

2

1

2

1,

2

1

,

22

6

1

kyhxfhk

kyhxfhk

kyhxfhk

yxfhk

kkkkk

++=

++=

++=

=

+++=


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CONTINUEDCONTINUED

xi

x +/2

x + h

f 1

f 2

f 3f 4

( )432122

6

1

fffff +++=

f


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ExampleExample

Let f=y=y


The analytical solution is

The step size & co-efficient

( )x

exy =

1.0=h


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CONTINUEDCONTINUED:The example of a single step

( )[ ] ( ) ( )

( )

( )

( )[ ] ( )

[ ] 105170833.1226

1

110525.010525.1,1.01.0,

10525.02/105.01,05.01.0

2

1,

2

1

105.005.1,05.01.021,

21

1.011.01,01.0,

4321n1n

34

23

12

1

=++++=

==++=

=+=

++=

==

++=

====

+kkkkyy

fkyhxfhk

fkyhxfhk

fkyhxfhk

fyxfhk


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CONTINUEDCONTINUEDxi yi k k1 k2 k3 k4 exact

0 1 0.105170833 0.1 0.105 0.10525 0.110525 1

0.1 1.105170833 0.116231738 0.110517083 0.116043 0.116319 0.122149 1.105171

0.2 1.221402571 0.128455926 0.122140257 0.128247 0.128553 0.134996 1.221403

0.3 1.349858497 0.141965743 0.13498585 0.141735 0.142073 0.149193 1.349859

0.4 1.49182424 0.156896399 0.149182424 0.156642 0.157015 0.164884 1.491825

0.5 1.648720639 0.173397323 0.164872064 0.173116 0.173528 0.182225 1.648721

0.6 1.822117962 0.191633665 0.182211796 0.191322 0.191778 0.20139 1.822119

0.7 2.013751627 0.211787937 0.201375163 0.211444 0.211947 0.22257 2.013753

0.8 2.225539563 0.23406185 0.222553956 0.233682 0.234238 0.245978 2.225541

0.9 2.459601414 0.25867833 0.245960141 0.258258 0.258873 0.271847 2.459603

1 2.718279744 0.285883746 0.271827974 0.285419 0.286099 0.300438 2.718282

1.1 3.00416349 0.315950378 0.300416349 0.315437 0.316188 0.332035 3.004166

1.2 3.320113868 0.349179142 0.332011387 0.348612 0.349442 0.366956 3.320117

1.3 3.66929301 0.385902604 0.366929301 0.385276 0.386193 0.405549 3.669297

1.4 4.055195614 0.426488302 0.405519561 0.425796 0.426809 0.4482 4.0552

1.5 4.481683916 0.471342432 0.448168392 0.470577 0.471697 0.495338 4.481689

1.6 4.953026348 0.520913909 0.495302635 0.520068 0.521306 0.547433 4.953032

1.7 5.473940256 0.575698858 0.547394026 0.574764 0.576132 0.605007 5.473947

1.8 6.049639115 0.636245587 0.604963911 0.635212 0.636725 0.668636 6.049647

1.9 6.685884702 0.703160066 0.66858847 0.702018 0.703689 0.738957 6.685894

2 7.389044767 0.777111996 0.738904477 0.77585 0.777697 0.816674 7.389056


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CONTINUEDCONTINUED

The values are equivalent to.those of the exact solution If

= .we were to go out to x 5

( ) =y 5 .148 4131591( .148 4125901 )

The error is small relative to.the exact solution


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CONTINUEDCONTINUEDA comparison between the 2ndorder and the 4th -order Runge Kutta

.methods show a slight difference


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Methods ComparisonMethods Comparison

Metho

Euler

( )[ ]yxfhk

kk

,1

1

==

[ ]

( )[ ]

( )[ ]12

1

21

,

,

2

1

kyhxfhk

yxfhk

kkk

++==

+=

[ ]

( )[ ]

++=

++=

=

+=

23

12

1

31

3

2,

3

2

3

1,

3

1

,

34

1

kyhxfhk

kyhxfhk

yxfhk

kkk

[ ]

( )[ ]

( )[ ]34

23

12

1

4321

,

2

1,

2

1

2

1,

2

1

,

226

1

kyhxfhk

kyhxfhk

kyhxfhk

yxfhk

kkkkk

++=

++=

++=

=

+++=

Stiffness & StiffStiffness & Stiff


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Stiffness & StiffStiffness & StiffSystemsSystems

A stiff system is one involving rapidly changingcomponents together with slowly changingones.

An example of a single stiff ODE is:

whose solution if y(0)=0 is:

dy

dt=1000 y+3000 2000 e

t

y=30.998e1000t2.002e

t

Transient StabilityTransient Stability


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Transient StabilityTransient StabilityAnalysisAnalysis

For transient stability analysis weneed to consider three systems

1.Prefault - before the fault occurs thesystem is assumed to be at anequilibrium point

2.Faulted - the fault changes the

system equations, moving thesystem away from its equilibriumpoint

3.Postfault - after fault is cleared the

Transient Stability SolutionTransient Stability Solution


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Transient Stability SolutionTransient Stability SolutionMethodsMethods

There are two methods for solving thetransient stability problem

1.Numerical integration

l this is by far the most common technique,particularly for large systems; during thefault and after the fault the powersystem differential equations are solvedusing numerical methods

2.Direct or energy methods; for a two bussystem this method is known as theequal area criterial mostly used to provide an intuitive insight

into the transient stability problem


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ExampleExample

A 60 Hz generator is supplying 550 MW to an infinite bus(with 1.0 per unit voltage) through two paralleltransmission lines. Determine initial angle change for afault midway down one of the lines.H = 20 seconds, D = 0.1. Use t=0.01 second.


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CONTINUEDCONTINUED

We first need to determine the Pre-faultvalues.

Since P= 550 MW (5.5 pu) I=5.5

Next to get we need to determinethe thevenin equivalent during the fault

looking into the network from thegenerator

8.28141.15.5*1.00.1 =+= jEa

)(eP

08333.01.0||05.005.0 jjjjZth =+=03333.0 =thV


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CONTINUEDCONTINUED

Therefore prefault we have andAlso during the fault

Let and The Equation tointegrate are

and


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CONTINUEDCONTINUED

Now

With Eulers Method we get


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CONTINUEDCONTINUED

0 0.5 1 1.5 2

Simulation time in seconds

0

60

120

180

240

Gen

eratoranglein

degrees

clearing at 0.3 seconds




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CONCLUSIONCONCLUSION


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BooksBooks

Numerical Methods for Engineers-FifthEdition

. . & . . . -S C Chapra R P Canale McGraw Hill College Applied Numerical Analysis-Second Edition . . & . . ,C F Gerald P O Wheatley Addison Wesley

Boston Elementary Numerical Analysis-Third

Edition

. . . -S D Conte and C de Boor McGraw Hill College Numerical-Methods-Third Edition . & . , -J Douglas Richard L Burden Youngtown State

.University Foundations of Differential Calculus

Euler -


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- William Shakespeare

All's Well, that Ends Well

70 numerical integration techniques

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