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9/2/06 EE532 Lecture 3(Ranade) 1

EXPERT SYSTEMS AND SOLUTIONS

Email: [email protected]

[email protected]

Cell: 9952749533www.researchprojects.infoPAIYANOOR, OMR, CHENNAI

Call For Research Projects Finalyear students of B.E in EEE, ECE, EI,

M.E (Power Systems), M.E (AppliedElectronics), M.E (Power Electronics)

Ph.D Electrical and Electronics.Students can assemble their hardware in our

Research labs. Experts will be guiding theprojects.

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EE532

Power System Dynamics and

Transients

Satish J Ranade

Classical AnalysisNumerical Solution & Multi-machine

Systems

Lecture 5

EUMP Distance Education Services

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First swing stability-Numerical Solution

A generator connected to an infinite bus through a line.

Initially Pm=PePm Pe

jXL

+

E/

Pe

jXd

V/0

Fixed (Infinite Bus)

Pm

Stability is governed by the Swing Equation

d2/dt2 = (f/H) (Pm-Pe)

d /dt = -syn

Pe = E V sin () /(X+XL)

Swing Equation

Power Angle

Equation

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Nonlinear ODE in state variable formd /dt = (f/H) (Pm-Pmax sin )

d /dt = -syn

Pmax=EV/(Xd+XL)

Usually cannot get a closed form solution

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Numerical solution find (t) and (t)d /dt = (f/H) (Pm-Pmax sin )

d /dt = -syn

tn-2 tn-1 tn t

nn-1

n-2

Step sizeDivide time into intervals

tn-2,tn-1,tn,

Predict n from n-1,

n-2,

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Numerical solution find (t) and (t)d /dt = (f/H) (Pm-Pmax sin )

d /dt = -syn

tn-1 tn-2 tn+1 t

n-1

Euler Method

Uniform time step h

n = n-1+ h d/dt|t=tn-1h

d/dt|t=tn-1

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First swing stability-Equal Area Criterion

Application

1. Establish initial conditions

2. Define sequence of events and network

for each event

3. Develop Power angle curves

4. Apply EAC

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First swing stability-Equal Area Criterion

Example 1

Stability under small change in mechanical power

A 10 MVA, 0.8 pf lagging, 4160 V, 60Hz, three-phase

generator supplies 50% rated power at .8 pf lagging

to a 4160 V infinite bus.Determine if the generator is first-swing stable if the

prime mover power is increased by 10%

jXL

+

E/Pe

jXd

V/0

Pm

S

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First swing stability-Numerical Solution-Small change in Pmf 60:= syn 2 f:= H 2:= sec D 0:=

Pe delta( ) 2.828 sin delta( ):=

0 377:=rad

s

0 8.133 deg:= h .001:=

Pm n( ) .4 n h .1

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First swing stability-Numerical Solution-Small change in Pm

0 200 400 600 800 10008

9

10

n

deg

n

0 200 400 600 800 1000376.6

376.8

377

377.2

377.4

n

n

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Pm

Pe

P

o

APm1

C E

2x 0 1 2 3 40

0.5

1

1.5

2Pe

Pm

Pm1

Guess 2 10deg:=Given

0

2

Pm1 Pe ( )( )

d

0

Find 2( ) 9.77 deg=

x asinPm1

Pmax

:=

x 8.949 deg=

Remember 0 8.13 deg:=

EAC

Equal Area Criterion-Small change in mechanical power

Fi i bili N i l S l i S ll h i P

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0 200 400 600 800 10008

9

10

n

deg

The EAC in the previous slide says angle swings to 9.77 deg

and then swings back

Oscillates around the new equilibrium of 8.949 deg

( Step size of 0.001 is a little big oscillation is growing

Due to numerical instability)

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Effect of Damping Damper windings provide relative speed

damping. Other effects provide absolute damping.

This will make swing settle

0 200 400 600 800 10008

8.5

9

9.5

n

deg

n

0 200 400 600 800 1000376.9

377

377.1

n

n

Swing equation with relative speed damping

d /dt = (f/H) (Pm-Pmax sin -D (- syn)

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First swing stability-Numerical Solution-Fault

Example 2-Fault

1 2

3

The infinite bus receives 1 pu real power at 0.95 power

factor lagging

A fault at bus 3 is cleared by opening lines from 1-3 and 2-3

when the generator power angle Reaches 40 deg.

Is the system first swing stable?

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Example 2-Fault

Pm

Pe

P

o mcl

0]d)sin(14.2[Pm]d)sin(915.0[Pmm

cl

cl

0

=+

Apply EAC

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First swing stability-Numerical Solution-Faultf 60:= syn 2 f:= H 3:= sec D 0.:=

Pe delta( ) 2.828 sin delta( ):=

0 377:= rads 0 23.946deg:= h .001:=

Pm n( ) 1:=

Pe delta n,( ) 2.4638sin delta( )( ) n h .1

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First swing stability-Numerical Solution-Fault

0 200 400 600 800 10000

20

40

60

ndeg

n

0 200 400 600 800 1000370

375

380

385

n

n

E l A C i i

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Equal Area Criterion

Example 2-Fault

Pm

Pe

P

o mcl

23.95 40 55 554024

Rotor swings to 55 degrees then swings back- STABLE

Fi t i t bilit N i l S l ti F lt t>t i t 35

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First swing stability-Numerical Solution-Fault t>trict=.35

0 200 400 600 800 10000

200

400

600

ndeg

n

0 200 400 600 800 1000360

380

400

420

n

n

E l A C it i

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Example 2-Fault

Pm

Pe

P

o mcl

23.95 40 55 15612024

Rotor swings past 156 degrees UN STABLE

E l A C it i

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Example 2-Fault- Critical Clearing

Pm

Pe

P

0 1 m=- 1cl=112.9

23.95 15611224

Rotor swings past 156 degrees UN STABLE

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Stability of Numerical Solutions Can become unstable due to

Roundoff

Approximation

tn-1 tn-2 tn+1 t

n-1

h

d/dt|t=tn-1Approximation Error

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Stability of Numerical SolutionsCan control

Roundoff ( Reduce airthmetic operations)

Approximation(Use more advanced method, but will increase arithmetic operations)

Need to chose a reasonable step size ( or adaptively vary step size)

Numerical stability Means the the global error remains bounded

(See Crows Text for EE531/ Also see Kundur)

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Stability of Numerical Solutions

For the small change in Pm(Example 1) Euleris unstable even with h=.001

0 2 4 6 8 10

6

8

10

1211.846

6.136

ndeg

101 103 n h

n

Note x- axis units: n h = time in seconds

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Additional Methods

Taylor Series-based

See Crow Text

...)tt(2

1)tt()t(x)t(x

)t),t(x(f:ODE

2

n2

n

n

dt

)t(xd

n1ndt

)t(dx

n1nn1n

nndt)t(dx

+++==

+++

First order approximation gives the Euler method

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Additional Methods

Taylor Series-based Second Order

...)tt(2

1)tt()t(x)t(x

)t),t(x(f:ODE

2

n2

n

n

dt

)t(xd2

n1ndt

)t(dx

n1nn1n

nndt

)t(dx

+++=

=

+++

))t),t(x(f)t),t(x(f(2

1)tt()t(x)t(x

)tt/()]t),t(x(f)t),t(x(f[

nn1n1nn1nn1n

n1nnn1n1ndt

)t(xd2

n2

++=

=

++++

+++

Approximate the second derivative to get a second order

method

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Additional Methods

Taylor Series-based Second order method

))t),t(x(f)t),t(x(f(2

1)tt()t(x)t(x

nn1n1nn1nn1n ++= ++++

We do not know f(x(tn+1 ),tn+1 ) so approximate it by theEuler formula

))t),t(x(f))}tt).(t),t(x(f)t(x(f({

2

1)tt()t(x)t(x

nnn1nnnnn1nn1n+++= +++

Euler formula for x(tn+1 )

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Additional Methods Taylor Series-based Second order method

))t),t(x(f))}tt).(t),t(x(f)t(x(f({2

1

)tt()t(x)t(x nnn1nnnnn1nn1n+++=

+++

Kundur and GS call this modified Euler. It is also a second order

Runge Kutta. Our texts write the formula as follows:

At time tn we know x(tn)

Calculate derivative dx/dt=f(x(tn), tn)Now estimate x(tn+1 ), cal the estimate x

~

x~(tn+1 ) = x(tn)+h f(x(tn), tn), h= tn+1 tnNext estimate a new value for the derivative, call it dx~/dt

dx~/dt=f(x~ (tn+1 ), tn+1 )

Average the derivatives dx/dt and dx~/dt

Finally, the next value x(tn+1 ) = x(tn)+h (dx/dt + dx~/dt)/2

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Additional Methods Taylor Series-based Second order method

))t),t(x(f))}tt).(t),t(x(f)t(x(f({2

1

)tt()t(x)t(x nnn1nnnnn1nn1n+++=

+++

h

Actual

Estimated

Slope 1 Slope 2

Average

First Estimate

Final Estimate

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Additional Methods Taylor Series-Fourth Order Runge Kutta

)ht,3hK)t(x(f2K)2/ht,2/2hK)t(x(f3K)2/ht,2/1hK)t(x(f2K )t),t(x(f1K

tth6/)4K3K22K21K(h)t(x)t(x

nn

nn

nn

nn

n1n

n1n

++=++=++=

=

=++++=

+

+

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Additional Methods Taylor Series-Fourth Order Runge Kutta

)ht,3hK)t(x(f2K)2/ht,2/2hK)t(x(f3K)2/ht,2/1hK)t(x(f2K )t),t(x(f1K

tth6/)4K3K22K21K(h)t(x)t(x

nn

nn

nn

nn

n1n

n1n

++=++=++=

=

=++++=

+

+

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Additional Methods Multistep

2

n1n0n )tt(a)tt(a)t(x)t(x ++=)t),t(x(f)tt(a2adt/dx

n10 =+=

Assume a polynomial fit, e.g.,

Then

Next for t=tn )t),t(x(fa nn0 =

And for t=tn+1 = tn+h )t),t(x(fhaa 1n1n10 ++=+So h2/))t),t(x(f)t),t(x(f(a 1n1n1n1n1 ++++ =

h2/))t),t(x(f)t),t(x(fh)t),t(x(f(h)t(x)t(x 1n1n1n1n2

nnn1n +++++ ++=

2/))t),t(x(f)t),t(x(f(h)t(x)t(x 1n1nnnn1n +++ ++=

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Additional Methods Multistep Assume a polynomial fit, e.g.,


If we approximate f(x(tn+1 ),tn+1 ) as before we get the

modified Euler or RK2

Alternatively, if we solve for x(tn+1 ) we call the method

The Trapezoidal rule.

Nonlinear case, must solve iteratively

Also called an Implicit Method

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Additional Methods Trapezoidal Rule


h

Trapezoid approximates

Area under the curve

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