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9/2/06 EE532 Lecture 3(Ranade) 1
EXPERT SYSTEMS AND SOLUTIONS
Email: expertsyssol@gmail.com
expertsyssol@yahoo.com
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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9/2/06 EE532 Lecture 3(Ranade) 3
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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9/2/06 EE532 Lecture 3(Ranade) 4
First swing stability-Numerical Solution
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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9/2/06 EE532 Lecture 3(Ranade) 5
First swing stability-Numerical Solution
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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9/2/06 EE532 Lecture 3(Ranade) 6
First swing stability-Numerical Solution
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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9/2/06 EE532 Lecture 3(Ranade) 7
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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9/2/06 EE532 Lecture 3(Ranade) 8
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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9/2/06 EE532 Lecture 3(Ranade) 9
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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9/2/06 EE532 Lecture 3(Ranade) 10
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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9/2/06 EE532 Lecture 3(Ranade) 11
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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9/2/06 EE532 Lecture 3(Ranade) 12
First swing stability-Numerical Solution-Small change in Pm
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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9/2/06 EE532 Lecture 3(Ranade) 13
First swing stability-Numerical Solution-Small change in Pm
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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9/2/06 EE532 Lecture 3(Ranade) 14
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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9/2/06 EE532 Lecture 3(Ranade) 15
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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9/2/06 EE532 Lecture 3(Ranade) 16
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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9/2/06 EE532 Lecture 3(Ranade) 17
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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9/2/06 EE532 Lecture 3(Ranade) 18
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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9/2/06 EE532 Lecture 3(Ranade) 19
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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9/2/06 EE532 Lecture 3(Ranade) 20
Equal Area Criterion
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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9/2/06 EE532 Lecture 3(Ranade) 21
Equal Area Criterion
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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9/2/06 EE532 Lecture 3(Ranade) 22
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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9/2/06 EE532 Lecture 3(Ranade) 23
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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9/2/06 EE532 Lecture 3(Ranade) 24
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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9/2/06 EE532 Lecture 3(Ranade) 25
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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9/2/06 EE532 Lecture 3(Ranade) 26
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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9/2/06 EE532 Lecture 3(Ranade) 27
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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9/2/06 EE532 Lecture 3(Ranade) 28
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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9/2/06 EE532 Lecture 3(Ranade) 29
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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9/2/06 EE532 Lecture 3(Ranade) 30
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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9/2/06 EE532 Lecture 3(Ranade) 31
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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9/2/06 EE532 Lecture 3(Ranade) 32
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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9/2/06 EE532 Lecture 3(Ranade) 33
Additional Methods Multistep Assume a polynomial fit, e.g.,
2/))t),t(x(f)t),t(x(f(h)t(x)t(x 1n1nnnn1n +++ ++=
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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9/2/06 EE532 Lecture 3(Ranade) 34
Additional Methods Trapezoidal Rule
2/))t),t(x(f)t),t(x(f(h)t(x)t(x 1n1nnnn1n +++ ++=
h
Trapezoid approximates
Area under the curve
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9/2/06 EE532 Lecture 3(Ranade) 35
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