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# Multimachine Simulation. Lecture 20

## 1.

ECE 576 – Power SystemDynamics and Stability

Lecture 20: Multimachine Simulation

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

[email protected]

1

## 2.

Announcements• Read Chapter 7

• Homework 6 is due on Tuesday April 15

2

## 3.

Simultaneous Implicit• The other major solution approach is the simultaneous

implicit in which the algebraic and differential

equations are solved simultaneously

• This method has the advantage of being numerically

stable

3

## 4.

Simultaneous Implicit• Recalling the first lecture, we covered two common

implicit integration approaches for solving x f (x)

– Backward Euler

x(t t ) x(t ) tf x(t t )

For a linear system we have

x(t t ) I t A x(t )

1

– Trapezoidal

t

x(t t ) x(t ) f x(t ) f x(t t )

2

For a linear system we have

x(t t ) I t A

1

t

I 2 A x(t )

• We'll just consider trapezoidal, but for nonlinear cases 4

## 5.

Nonlinear Trapezoidal• We can use Newton's method to solve x f (x) with

the trapezoidal

Right now we

are just

considering

We are solving for x(t+ t); x(t) is known the differential

equations;

The Jacobian matrix is

we'll introduce

f1

the algebraic

f1

x

equations

x

1

n

t

shortly

J x(t t )

I

2

f

f

1

n

x1

xn

t

x(t t ) x(t ) f x(t t ) f x(t ) 0

2

5

## 6.

Nonlinear Trapezoidal usingNewton's Method

• The full solution would be at each time step

– Set the initial guess for x(t+ t) as x(t), and initialize the

iteration counter k = 0

– Determine the mismatch at each iteration k as

t

h x(t t ) x(t t ) x(t )

f x(t t ) ( k ) f x(t )

2

– Determine the Jacobian matrix

– Solve

(k )

x(t t )

( k 1)

(k )

x(t t )

(k )

J (x(t t ) h x(t t )( k )

(k )

1

– Iterate until done

6

## 7.

Infinite Bus GENCLS Implicit Solution• Assume a solid three phase fault is applied at the

generator terminal, reducing PE1 to zero during the

fault, and then the fault is self-cleared at time Tclear,

resulting in the post-fault system being identical to the

pre-fault system

– During the fault-on time the equations reduce to

d 1

1, pu s

dt

d 1, pu

1

1 0

dt

2 3

That is, with a solid

fault on the terminal of

the generator, during

the fault PE1 = 0

7

## 8.

Infinite Bus GENCLS Implicit Solution• The initial conditions are

(0 ) 0.418

x(0 )

(

0

)

0

pu

• Let t = 0.02 seconds

• During the fault the Jacobian is

1 3.77

0.02 0 s

J x(t t )

I

0

1

2 0 0

• Set the initial guess for x(0.02) as x(0), and

0

f x 0

0

.

1667

8

## 9.

Infinite Bus GENCLS Implicit Solution• Then calculate the initial mismatch

h x(0.02)(0 )

x(0.02) (0 ) x(0)

0.02

f x(0.02) (0 ) f x(0)

2

• With x(0.02)(0) = x(0) this becomes

0.418 0.418 0.02 0 0 0

h x(0.02)

2 0.167 0.167 0.00334

0 0

(0 )

• Then

1

x(0.02)

( 1)

0.418 1 3.77 0 0.4306

0

0

1

0

.

00334

0

.

00334

9

## 10.

Infinite Bus GENCLS Implicit Solution• Repeating for the next iteration

f x 0.02

( 1)

1.259

0

.

1667

0.4306 0.418 0.02 1.259 0

h x(0.02)

0

.

00334

0

0

.

167

0

.

167

2

( 1)

0.0

0.0

• Hence we have converged with

0.4306

x(0.02)

0

.

00334

10

## 11.

Infinite Bus GENCLS Implicit Solution• Iteration continues until t = Tclear, assumed to be 0.1

seconds in this example

0.7321

x(0.10 )

0

.

0167

• At this point, when the fault is self-cleared, the equations

change, requiring a re-evaluation of f(x(Tclear))

d

pu s

dt

d pu 1 1.281

1

sin

dt

6

0.52

6.30

f x 0 .1

0

.

1078

11

## 12.

Infinite Bus GENCLS Implicit Solution• With the change in f(x) the Jacobian also changes

s

3.77

1

0.02 0

J x(0.12 )

I

0

.

00305

1

2 0.305 0

(0 )

• Iteration for x(0.12) is as before, except using the new

function and new Jacobian

h x(0.12)(0 )

x(0.12) (0 ) x(0.01)

0.02

f x(0.12) (0 ) f x(0.10 )

2

1

x(0.12)

( 1)

3.77 0.1257 0.848

0.7321 1

0

.

0167

0

.

00305

1

0

.

00216

0

.

0142

This also converges quickly, with one or two iterations

12

## 13.

Computational Considerations• As presented for a large system most of the

computation is associated with updating and factoring

the Jacobian. But the Jacobian actually changes little

and hence seldom needs to be rebuilt/factored

• Rather than using x(t) as the initial guess for x(t+ t),

prediction can be used when previous values are

available

x(t t )(0 ) x(t ) x(t ) x(t t )

13

## 14.

Two Bus Results• The below graph shows the generator angle for varying

values of t; recall the implicit method is numerically

stable

14

## 15.

Adding the Algebraic Constraints• Since the classical model can be formulated with all the

values on the network reference frame, initially we just

need to add the network equations

• We'll again formulate the network equations using the

form

Ι (x, y ) Y V or Y V Ι (x, y ) 0

• As before the complex equations will be expressed

using two real equations, with voltages and currents

expressed in rectangular coordinates

15

## 16.

Adding the Algebraic Constraints• The network equations are as before

VD1

V

Q1

VD 2

y

VDn

VQn

n

G

V

B

V

I

(

x,y

)

0

1k QK

ND1

1k Dk

k 1

n

GikVQk BikVDK I NQ1 (x,y ) 0

k 1

n

G2 kVDk B2 kVQK I ND 2 (x,y ) 0

g (x, y ) k 1

n

G

V

B

V

I

(

x,y

)

0

nk QK

NDn

nk Dk

k 1

n

G

V

B

V

I

(

x,y

)

0

nk DK

NQn

nk Qk

k 1

16

## 17.

Classical Model Couplingof x and y

• In the simultaneous implicit method x and y are

determined simultaneously; hence in the Jacobian we

need to determine the dependence of the network

equations on x, and the state equations on y

• With the classical model the Norton current depends on

1

x as I Ei i , G jB

Ni

Rs ,i jX d ,i

i

i

Rs ,i jX d ,i

I Ni I DNi jI QNi Ei cos i j sin i Gi jBi

EDi jEQi Ei cos i j sin i

I DNi EDi Gi EQi Bi

I QNi EDi Bi EQi Gi

17

## 18.

Classical Model Couplingof x and y

• The in the state equations the coupling with y is

recognized by noting

PEi EDi I Di EQi I Qi

I Di jI Qi EDi VDi j EQi VQi Gi jBi

I Di EDi VDi Gi EQi VQi Bi

I Qi EDi VDi Bi EQi VQi Gi

PEi EDi EDi VDi Gi EQi VQi Bi EQi EDi VDi Bi EQi VQi Gi

PEi EDi2 EDiVDi Gi EQi2 EQiVQi Gi EDiVQi EQiVDi Bi

18

## 19.

Variables and Mismatch Equations• In solving the Newton algorithm the variables now

include x and y (recalling that here y is just the vector

of the real and imaginary bus voltages

• The mismatch equations now include the state

integration equations

h x(t t )( k )

x(t t )

(k )

t

x(t )

f x(t t ) ( k ) , y (t t ) ( k ) f x(t ), y (t )

2

• And the algebraic equations

g x(t t ) , y (t t )

(k )

(k )

19

## 20.

Jacobian Matrix• Since the h(x,y) and g(x,y) are coupled, the Jacobian is

J x(t t ) , y (t t )

h x(t t ) , y (t t ) h x(t t ) , y (t t )

(k )

(k )

(k )

(k )

x

(k )

(k )

g

x

(

t

t

)

,

y

(

t

t

)

x

(k )

(k )

y

(k )

(k )

g x(t t ) , y (t t )

y

– With the classical model the coupling is the Norton current at

bus i depends on i (i.e., x) and the electrical power (PEi) in

the swing equation depends on VDi and VQi (i.e., y)

20

## 21.

Jacobian Matrix Entries• The dependence of the Norton current injections on is

I DNi Ei cos i Gi Ei sin i Bi

I QNi Ei cos i Bi Ei sin i Gi

I DNi

Ei sin i Gi Ei cos i Bi

i

I QNi

i

Ei sin i Bi Ei cos i Gi

– In the Jacobian the sign is flipped because we defined