ECE576_Spring2014_Lect20

Содержание

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Announcements Read Chapter 7 Homework 6 is due on Tuesday April 15

Announcements

Read Chapter 7
Homework 6 is due on Tuesday April 15

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Simultaneous Implicit The other major solution approach is the simultaneous implicit

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

Simultaneous Implicit

Recalling the first lecture, we covered two common implicit integration

approaches for solving
Backward Euler
Trapezoidal
We'll just consider trapezoidal, but for nonlinear cases
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Nonlinear Trapezoidal We can use Newton's method to solve with the

Nonlinear Trapezoidal

We can use Newton's method to solve with the trapezoidal
We

are solving for x(t+Δt); x(t) is known
The Jacobian matrix is

Right now we are just considering the differential equations; we'll introduce the algebraic equations shortly

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Nonlinear Trapezoidal using Newton's Method The full solution would be at

Nonlinear Trapezoidal using Newton's Method

The full solution would be at each time

step
Set the initial guess for x(t+Δt) as x(t), and initialize the iteration counter k = 0
Determine the mismatch at each iteration k as
Determine the Jacobian matrix
Solve
Iterate until done
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Infinite Bus GENCLS Implicit Solution Assume a solid three phase fault

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

That is, with a solid fault on the terminal of the generator, during the fault PE1 = 0

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Infinite Bus GENCLS Implicit Solution The initial conditions are Let Δt

Infinite Bus GENCLS Implicit Solution

The initial conditions are
Let Δt =

0.02 seconds
During the fault the Jacobian is
Set the initial guess for x(0.02) as x(0), and
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Infinite Bus GENCLS Implicit Solution Then calculate the initial mismatch With

Infinite Bus GENCLS Implicit Solution

Then calculate the initial mismatch
With x(0.02)(0) =

x(0) this becomes
Then
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Infinite Bus GENCLS Implicit Solution Repeating for the next iteration Hence we have converged with

Infinite Bus GENCLS Implicit Solution

Repeating for the next iteration
Hence we have

converged with
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Infinite Bus GENCLS Implicit Solution Iteration continues until t = Tclear,

Infinite Bus GENCLS Implicit Solution

Iteration continues until t = Tclear, assumed

to be 0.1 seconds in this example
At this point, when the fault is self-cleared, the equations change, requiring a re-evaluation of f(x(Tclear))
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Infinite Bus GENCLS Implicit Solution With the change in f(x) the

Infinite Bus GENCLS Implicit Solution

With the change in f(x) the Jacobian

also changes
Iteration for x(0.12) is as before, except using the new function and new Jacobian

This also converges quickly, with one or two iterations

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Computational Considerations As presented for a large system most of the

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+Δt), prediction can be used when previous values are available
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Two Bus Results The below graph shows the generator angle for

Two Bus Results

The below graph shows the generator angle for varying

values of Δt; recall the implicit method is numerically stable
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Adding the Algebraic Constraints Since the classical model can be formulated

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
As before the complex equations will be expressed using two real equations, with voltages and currents expressed in rectangular coordinates
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Adding the Algebraic Constraints The network equations are as before

Adding the Algebraic Constraints

The network equations are as before

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Classical Model Coupling of x and y In the simultaneous implicit

Classical Model Coupling of 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 x as
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Classical Model Coupling of x and y The in the state

Classical Model Coupling of x and y

The in the state equations

the coupling with y is recognized by noting
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Variables and Mismatch Equations In solving the Newton algorithm the variables

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
And the algebraic equations
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Jacobian Matrix Since the h(x,y) and g(x,y) are coupled, the Jacobian

Jacobian Matrix

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

the classical model the coupling is the Norton current at bus i depends on δi (i.e., x) and the electrical power (PEi) in the swing equation depends on VDi and VQi (i.e., y)
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Jacobian Matrix Entries The dependence of the Norton current injections on

Jacobian Matrix Entries

The dependence of the Norton current injections on δ

is
In the Jacobian the sign is flipped because we defined

 

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Jacobian Matrix Entries The dependence of the swing equation on the generator terminal voltage is

Jacobian Matrix Entries

The dependence of the swing equation on the generator

terminal voltage is
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Two Bus, Two Gen GENCLS Example We'll reconsider the two bus,

Two Bus, Two Gen GENCLS Example

We'll reconsider the two bus, two

generator case from Lecture 18; fault at Bus 1, cleared after 0.06 seconds
Initial conditions and Ybus are as covered in Lecture 18

PowerWorld Case B2_CLS_2Gen

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Two Bus, Two Gen GENCLS Example Initial terminal voltages are

Two Bus, Two Gen GENCLS Example

Initial terminal voltages are

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Two Bus, Two Gen Initial Jacobian

Two Bus, Two Gen Initial Jacobian

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Results Comparison The below graph compares the angle for the generator

Results Comparison

The below graph compares the angle for the generator at

bus 1 using Δt=0.02 between RK2 and the Implicit Trapezoidal; also Implicit with Δt=0.06
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Four Bus Comparison

Four Bus Comparison