project 103 - nonlinear equations++

8/12/2019 Project 103 - Nonlinear Equations++

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ENGR 103 Computer-Aided Analysis Tools for Engineers

Department of Electrical EngineeringThe City College of The City University of New

York (CCNY)

Spring 2014Session 5

ProjectNon Linear Equations Numerical Analysis

Dr. Nidal Khrais


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y4 y3 y1

y2

I4

I3 I2

I1 43

2

1y34

y23

y13

y12

Kirchoffs Current Law (KCL) requires that each of the current injections beequal to the sum of the currents flowing out of the bus and into the linesconnecting the bus to other buses, or to the ground. Therefore, recallingOhms Law, I=V/Z=VY, the current injected into bus 1 may be written as: I1=(V1-V2)y12 + (V 1-V3)y13 + V 1y1


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Similarly, we may develop the current injections at buses 2, 3,and 4 as:

I2= V 1(-y21) + V 2( y2 + y 21 + y 23 + y24) + V 3(-y23) + V 4(-y24)

I3= V 1(-y31)+ V 2(-y32) + V 3( y3 + y31 + y 32 + y 34) + V 4(-y34)

I4= V 1(-y41)+ V 2(-y42) + V 3(-y34)+ V 4( y4 + y41 + y42 + y 43)

where we recognize that the admittance of the circuit from bus k to bus i is the same as the admittance from bus i to bus k, i.e., yki=yik


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current injections are linear functions of the nodal voltages.Therefore, we may write these equations in a more compact formusing matrices according to:

4

3

2

1

4342414434241

3434323133231

2423242321221

1413121413121

4

3

2

1

V

V

V V

y y y y y y y

y y y y y y y

y y y y y y y y y y y y y y

I

I

I I

4342414434241

3434323133231

2423242321221

1413121413121

y y y y y y y

y y y y y y y

y y y y y y y

y y y y y y y

Y


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Denoting the element in row i, column j, as Y ij, we rewrite as:

4 44 34 24 1

3 43 33 23 1

2 42 32 22 1

1 41 31 21 1

Y Y Y Y

Y Y Y Y

Y Y Y Y Y Y Y Y

Y

where the terms Y ij are not admittances but ratherelements of the admittance matrix. Therefore, eq. becomes:


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4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

V

V V

V

Y Y Y Y

Y Y Y Y Y Y Y Y

Y Y Y Y

I

I I

I

4

3

2

1

4

3

2

1

,

I

I

I I

I

V

V

V V

V

Defining the vectors V and I, we may write in compact form:

V Y I


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several observations about the admittance matrix1) The matrix is symmetric, i.e., Y ij=Y ji.

2) A diagonal element Y ii is obtained as the sum of admittancesfor all branches connected to bus i,i.e., N

ik k ik iii y yY

,1

3) The off-diagonal elements are the negative of the admittances

connecting buses i and j, i.e., Y ij=-y ji.


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iQ P VI S *2424323222121 I V Y V Y V Y V Y


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Relationship between real Power and Reactive powersupplied to the system at bus i and the current

injected into the system at that bus:

where Vi is the voltage at the bus; Ii* - complex conjugateof the current injected at the bus; Pi and Qi are real andreactive powers. Therefore,

iiii I V jQ P *


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Given that:

j

n

j iji

jijiii

bus

V Y I

V Y V Y V Y I

I V Y

1

2211 ......

Therefore:

iiii I V jQ P

slide previous from*

: _ _


12/38Non Linear Equations- Numerical Analysis


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There are 4 variables that are associated with each bus:o P

o Qo Vo Meanwhile, there are two power flow equations

associated with each bus. In a power flow study, two of the four variables aredefined an the other two are unknown.That way, we have the same number of equations asthe number of unknown. The known and unknown variables depend on thetype of bus.


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Bus Types1)Load Bus PQ Bus:a-Active Power P is knownb-Reactive Power Q is knownc-voltage magnitude |v| is unknownD- voltage angle is unknown

2) Generator Bus (Voltage Bus) PV Bus: a-P is knownb-|V| is knownc-Voltage angle is unknownd-Q is unknown.

3) Slack Bus: choose a reference bus (must beA generator bus) . Assume :|V| = 1.0

= 0

Mechanical Pow erP: ElectricalPower

|V|Generator ExcitationField Current

Electric Generator/Generator Bus

Q & are unknown


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Note that the power flow equations are non-linear, thuscannot be solved analytically. A numerical iterativealgorithm is required to solve such equations. A standard

procedure follows:1.Create a bus admittance matrix Ybus for the powersystem;

2.Make an initial estimate for the voltages (bothmagnitude and phase angle) at each bus in the system;3.Substitute in the power flow equations and determinethe deviations from the solution.

4.Update the estimated voltages based on somecommonly known numerical algorithms (e.g., Gauss-Seidel, Newton-Raphson).5.Repeat the above process until the deviations from thesolution are minimal.


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The power flow Analysis equations

I V Y bus

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

V

V

V

V

Y Y Y Y

Y Y Y Y

Y Y Y Y

Y Y Y Y

I

I

I

I

2424323222121 I V Y V Y V Y V Y


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The current injected at bus 2 can be found as:

*

2

222

2

22*

2

22

*

22

V jQ P

I

V jQ P

I

jQ P I V


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)]([1

)]([1

)]([1

: _ _

)]([1

: _ _

)1(343

)1(242

)0(141)0*(

4

44

44

)1(4

)0(434

)1(232

)0(131)0*(

3

33

33

)1(3

)0(424

)0(323

)0(121)0*(

2

22

22

)1(2

424323121*2

22

222

2

*2

224243232221212

V Y V Y V Y V

jQ P Y

V

V Y V Y V Y V jQ P

Y V

V Y V Y V Y V

jQ P Y

V iterationoneafter

V Y V Y V Y V

jQ P Y

V

V for solving

V jQ P

V Y V Y V Y V Y I

(Assume V1 is a slack bus:|V1| = 1.0 ; = 0 )


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Gauss-Seidel MethodThis method is also known as the method of successive

displacements. Consider the nonlinear equation f (x) = 0 . Re-write the equation in different form: x = g( x) . We assume x(0) is an initial "guess" of the solution, then "refine" the solution using:


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If this process is convergent, then the successivesolutions approach a value which is declared asthe solution. Thus if at some step k +1 we have:

where is the desired "accuracy",


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Example: Use the Gauss-Seidel method to obtainthe roots of the equation:

1) First using analytical solution:Roots:4.00001.0000 + 0.0000i1.0000 - 0.0000i

Note the doubleroot at x = 1


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2) Using Gauss-Seidel Technique: First the equation isexpressed in a different form thus:


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Next the process is generalized for n equations in nvariables:

Solving for one variable from each equation, the systemof equations car be represented as follows:


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First a starting solution is assumed to be: Now there are two ways to proceed:


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Acceleration factor:An acceleration factor can be used as before, thus at

each step upgrade the estimate asFollows:

Clearly if = 1 then the iteration is the same asbefore. Usually but not muchlarger than unity.

1


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Newton-Raphson MethodConsider the Taylor-series expansion of the function f(x)

about a value x = xo:

Using only the first two terms of the expansion, a firstapproximation to the root of the equation f(x) = 0

can be obtained from:

...))(!2/)(''())((')()( 200000 x x x f x x x f x f x f


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Such approximation is given by:

The Newton-Raphson method consists in obtainingimproved values of the approximate root through the

recurrent application of equation above. For example, thesecond and third approximations to that root will be given by:


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This iterative procedure can be generalized by writingthe following equation, where i represents the iteration

number:

After each iteration the program should check to see ifthe convergence condition, below, is satisfied:

|)()1(| ii x f x f


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The figure below illustrates the way in which the solution is found by using the Newton-Raphson method.At x = x0, the equation f(x) = 0 f(xo)+f'(xo)(x1 -xo) represents a straight linetangent to the curve y = f(x). The tangent line intersects the x-axis

(i.e., y =f(x) = 0) at the point x1 as given by x1 = xo - f(xo)/f'(xo).

At x = x1, the equation f(x) = 0 f(x1)+f'(x1)(x2 x1) represents a straight linetangent to the curve y = f(x). The tangent line intersects the x-axis(i.e., y =f(x) = 0) at the point x2 as given by x2 = x1 - f(x1)/f'(x1).


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The Newton-Raphson method converges relatively fast formost functions regardless of the initial value chosen. The

main disadvantage is that you need to know not only thefunction f(x), but also its derivative, f'(x), in order toachieve a solution.

Example: find the solution for f(x) = 0 using 3 as the initialestimate where f(x) is shown below:


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Note: Analytical solution gives:

16227766.310 x

100)( 2 x x f


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The Project:1) Given the following simple power system that has 4busses, 5 transmission lines, 1 generator, and 3 loads.

iQ P VI S

I V Y bu s*


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2) given that the corresponding is shown below:busY

puQ

pu P

puQ

pu P puQ

pu P

15.0

2.0

05.0

1.02.0

3.0

4

4

3

3

2

2

3) Given the following power values in per unit (pu) at the load buses 2, 3, 4 are:


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Find the node (bus) voltages using:1) Gauss iteration

2) Gauss-Seidel iteration3) Gauss-Seidel using an acceleration factor = 1.64) Using Newton-Raphson Method5) For each method, draw the flowchart

6) For each method, write the pseudocode


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Example:


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project 103 - nonlinear equations++

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