engm engm 541, engm 670 541, engm 670-xx55 && mece 758mece...

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ENGM ENGM 541, ENGM 670541, ENGM 670--X5X5

& & MECE 758MECE 758--X5X5Modeling and Simulation of Engineering SystemsModeling and Simulation of Engineering Systems

Winter Winter 20112011

Lecture 2:Lecture 2:

ExtremumExtremum Approach; Approach;

General Formulation of General Formulation of

LumpedLumped--Parameter Equilibrium Systems;Parameter Equilibrium Systems;

Solution MethodsSolution Methods

M.G. LipsettM.G. Lipsett

Department of Mechanical EngineeringDepartment of Mechanical Engineering

University of AlbertaUniversity of Albertahttp://www.ualberta.ca/~mlipsett/ENGM541/ENGM541.htm

© MG Lipsett, 2011 2

Department of Mechanical EngineeringEngineering Management Group

ENGM 541, 670-X5, MECE 758 – Modeling and Simulation of Engineering Systems

Lecture 2: Extremum Approach; General Formulation for Lumped-Parameter Equilibrium

Review of Modeling & SimulationReview of Modeling & Simulation

• A model is a representation of knowledge

– Rules, physical analogs, algebraic equations of physical laws

• A system is a bounded region comprising known elements

that each interact in understandable ways

• The types of systems of interest in this course include:

– Models of physical systems

• Mechanical, electrical, thermal, structural, hydraulic, etc.

– Models of material, energy, and information flow for engineering

decisions

• Production systems, Economics, Scheduling, Inventory, …

• Simulations are solutions of differential equations that are

functions of time

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Equilibrium Problems for LumpedEquilibrium Problems for Lumped--Parameter SystemsParameter Systems

• Steady-state solutions (no change over time)

• In a lumped-parameter problem, the continuous system has been modeled using systems of interconnected elementselementsconnected at nodesnodes

•• LoopLoop variables describe the path around the loop of some

or all of the elements

Node variables describe variables that go through an

element and which come together at a node.

The loop and node variables are related by the constitutive

relationships of the elements.





Formulating Constitutive RelationshipsFormulating Constitutive Relationships

1. State the variables

2. Describe the element

3. Sketch the constitutive relationship.

4. Use an analytic expression for the relationship

Loop variable

Node variable

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General Procedure for Setting Up A ProblemGeneral Procedure for Setting Up A Problem

1. Choose the variable in which you want your final equations expressed

2. Choose variables so as to satisfy the pertinent admissibility requirement

3. Choose other variable type & write as many equations as necessary to check that admissibility is satisfied.

4. Relate the loop and node variables using the constitutive relationships.

5. Eliminate all but the chosen variables (all of one type) from the equations). Substitute in the equations and group terms.

6. Non-dimensionalise the variables.





Extremum FunctionsExtremum Functions

• The other way of formulating the equations governing systems is to use extremum functions. This includes energy methods.

• We make up a scalar function from the constitutive relationships of all the elements in the system, and search for an extreme value of the function (e.g. minimum

potential energy).

• We go back to our original definition of a constitutive

relationship to define two quantities:

1. Content U (energy)

2. Co-Content U* (co-energy)

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EnergyEnergy

• Area under the curve is the energy U in the element:

• We write p (which is a node variable) as a function of q(loop variable) and U becomes a function of q only.





CoCo--EnergyEnergy

• Similarly to energy, with co-energy U* as a function of p only

• For all sets of state variables satisfying node (loop) admissibility, those also satisfying loop (node) admissibility will render the co-energy (energy) an extreme value.

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Example: Extremum ApproachExample: Extremum Approach

Recall our system of carts and springs.

There are two types of elements in this system:

1) Spring

2) Force source (which yields a “through” variable)

K/6

K/6

K/3 K/2

2P P





Example (2): Spring Constitutive RelationshipExample (2): Spring Constitutive Relationship

• Consider the spring constitutive relationship:

• Force f through distance δ does positive work on the spring,

δ

loop variable

node variable

f

f = k δ w.r.t loop variable; or

δ = (1/k) f w.r.t node variable

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Example (3): Energy & CoExample (3): Energy & Co--Energy for SpringEnergy for Spring

• Energy:

• Co-Energy:





Example (4): Force SourceExample (4): Force Source

• A specified force regardless of the deflection, visualised as:

• Like a potential energy term for a hanging mass

• Once again we must be careful about signs

• Question: What’s U* for a force source?

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Example (5):Displacement SourceExample (5):Displacement Source

• In other problems, we might have a displacement source,

which is a prescribed displacement regardless of the force

applied – physical constraint such as a stop, a track, or a

wall. In other types of systems this would be analogous to a

voltage source, a temperature source, or a pressure source.

• The constitutive relationship for a displacement source is

visualised as follows:





Example (6): Example (6): Spring Force Equations & CoSpring Force Equations & Co--EnergyEnergy

1) Choose an admissible set of node variables (forces at nodes).

To be admissible, the variables need to satisfy the node

admissibility requirements:

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Example (7): CoExample (7): Co--Energy ExpressionEnergy Expression

• Now we write an expression for the co-energy of the system,

with one term for each element in the system:

• For solution, U* will have an extreme value:

• We first differentiate U* by f1 :





Example (7): NonExample (7): Non--DimensionaliseDimensionalise

• To non-dimensionalise:

• And get

• Next, we differentiate U* by f2 to get

• And put the two expressions into matrix form as

• Recall from the direct formulation that the solution is

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Example (8): Symmetric Matrix FormExample (8): Symmetric Matrix Form

• Note that the coefficient matrix resultant from an extremum

function approach is symmetricsymmetric.

• This is always true.

• A symmetric coefficient matrix is a good check for the

solution (but it doesn’t guarantee the right answer!)





Example #2: Fluid Flowing in PipesExample #2: Fluid Flowing in Pipes

Given: P0, P4

Elements are pipe sections

and pressure sources

Consider the constitutive relationship for a pipe section:

P0

P4P4

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Example #2 (2): Pressure SourceExample #2 (2): Pressure Source

∆ P = Pi – Pj = constant (regardless of flow q)

Co-energy = q ( Pi – Pj )

Now, construct in terms of flows in the pipe segments

(elements) a node variable formulation using co-energy:

Pi Pj





Example #2 (3): Define Admissible Node Variables Example #2 (3): Define Admissible Node Variables

• Node 1:

• Node 2:

• Node 3:

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Example #2 (4): Write Total CoExample #2 (4): Write Total Co--Energy of SystemEnergy of System

To eliminate variables, express them in terms of other variables

Write the total co-energy of the system:

Substitute for q1 , q5 , & q6 (from node admissibility):





Example #2 (5): Find Extreme ValueExample #2 (5): Find Extreme Value

For solution, U* has extreme value, so we generate a set of three equations (one for each admissible variable):

And solve to get

where

=

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Formulating the General Equilibrium ProblemFormulating the General Equilibrium Problem

• Any formulation we develop will start with a set of variables

xi , i = 1, …, N ; they represent the state of the system. We

determine N functions of xi , each equal to some forcing

value Ci.

• If any of the xi appears to a power other than 1, or is combined in a multiplicative sense with any other xi , then the entire system is nonlinear.

• Very often, however, we find that this is not the case and that the equations are linear.





Review of Form of Linear EquationsReview of Form of Linear Equations

Linear equations can be written as:

which we can write in matrix form as

known, constant coefficient matrix knowns

We typically write this as

The solution will often be written as:

=

=

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Solutions to Linear EquationsSolutions to Linear Equations

• This notation is usually restricted to analytical work.

Numerical solutions may or may not invert the matrix,

depending on the efficiency.

• An efficient algorithm will minimise the number of

operations &/or the storage requirements.

• We will consider different methods for solving [A]{x} = {c}

Cramer’s Rule: Use Determinants

• If we want to find xi , we find the determinant of the coefficient matrix, then we replace the i th column of [A] with the vector {c}, find another determinant, and divide it by the first determinant.





Example: Determinants to Solve for {Example: Determinants to Solve for {xx}}

Say we have the system

and we want to find x2.

=

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Computational RequirementsComputational Requirements

The computational effort (number of

operations, counting multiplications &

divisions, not additions & subtractions)

involved in taking the determinant

grows as N! (N-factorial), which is the

fastest growing function known.

N N !

1 1

2 2

3 6

4 24

5 120

… …

10 3,628,800

… …

20 2.43 x 1018





Gaussian Elimination Method for Solving {Gaussian Elimination Method for Solving {xx}}

The most familiar method, comprising two parts:

An elimination phase to transform [A]{x} = {c} into [U]{x}={d}

where

followed by a solution phase to find xi , i = 1, …, N

by back substitution

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Gaussian Elimination Method (2)Gaussian Elimination Method (2)

• In the first phase, each row (equation) is first divided by its first element, and then the first equation is subtracted from

all the other equations, leaving us with

• Zeroing the first column requires ~ N2 operations. We

eventually want to get all zeros below the diagonal.

• We repeat this process again using the remaining elements in the lower right (N-1)x(N-1) submatrix of [A] to give

=

=





Gaussian Elimination Method (3)Gaussian Elimination Method (3)

After decomposing N rows we get the upper diagonal matrix [U]. From

• We find xN , then back substitute to get xN-1 , & so on, & so on.

• The overall number of operations varies on the order of N3

N N3

10 1000

… …

100 106

Size of system Number of operations

(which is still significantly smaller than for Cramer’s Rule when N <6)

[ ]

=

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Systems of Equations with Banded FormSystems of Equations with Banded Form

• We often find that our equations have a banded form, which

means that all of the non-zero coefficients are near the

diagonal, and all the terms far away from the diagonal are

zero.

• If we do Gaussian elimination on this banded matrix we only need M2 operations; total number of operations ≈ M2N .





LimitedLimited--Bandwidth Equilibrium ProblemsBandwidth Equilibrium Problems

• Coefficient matrices with limited bandwidth are very common in engineering systems, and typically arise when a node is only acted upon by its nearest neighbouring nodes

through the neighbouring elements.

• We can defeat the advantages of bandwidth if we choose an inappropriate numbering system for the variables of the

problem.

• Example: We want to find the displacements of all the carts.

• Choose unique (admissible) loop variables u1, u2 , u3 , u4 , u5

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LimitedLimited--Bandwidth Equilibrium Problems (2)Bandwidth Equilibrium Problems (2)

Node 1:

Node 2:

Etc. …, yielding

=





LimitedLimited--Bandwidth Equilibrium Problems (3)Bandwidth Equilibrium Problems (3)

• If, on the other hand, this were the case:

• Then the non-zero elements of the coefficient matrix would look like this:

• Which is a full matrix, and so we can’t use the more efficient limited bandwidth solution. (If we had numbered u2 as u5, or u5 as u2, then we would have ended up with a full matrix as well. Try to number sequentially.)

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Another Example of Node NumberingAnother Example of Node Numbering

• Steady-state heat conduction into a hollow cylinder, considered as a two-dimensional problem

• We assume a set of nodes, each with a unique temperature. This satisfies loop admissibility automatically.

• The number of unknowns equals the number of nodes.

• Notice there is some symmetry. The system is symmetric about the vertical axis through its centre, so we only have to analyse half the cylinder.

Equilibrium problem: what is the

temperature distribution in the wall?





Node Numbering Example (2)Node Numbering Example (2)

• Decide on a temperature profile of interest, say radial lines 10° apart (19 lines: 0° to 180°) with 10 nodes within the wall on each radial line

• Total of 190 nodes & 190 unknown temperatures

• The problem could thus be formulated as

• If we use a random numbering for the nodes, then we could fill the matrix. In that case, the effort to solve using Gaussian elimination would be ≈ (190)3 ≈ 8 x 106 operations

=

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Node Numbering Example (3)Node Numbering Example (3)

• Consider a couple of radial lines and choose an efficient

numbering scheme to reduce the bandwidth of the problem

BC

D E

A F

0°

10°20°

30°

G

We have a couple of options:

•Proceed from A along circumference along half-circle;

•Return along next higher radius set of nodes;

•Include next radius node out for connection point.

Bandwidth is 19 + 19 + 1 = 39

Or:

•Proceed along radial line from A to D;

•Return along neighbouring radius to F;

•Index to include G.

Bandwidth is 10 + 10 + 1 = 21. Effort in solving is

M2N ≈ (21)2(190) ≈ 8 x 104, which is 100 x more efficient than

solving the full matrix





Matrix InversionMatrix Inversion

• Solving the system of linear equations [A]{x} = {c} by inverting the matrix yields

• which involves a matrix inversion and a matrix multiplication. We can think of matrix inversion as a set of Nequilibrium problems, in which each column of [A]-1 is a

vector of unknowns & each column of [I] is a different right-hand side:

• So there are N such problems. Matrix inversion is only efficient if solving significantly more than N problems with

the same coefficient matrix.

[ ] { } { }=

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Matrix InversionMatrix Inversion

• Solving the system of linear equations [A]{x} = {c} by

inverting the matrix yields

• which involves a matrix inversion and a matrix

multiplication. We can think of matrix inversion as a set of N

equilibrium problems, in which each column of [A]-1 is a

vector of unknowns & each column of [I] is a different right-

hand side: Expanding, this becomes

• So there are N such problems. Matrix inversion is only efficient if solving significantly more than N problems with the same coefficient matrix.

[ ]

=

[ ] { } { }=

[ ]

=

[ ] { } { }=

[ ] { } { }=





Iterative TechniquesIterative Techniques

• Iterative solution methods are indirect, theoretically taking an infinite number of steps to find a solution from an initial guess.

• Practically, they will be “close enough” after a finite number of steps; but we can’t predict exactly the number of steps required to get close enough.

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Iteration by Total StepsIteration by Total Steps

• Consider the following expression:

(which has N equations of this type)

• We start by picking an initial set of guesses for the

unknowns

• When we substitute these guesses into the equations, typically they won’t provide a satisfactory solution

• So we need to refine the set of guesses





Iteration by Total Steps (2)Iteration by Total Steps (2)

• We can refine the set of guesses to find a new (and hopefully improved) set

• using the equations themselves, according to

Example: Given the set of equations in three unknowns:

Solve for x1, x2, and x3

=

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Method of Iteration by Total StepsMethod of Iteration by Total Steps

• Assume an initial set of guesses (and then at each

successive step use the most recent guesses):

• Having calculated for x1(1), x2

(1), and x3(1), we check the

equations.

• If they are not satisfied, we repeat, using the most recent guesses x1

(1), x2(1), and x3

(1) to generate another set of guesses x1

(2), x2(2), and x3

(2)





Notes on Total Steps Calculation MethodNotes on Total Steps Calculation Method• The calculation method takes all the terms but the diagonal on to the

RHS, then divides by the diagonal coefficient to get an expression for

the variable associated with this row.

•• ConvergenceConvergence is met when the values of the set of guesses changes

very little after a step. We can check in advance whether we can expect

convergence using a row-sum criterion:

• Add up the absolute values of the coefficients in a row of [A]

• Divide the sum by the absolute value of the diagonal element

• Generate the terms

• Do a test to see whether all

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SummarySummary

• We have developed methods for generating governing

equations for equilibrium systems using extrema.

• We have seen a few of the many methods for the solving

the equations for linear systems.

Next lecture:

• We will look at other iterative solution methods for (big)

linear systems, and then consider how to deal with

nonlinearities.

• We will also look at what kinds of equilibrium systems arise in engineering.

• Then we will move to modeling time-varying systems.





Break Time: Optical IllusionsBreak Time: Optical Illusions

Source: http://www.mcescher.com

M.C. Escher breaks the loop admissibility law.

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