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

1

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 1:Lecture 1:

Introduction; Course Overview; Introduction; Course Overview;

Modeling Physical Systems, Modeling Physical Systems,

LumpedLumped--Parameter Equilibrium SystemsParameter Equilibrium Systems

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

Lecture 1: Course Introduction; Lumped-Parameter Equilibrium Systems

IntroductionIntroduction

• Engineering systems often comprise complicated assemblies of

components, which can have complex behaviours that are difficult to predict

Internet Sources: www.coolestgadgets.com; www.nasa.gov; www.microway.com.au; www.pbs.org; www.emercedesbenz.com; www.syncrude.com

2





Mathematical Mathematical Analysis Analysis in Engineeringin Engineering

• Engineering analysis: formulating governing equations that

describe the behaviour of physical and technological

systems, for the purpose of analysis and design

• Numerical analysis: solving mathematical equations using

algorithms

• Scientific computing: development of reliable numerical

models that can be tested in a range of cases (including

known benchmarks)





What is Modeling?What is Modeling?

• 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

• Applied numerical modeling has joined empirical experimentation and analytical methods for solving problems of mathematical physics

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

• Models of physical systems– Mechanical, electrical, thermal, structural, hydraulic, etc.

– Combinations of different physical systems (mixed systems)

• Models of material, energy, and information flow for engineering decisions– Production systems

– Economics

– Scheduling

– Inventory, and so on, and so on,…

3





What is Simulation?What is Simulation?

• Simulations are solutions of equations that are functions of time

• For continuous systems, we develop (and solve) differential equations

• Examples:

– Vehicle dynamics

– Thermofluid interactions

– Industrial processes

– Biological processes

– Climate change, and so on, and so on,…

• Often the equations can not be solved in closed-form

• Sometimes simulations are based on empirical understanding of time-varying behaviour that is not expressed as differential equations (correlations, discrete events, etc.). These are valuable for systems that are not characterised well by differential equations.





Scientific Computing at a GlanceScientific Computing at a Glance

(Adapted from A. Quarteroni, “Mathematical Models in Science and Engineering,

Notices of the American Mathematical Society Jan 2009)

Interesting problem

Data from the problem

Understanding of the problem

•Defining the system

•Uncertainty

•Sensitivity

•Parameter identification

•Statistical analysis

Modeling the system

•Geometry and mesh/network

•Governing equations & analysis

•Numerical approximation

•Algorithms for solving

Computer simulation &

post-processing

•Visualisation of results

Validation /

Verification •Comparison to known results

•Benchmark cases

•Experiments

4





Decision

Engineering Analysis for Design at a GlanceEngineering Analysis for Design at a Glance

Possible

solution

Design

Performance

specifications

Model of

System behaviour

Problem

definition

Assessment of

Performance

of proposed solution

Modeling to predict

how a design will perform

is key to a successful

solution





ENGM 541 Course IntroductionENGM 541 Course Introduction

• Why do engineers need to learn about modeling and simulation?

• Most engineering problems are too complicated or complex

to solve analytically

• Engineers rely on numerical modeling and simulation to analyse and design systems that have time-varying aspects

• Engineering managers use models of technologies and business processes for decision making

• You may want do develop models to solve a technical or

business problem, by designing a solution and modeling how you expect it to perform

• You may need to interpret the results of models created by

others

5





General Course OutlineGeneral Course Outline

• Understanding concepts of formulating mathematical models

based on physics (and other rules of interaction) between the

elements of a system

• Formulating governing equations and choosing solution

methods for different types of analyses of physical systems

• Understanding advantages and limitations of numerical

solution methods

• Understanding simple models for financial decisions and

technological systems that have event-based dynamics

• Using modeling and simulation for design

• Presenting and interpreting analysis and simulation results

• Analysing engineering systems and processes using general purpose programs: MATLAB® and SIMULINK®





ENGM 541 Course Overview (1)ENGM 541 Course Overview (1)

• Lecture Room:

• Time Slots:

• Instructor:

• Office:

• Office Hours:

• TA:

• Course Text:

• E-Class & Course Web Site:

ETLE 2-001

Lectures: Wednesdays 5:00 pm – 8:00 pm

Laboratories: Thursdays 5:00 pm – 8:00 pm in ETLE 2-005

(required for ENGM 541 only)

MG Lipsett ([email protected])

Room 5-8J, Mechanical Engineering Building

(5th Floor West)

Wednesdays 1:00–3:00 pm (other times by appointment)

Masoud Mashkournia

Modeling and Analysis of Dynamic Systems,

by R. Esfandiari & B. Lu (CRC Press)

http://www.ualberta.ca/~mlipsett/ENGM541/ENGM541.htm- Lecture slides

- Assignments

- FAQ and announcements

- Worked examples and sample test questions

CHECK ECLASS & THE WEB SITE OFTEN !!

6






Marks:Marks:• Assignments: 25%

Will be due in class and cannot be accepted after solutions are posted

• ENGM 541 Labs: 5%

• ENGM 541 Project: 15% (ENGM 670 & MECE 758: 20%) Individual, criteria to be announced, due April 6 2011 (before the exam)

• Midterm Examination: 20%Wednesday March 2, 2011, 5:00 pm – 7:00 pm in ETLE 2-001

• Final Examination: 30%Wednesday April 13, 2011, 5:00 pm – 7:30 pm in ETLE 2-001

• Examinations will be open book & open notes

• Calculators are allowed but communication features must be turned off (no computers)





ENGM 670 & MECE 758 Course OutlineENGM 670 & MECE 758 Course Outline

• Lectures will be the same for ENGM 541, ENGM 758, and ENGM 670

• But there are additional requirements for grad students:

• Supplementary readings

– MECE 758: more on physical systems

– ENGM 670 more on technological systems

• More assignment problems

• Additional scope for the individual project

• Different exam questions

7





ENGM ENGM 670 & MECE 758 670 & MECE 758 Course Overview (2)Course Overview (2)

Marks:Marks:• Assignments: 25%

Will be due in class and cannot be accepted after solutions are posted

• Lab attendance is not required; but you are responsible for being able to do the Matlab coding covered in the labs

• Project: 20% Individual, criteria to be announced, due April 6 2011 (before the exam)

• Midterm Examination: 25%Wednesday March 2, 2011, 5:00 pm – 7:00 pm in ETLE 2-001

• Final Examination: 30%Wednesday April 13, 2011, 5:00 pm – 7:30 pm in ETLE 2-001

• Examinations will be open book & open notes

• Calculators are allowed but communication features must be turned off (no computers)





General General Course Course Success FactorsSuccess Factors

• Keys to success:

– Do the homework to master model building

– Try the examples in MATLAB

– Check E-Class and the web site often

• FAQ, worked examples, sample tests…

– Ask questions! (but think first…)

• This is a demanding course – but you will gain a valuable approach to analysis and design

• We have to “unlearn” some things to do general systems analysis correctly

• We will also learn by doing

• I’ll do my best to be interesting

8






• University policy: suspected cheating or plagiarism will be

reported and investigated

• Professional ethics and integrity

Do the right thing. It will gratify some people

and astonish the rest. -Mark Twain





Your Instructor: MG LipsettYour Instructor: MG Lipsett

• Professional Engineer since 1986

• Research

– Reliability of complex systems (anomalies, machinery diagnostics)

– Robotics and automation (excavation, remote embedded sensing)

– More sustainable processes for oilsands bitumen production and

reclamation

• Industrial Experience

– Atomic Energy of Canada Ltd (R&D in robotic inspection, hazardous

waste site remediation, reliability)

– Syncrude Canada Ltd (mining automation & space robotics

teleoperation, extraction process R&D, mine maintenance & reliability)

– Seven years in leadership and management roles (Operations, R&D,

Projects)

9





Engineering AnalysisEngineering Analysis

• Types of analysis:

• Two means of modeling physical systems:

• Once a model has been developed, then numerical procedures can be used to study system behaviour using computers





Modeling Physical SystemsModeling Physical Systems

• Consider a beam:

• This is an inherently continuous structure. When we

analyse this beam for deflections, natural frequencies, etc., we can start from one of two approaches.

10





LumpedLumped--Parameter ModelParameter Model

• The properties of the continuous system are visualised as being separate from one another

• The beam is modeled as a linkage mechanism

• We find a set of algebraic equations from which we can determine the deflections

• The price we pay is one of approximating the physical system at the modeling level.





Continuous ModelContinuous Model

• Alternatively, the beam is modeled by deriving differential equations that represent the continuous system

• The solution to the differential equations requires that they

be approximated by algebraic equations (e.g. finite difference expressions), for almost all non-trivial cases

11





Solving Algebraic Equations of the ModelSolving Algebraic Equations of the Model

• In either case, we are solving algebraic equations.

• After the modeling is complete, we choose the type of solution:

• We want to have a consistent way to set up problems – and to solve them.





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

• We are looking for steady-state solutions to problems where the continuous system has been modeled using lumped parameters.

• We are concerned with systems of interconnected elementselements. The elements within the problem have properties that we must know before we can proceed.

• Elements are connected at nodes. nodes. Here is an example of a system network:

• Loops are paths that start at a particular node, pass through a number of elements, and return to the original node.

12





Loop and Node VariablesLoop and Node Variables

• A system will have both loop and node variables.

• Loop variables describe the path around the loop.

Examples:

• Node variables describe variables that come together at a

node.

Examples:

• Loop and node variables:

• 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

13





Constitutive Relationship Example #1Constitutive Relationship Example #1

1. State variables:

2. Describe element:

3. Sketch:

4. Write analytical relationship:





Constitutive Relationship Example #2Constitutive Relationship Example #2

1. State variables:

2. Describe element:

3. Sketch:

4. Write analytical relationship:

14





Admissibility LawsAdmissibility Laws

• The node laws satisfy the admissibility requirement that the

node variable is conserved at a node

• The loop laws are similar (but different). Loop variables are

governed by loop admissibility laws that require the value of

the loop variable at a node to have only one value





GeneralisingGeneralising Kirchoff’sKirchoff’s LawLaw

• We use a general approach for system networks using the principles of Kirchoff’s Laws.

• Kirchoff’s Laws for electrical circuits use the physical laws

of conservation of charge (node law) and conservation of energy added or taken by a potential field (around loops, mesh law), including dissipation. Gain or loss around an

entire loop has to be zero (because there is no net change

in the location with respect to the field).

• For other types of physical systems, we construct our variable assignments so that we can exploit similar physical

laws:

– Conservation of momentum law (D’Alembert’s law for forces)

– Conservation of mass law for flows, etc., etc.

• For non-physical systems, we need similar loop & node

laws

15





Example: LumpedExample: Lumped--Parameter Electrical NetworkParameter Electrical Network

C

L

R1

R2





Some Equilibrium Element TypesSome Equilibrium Element Types

Type Node Variable Loop Variable

Mechanical

Electrical

Fluid Flow

Heat Transfer

16





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.





Example: LumpedExample: Lumped--Parameter Mechanical SystemParameter Mechanical System

To model this system, we have two possible approaches:

1) Find the forces in the springs (node variables)

2) Find the displacements of the carts (loop variables)

K/6

K/6

K/3 K/2

2P P

17





Case 1: Find the Forces in the SpringsCase 1: Find the Forces in the Springs

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

2) Satisfy node admissibility.





Case 1: Forces in Springs (2)Case 1: Forces in Springs (2)

3) Choose loop variables (displacements across elements) and ensure they satisfy loop admissibility.

K/6

K/6

K/3 K/2

2P P

18






4) Apply constitutive relationships. For linear spring element,

this will be: fi = ki δinode variable loop variable






5) Substitute into the loop equations.

19






6) Try to express in non-dimensional form.





Case 2: Find the Displacements in the NodesCase 2: Find the Displacements in the Nodes

1) Choose a set of loop variables.

2) Satisfy loop admissibility.

K/6

K/6

K/3 K/2

2P P

20





Case 2: Displacement of Nodes (2)Case 2: Displacement of Nodes (2)

3) Choose node variables (forces at nodes) and ensure they

satisfy node admissibility.

4) Apply constitutive relationships.






5) Substitute into node equations.

21






Are we done yet? Well, not quite.

From the solution for y1, y2, go back to the definition of the non-

dimensional variables to solve for the displacement (the

loop variables); then, from their solution, we can find forces

using the constitutive relationships.

These two methods are called Direct ApproachesDirect Approaches.





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)

22





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.

23





Break Time: Flexibility of Thinking ProblemsBreak Time: Flexibility of Thinking Problems

• 8D – 24H = 1W

• 3P = 6

• HH & MH @ 12 = N or M

• 4J+4Q+4K = All the FC

• S&M&T&W&T&F&S are D of W

• 23Y – 3Y = 2D

• E – 8 = Z

• Y + 2D = T

• C + 6D = NYE

• Y – S – S – A = W

• NN = GN

• N + P + SM = S of C

• 1 + 6Z = 1M

• R = R = R

• 1B in the H = 2 in the B

Each problem is an equation, which can be solved by substituting the appropriate words for the letters. Examples:

3F = 1Y (3 Feet = 1 Yard)4LC = GL (4 Leaf Clover = Good Luck)

Source: A Whack on the Side of the Head, R.von Oech

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

Documents