2 141 Fall 2002 Modeling and Simulation of Dynamic Systems Volume III

8/13/2019 2 141 Fall 2002 Modeling and Simulation of Dynamic Systems Volume III

1/289

2.141 Fall 2002Modeling and Simulation of DynamicSystems

Volume III

Neville Hogan


2/289


3/289

2.141 Fall 2002Modeling and Simulation of DynamicSystems

Neville Hogan

MIT OpenCourseWareCambridge, MAhttp://ocw.mit.edu/


4/289

2006 Massachusetts Institute of Technology

The material in this book is provided under a Creative Commons license that grantsyou certain privileges to use, copy, or adapt the contents for non-commercialeducational purposes as long as you give credit to MIT OpenCourseWare and to thefaculty author, and you make any derivative works freely and openly available toothers under the same terms as our license. Please refer to the full text of thelicense and other notices at the back of this book.

Printed and bound in the United States of America.

MIT OpenCourseWareBuilding 9-213

77 Massachusetts AvenueCambridge, MA 02139-4307USA

ISBN

2006XXXXXX

10 9 8 7 6 5 4 3 2 1


5/289

What is MIT OpenCourseWare?

MIT OpenCourseWare (OCW) is a remarkable story of an institution rallying aroundan ideal, and then delivering on the promise of that ideal. It is an ideal that flowsfrom the MIT Faculty's passionate belief in the MIT mission, based on the conviction

that the open dissemination of knowledge and information can open new doors tothe powerful benefits of education for humanity around the world.

Available online at http://ocw.mit.edu, MIT OCW makes the MIT Faculty's coursematerials used in the teaching of almost all of MIT's undergraduate and graduatesubjects available on the Web, free of charge, to any user anywhere in the world.MIT OCW is a large-scale, Web-based publication of educational materials.

With 1300 courses now available, MIT OCW delivers on the promise of open sharingof knowledge.

Educators around the globe are encouraged to utilize the materials forcurriculum development;

Self-learners and students may draw upon the materials for self-study or

supplementary use.

Course materials contained on the MIT OCW Web site may be used, copied,distributed, translated, and modified by anyone, anywhere in the world. All that isrequired of adopters of the materials is that the use be non-commercial, that theoriginal MIT faculty authors receive attribution if the materials are republished orreposted online, and that adapters openly share the materials in the same manner asOCW.

MIT OCW differs from other Web-based education offerings: It is free and open; It offers a unique depth and breadth of content; and,

It takes an institutional approach to online course publication.

MIT OCW is not a distance-learning initiative. Distance learning involves the activeexchange of information between faculty and students, with the goal of obtainingsome form of a credential. Increasingly, distance learning is also limited to thosewilling and able to pay for materials or course delivery. MIT OCW is not meant toreplace degree-granting higher education or for-credit courses. Rather, the goal is toprovide the content that supports an education, for use by educators, students, andself-learners to supplement their teaching and learning activities.

Truly a global initiative, the MIT OCW site has received users from more than 215countries, territories, and city-states since the launch of the MIT OCW pilot site onSeptember 30, 2002. Materials have already been translated into at least 10

different languages.

MIT is committed to this project remaining free and openly available. MIT OCW is not

a degree-granting initiative, and there will not be a registration process required for

users to view course materials now, or in the future.


6/289

Contents

Highlights of this Course ...................................................................... 1

Course Description.............................................................................. 1

Syllabus ............................................................................................ 2

Calendar............................................................................................ 4

Lecture Notes..................................................................................... 8

Assignments .................................................................................... 11

Assignment 3: Gas-charged Accumulator, Solution andCommentary

Assignment 4: Control through Singularities, Solution andCommentaryAssignment 5: Simple Convection and Throttling, Solution andCommentary

Projects ........................................................................................... 75

Suggested Term Project Topics

Sample Student Projects

Study Materials................................................................................135

Introductory Modeling NotesOld Quiz for Study


7/289

Highlights of this Course

This course features an extensive collection of lecture notesplus background study materials on modeling.

Course Description

This course deals with modeling multi-domain engineering systems ata level of detail suitable for design and control system implementation.Topics covered include network representation, state-space models;multi-port energy storage and dissipation, Legendre transforms,

nonlinear mechanics, transformation theory, Lagrangian andHamiltonian forms and control-relevant properties. Applicationexamples may include electro-mechanical transducers, mechanisms,electronics, fluid and thermal systems, compressible flow, chemicalprocesses, diffusion, and wave transmission.

Course Meeting TimesLectures:

Two sessions / week1.5 hours / session

LevelGraduate

This bond graph models the free-flight and contact behaviors of a ball bouncing off ofa another ball. (Image courtesy of Prof. Neville Hogan.)

1


8/289

Syllabus

Prerequisites

2.151 or equivalent exposure to physical system modeling -- see meto clarify.

Textbook

Required

Brown, Forbes T. Engineering System Dynamics.New York: MarcelDekker, Inc., 2001. ISBN: 0824706161.

Course Description

This course is aboutModeling multi-domain engineering systems ata level of detail suitable for design and control system implementation.It also describesNetwork representation, state-space models,Multiport energy storage and dissipation, Legendre transforms,Nonlinear mechanics, transformation theory, Lagrangian and

Hamiltonian forms, Control-relevant properties. Theapplicationexamples may include electro-mechanical transducers, mechanisms,electronics, fluid and thermal systems, compressible flow, chemicalprocesses, diffusion, and wave transmission.

2


9/289

Grading Policy

Homework

About 6 homework problems will be assigned throughout the term at

approximately two-week intervals.

Term project

Students will be required to select a term project by lecture 8. A briefinterim progress report will be required by lecture 16. Term projectsmust be completed by the last week of the term, at which time a finalreport will be due. A brief oral presentation of each term project willalso be required.

ACTIVITIES PERCENTAGES

Homework 60%

Term Project 40%

Ethics Policy

Collaboration

Collaboration on and discussion of homework assignments isencouraged but each student must submit an individual solution.Collaboration on term projects is encouraged provided some means forclearly identifying individual contributions is proposed and approved bythe instructor.

Use of Material from Previous Years

Use of material from prior offerings of this subject in preparinghomework assignments defeats the purpose of the assignments and isforbidden.

3


10/289

Calendar

LEC # TOPICS KEY DATES

Week 1

1

Introduction

Multi-domain Modeling

Syllabus, Policies and Expectations

Assignment 1out

Week 2

2

Review: Network Models of Physical System

Dynamics

Bond-graph Notation

3

Review (cont.): Equivalent Behavior inDifferent Domains

Block Diagrams, Bond Graphs, Causality

Week 3

4Thvenin and Norton Equivalent Networks

Impedance Control and Applications

5

Energy-storing Coupling between Domains

Multi-port Capacitor

Maxwell's Reciprocity

Assignment 1due

Assignment 2out

Week 4

6

Energy, Co-energy, LegendreTransformation, Causal Assignment

Intrinsic stability

4


11/289

7Review Revisited: Magnetism and Electro-magnetism

Week 5

8Electro-magnetic-mechanical Transduction

Use of Co-energy Functions

Term projectproposal due

9 Linearized Energy-storing Transducer Models

Assignment 2due

Assignmnet 3out

Week 6

10

Cycle Processes

Work-to-heat Transduction

Thermodynamics of Simple Substances

11

Causal Assignments and Co-energyFunctions

Second Law for Heat Transfer

Multi-port Resistors

Week 7

12Nonlinear Mechanical Systems

Modulated Transformers and Gyrators

Assignment 3due

Week 8

13

Lagrangian Mechanics

Coordinates, State Variables andIndependent Energy Storage Variables

14 Nonlinear Mechanical Transformations and Assignment 4

5


12/289

Impedance Control out

Week 9

15

Hamiltonian Mechanics

Stable Interaction Control

Canonical Transformation Theory

16 Term Project Progress Report DiscussionTerm projectprogressreport due

Week 10

17

Identification of Physical and BehavioralParameters

Model structure

18

(Internally) Modulated Sources

Non-equilibrium Multi-port Resistors

Nodicity

Assignment 4due

Assignment 5out

Week 11

19Amplifying Processes

Small-signal and Large-signal Models

20

Thermodynamics of Open Systems

Convection and Matter Transport

Lagrangian vs. Eulerian Frames

Week 12

21Power Conjugates for Matter Transport

Second Law for Non-heat-transfer Processes

6


13/289

Throttling and Mixing

22

Bernoulli's Incompressible Equation

The "Bernoulli Resistor" and "Pseudo-bond-graphs"

Assignment 5due

Week 13

23Chemical Reaction and Diffusion Systems

Gibbs-Duhem equation

Week 14

24

Control-relevant Properties of PhysicalSystem Models

Causal Analysis, Relative Degree,Passivity and Interaction Stability

25Transmission Line Models

Term Project Presentations

Week 15

26Wrap-up Discussion

Term Project Presentations (cont.)

7


14/289

Lecture Notes and Topics

Lecture # Topic

Lecture 1 Introduction; Multi-domainModeling

Lecture 2 Review: Network Models ofPhysical System Dynamics; BondGraph Notation, Block Diagrams,Causality

Lecture 3 Review (cont.): EquivalentBehavior in Different Domains

Lecture 4 Thevenin and Nortan EquivalentNetworks; Impedance Control andApplications

Lecture 5 Energy-storing Coupling betweenDomains; Multi-port Capacitor;Maxwells Reciprocity

Lecture 6 Energy, Co-energy, LegendreTransformation, Causal

Assignment; Intrinsic stability

Lecture 7 Review Revisited: Magnetism andElectro-magnetism

Lecture 8 Electro-magnetic-mechanicalTransduction; Use of Co-energyFunctions

Lecture 9 Linearized Energy-storingTransducer Models

Lecture 10 Cycle Processes; Work-to-heatTransduction; Thermodynamics ofSimple Substances

8


15/289

Lecture 11 Causal Assignments and Co-energyFunctions; Second Law for HeatTransfer; Multi-port Resistors

Lecture 12 Nonlinear Mechanical Systems;

Modulated Transformers andGyrators

Lecture 13 Lagrangian Mechanics;Coordinates, State Variables andIndependent Energy StorageVariables

Lecture 14 Nonlinear MechanicalTransformations and ImpedanceControl

Lecture 15 Hamiltonian Mechanics; StableInteraction Control; CanonicalTransformation Theory

Lecture 16 Term Project Progress ReportDiscussion

Lecture 17 Identification of Physical andBehavioral Parameters; Model

Structure

Lecture 18 (Internally) Modulated Sources;Non-equilibrium Multi-portResistors; Nodicity

Lecture 19 Amplifying Processes; Small-signaland Large-signal Models

Lecture 20 Thermodynamics of Open Systems;Convection and Matter Transport;Lagrangian vs. Eulerian Frames

Lecture 21 Power Conjugates for MatterTransport; Second Law for Non-heat-transfer Processes; Throttlingand Mixing

9


16/289

Lecture 22 Bernoullis IncompressibleEquation; The Bernoulli Resistorand Pseudo-bond-graphs

Lecture 23 Chemical Reaction and Diffusion

Systems; Gibbs-Duhem equation

Lecture 24 Control-relevant Properties ofPhysical System Models; CausalAnalysis, Relative Degree, Passivityand Interaction Stability

Lecture 25 Transmission Line Models; TermProject Presentations

Lecture 26 Wrap-up Discussion; Term ProjectDiscussions (cont.)

10


17/289

Massachusetts Institute of TechnologyDepartment of Mechanical Engineering

2.141 Modeling and Simulation of Dynamic Systems

Assignment #3 Out: 10/3/02Due: 10/17/02

Gas-charged accumulator

Gas-charged accumulators are commonly used in hydraulic systems, primarily to reduce themagnitude of the pressure transients resulting from abrupt changes in flow rate. The attachedpaper considers the energy-dissipating effects of work-to-heat transduction and heat transfer in a

gas-charged accumulator. The authors point out that the common polytropic model (PVn=constant) used to characterize the pressure-volume relation in the gas fails to account for"thermal damping" and present a nonlinear model, a linear model and experimental data tosupport their point. You are to critique this paper by modeling and simulating the accumulator

system.

Background reading: A. Pourmovahed & D. R. Otis (1984) "Effects of Thermal Damping on theDynamic Response of a Hydraulic Motor-Accumulator System", J.D.S.M.C., 106:21-26.

1. Formulate a model of the thermofluid capacitive subsystem of P. & O. '84, fig. 5 andrepresent it by a bond graph. Assuming the flow rate Qaas an input and that nitrogen may be

modeled as an ideal gas, derive nonlinear dynamic equations suitable for simulating figs. 1, 3& 4 of P. & O. '84.

2. P. & O. '84 devote much of their paper to a linear analysis.

2a. Linearize your model about similar operating conditions. Comment on the usefulness ofthe "anelastic model" of P. & O. fig. 2.

2b. Develop a bond graph corresponding to your linearized model using single-port storageand dissipative elements and idealized transducers (e.g. gyrators, transformers) and showthat it yields the same linearized equations.

2c. Does your linearized model describe entropy production? Is entropy production anessentially nonlinear phenomenon?

3a. Using the two models developed above, determine parameters for simulations corresponding

to P. & O.'s experimental results and simulate the accumulator to obtain cross plotscorresponding to P. & O. '84 figs. 1, 3 & 4.

3b. Comment on the degree of (dis)agreement between these models and P. & O.'s experimentsand simulations; that is, how plausible is the ideal gas assumption in this case? (Bear inmind that your nonlinear model is different from the one P. & O. used -- see their reference#3.)

2.141 2002 page 1 assignment #311


18/289

4. Develop a model of the complete hydraulic system of P. & O. '84, fig. 5 and represent it by abond graph. Assuming the flow rate Qsand the force F as inputs and that nitrogen may be

modeled as an ideal gas, derive nonlinear dynamic equations suitable for simulating thetransient response of the system. Hint: Keep in mind that your choice of state variables caninfluence the complexity of your equations.

5. Assuming that both inputs are zero (Qs= 0, F = PoAm) simulate the pressure in the

accumulator in response to an initial velocity of the mass in the direction which wouldcompress the gas and large enough to cause a volume change comparable to that of P. & O.'84 fig. 1. Assume the gas is initially at equilibrium with ambient conditions. Chooseparameters so that the undamped natural frequency of the system is 0.01Hz (as in figs. 1, 3 &4.) Given an accumulator volume of 2 liters, the piston diameter of a corresponding linearactuator would probably be on the order of a couple of centimeters (i.e., a half to one inch).

Simulate three cases:

5a. isothermal conditions gas temperature constant at ambient temperature

5b. adiabatic conditions no heat transfer from the gas

5c. the conditions corresponding to P. & O.'s experimental data.

5d. Now choose parameters so that the undamped natural frequency of the system is ten timeshigher (i.e., 0.1Hz) and repeat 5c. above.

Based on this analysis, what would you conclude about the importance of the "thermal damping"phenomenon?



19/289



2.141 Assignment #3: GAS-CHARGED ACCUMULATOR

The figure below (after Pourmovahed & Otis, 1984) is a schematic diagram of the hydraulic andmechanical system.

gas

oil

motor

mF

ypiston or

diaphragm

QmQsAm

P

Qa

P

P, v

accumulator

Our first goal is to model the thermofluid subsystem assuming the flow rate Qaas an input. The

following bond graph is the simplest representation of the accumulator subsystem.

RTw

dSw/dtC

P

dV/dt dS/dtSf SeQa(t):

hydraulic domain thermal domain

T:Tw

The two-port capacitor on the left represents the reversible energy storage and work-to-heattransduction in the ideal gas in the accumulator. The two-port resistor on the right representsirreversible heat conduction through the wall of the accumulator with the concomitant entropyproduction. The outside of the wall is assumed to remain at a constant temperature (i.e., ambienttemperature).

The direction of the power bonds has been chosen to follow the usual convention that power ispositive into the storage element. However, the sign convention for the two-port resistor followsa more intuitive power in = power out convention but assumes for convenience that positiveheat flow is from the wall to the gas.

Nonlinear equations

Causal assignment indicates that both ports of the capacitor may be given the preferred integralcausal form. Note that the two-port resistor assumes its preferred causal form with temperaturesas inputs. Thus we may be confident that the model will properly reflect the second law.

We must choose state variables. There are many possible choices, but they are not equallysimple.

Gas-charged accumulator solution page 1 Neville Hogan13


20/289

Hydraulic side:

We could choose either pressure, P, or volume, V, as state variable but volume seems to be themost sensible choice as the corresponding state equation is simply

V = Qa(t)

The negative sign is due to the fact that positive flow rate compresses the gas.

Thermal side:

We might choose the energy variable, total entropy, S, but temperature, T, is more common. Wecould also use total internal energy, U, as a state variable. To choose between these, consider theavailable information.

Two-port capacitor constitutive equations are derived assuming nitrogen may be modeled as anideal gas. The ideal gas equation is

PV = m RT

where m is the mass of the gas and R is the gas constant for nitrogen (not the universal gas

constant). The second constitutive equation is derived by assuming

dU = m cvdT

where cvis the specific heat at constant volume. Integrating:

U - Uo= m cv(T To)

where subscript o denotes reference state.

The two-port resistor equations assume Fouriers-Law (i.e., linear) heat conduction.

Q = h Aw(Tw T)

where h is a heat transfer coefficient and Awand Tware the area and temperature of the wall.Consider using S and V as state variables.

T S = Q = h Aw(Tw T)

hence

S = h Aw(TwT 1)

However, I need T = T(S,V) and P = P(S,V) to form state equations. From class notes,

T = To VVo

R

cv exp S Somcv

P = Po

V

Vo

R

cv+ 1

exp

S So

mcv

These are output equations. State equations are



21/289


22/289

T = h Awm cv

T +h Awm cv

Tw+Rcv

TV Qa

V = Qa

These state equations are still coupled and nonlinear, but look a little simpler. Note that the

thermal time constant is again obvious but now the choice of reference temperature is lessimportant. The output equation for pressure is

P =m RT

V

The equations using energy as a state variable are similar to those using temperature as a statevariable. That shouldnt be surprising given the simple relation between temperature and internalenergy for and ideal gas. The point to remember is that the integral causal form does notconstrain your choice of state variables. State variables should be chosen for convenience and tomaximize insight. I will use temperature and volume as state variables.

Linearized equations

The second task is to linearize the model and compare to P&Os linearized model. Linearizingthe state equations is always helpful. In this case it can provide insight into the physical systemand the model and parameters used by P&O.

The two-port capacitor constitutive equation is derived from the ideal gas equation PV = m RT.To linearize, we need to write this in a causal formany of the four causal forms will serve. Usethe causal form with temperature and volume as (integrated) inputs, i.e.

CP

dV/dt dS/dt

T

P = m RTV

Denote the operating point by subscript o and linearize.

P Po m RVo

(T To) m RTo

Vo2(V Vo)

Write V = V Vo, T = T Toand P = P Po

m RVo

T +m RTo

Vo2(V)

Remember that the positive work compresses the gas, hence V is the appropriate displacementvariable on the hydraulic side.

This indicates that the hydraulic side of the linearized model may be represented by atransformer between thermal and hydraulic domains connected to a fluid capacitor by a 1-junction (common flow).

transformer modulus:m RVo



23/289

(inverse) fluid capacitance:m RTo

Vo2

The second constitutive equation is in the form S = S(T,V) It may be derived from the internalenergy equation and the first law.

dU = m cvdT = T dS P dV

dS =m cv

T dT +m RV dV

S =m cvTo

T m RVo

(V)

Again, remember that V is the appropriate flow variable of the hydraulic side given the powersign we have assumed.

This indicates that the thermal side of the linearized model may be represented by a thermalcapacitor connected to a transformer between thermal and hydraulic domains by a 0-junction

(common effort).

transformer modulus:m RVo

(as before)

thermal capacitance:m cvVo

A bond graph of the linearized two-port capacitor is as follows.

TFdV/dt T

1 0

CC : m cv/To

m R/Vo

: Vo2/m R To

Notethat this linearized model indicates that the strength of the coupling between the thermaland mechanical domains is proportional to the mass of gas and inversely proportional to thevolume it occupiesa useful insight.

The two-port resistor equations describe entropy flow as a function of temperature. For the gas

S =Q

T = h A

w

Tw

T 1

Linearizing

S S o

h AwTw

To2(To T)

S h Aw

Tw

To 1

h AwTw

To2(To T)



24/289

Assume the operating point (and also the reference thermal state) is at equilibrium with the walltemperature.

To= Tw

S

h Aw

To(To T)

Under the same operating conditions, the entropy flow rate from the wall is identical

S w=QTw

S w

h Aw

To(To T)

A bond graph of this linearization of the two-port resistor is as follows.

Se

dS/dt

1

R: To/h Aw

:Tw

Notethat this linearizedmodel does NOT reflect the second law of thermodynamicslinearization has eliminated entropy production due to heat transfer. This is due to the particularchoice of operating point. By choosing To= Twwe make the entropy flow from the wall

identical to the entropy flow to the gas.

S net= S S w= 0However, entropy production is notan essentially nonlinear phenomenonits the choice ofoperating point that matters. Linearization about equilibrium eliminates entropy production; amodel linearized about a non-equilibriumoperating point will describe entropy production.

A bond graph of the complete linearized system is as follows.

Se

dS/dt

1

R: To/h Aw

:TwTFdV/dt T

1 0

CC m cv/To:

m R/Vo

Vo2/m R To:

SfQa:

Using the causal assignment shown, linearized equations may be read from the graph.

T =To

m cv

S m RVo

V



25/289

S =h Aw

ToT

State equations

V = Qa

T = h Awm cv

T +Rcv

ToVo

Qa

Output equation

P =m RVo

T m RTo

Vo2 V

To compare with P&Os linearized equations, take the Laplace transform.

T

s +1

=Rcv

ToVo

Qa

V = Qas

P =m R To

Vo2

Rcv

+ s +1

(s +1

s) Qa

Substitute

m R To

Vo2 =

PoVo

Rcv

+ 1 =R + cv

cv =

PQa

=PoVo

s + 1s (s +1)

This is the same as derived by P&O.

The complete hydraulic system

Adding the rest of the system is straightforward. The piston can be modeled as a transformerlinking the fluid domain with the mechanical domain. The piston itself is simply a mass with a

force acting on it (though with an unusual sign convention). A bond graph is shown below.



26/289

Se

dS/dt

R

: F

:Tw

TFP

T

0 1

IC : 1/M

Am

SfQs:

Se

dV/dt

vm

The translational inertia is an ideal linear element. It makes little difference whether momentumor velocity is used as its state variable. For clarity I will use velocity with temperature andvolume as the gas state variables. State and output equations for the complete system are asfollows.

V = Amvm Qs

T =h Awm cv

( )Tw T +Rcv

TV ( )Amvm+ Qs

v m=1M

m R AmTV F

P =m RT

V

where Amis the piston area, vmis the velocity of the inertia and M its mass.



27/289

Parameters for simulation

For P&O fig. 1:

Po= 1014.5 psia (operating pressure in the accumulator)

Vo

= 103 cu. in. (operating volume of the accumulator)

m = 0.293 lbm (mass of gas)

V/Vo= 0.25

For P&O fig. 3:

Po= 438.5 psia

Vo= 122 cu. in.

m = 0.15 lbm

V/Vo= 0.05

For P&O fig. 4:

Po= 436.7 psia

Vo= 122 cu. in.

m = 0.149 lbm

V/Vo= 0.01

In all cases:

Tw= 540R (room temperature)

R = 660 in-lbf/lbm R (gas constant)cv= 1900 in-lbf/lbm R

=m cvh Aw

= 15.3 seconds

For the complete hydraulic system

To estimate the effective mass I assumed isothermal conditions and used the linearized model.Isothermal conditions keeps things simple because it essentially eliminates the thermal domainfrom participating in the dynamics. The undamped natural frequency is given by

n=P

oA

m

2

M Vo

hence the estimated mass is

M =PoAm2

Von2



28/289

Assuming a piston diameter of 0.75 inches, Amis 0.44 inches squared. Using the pressure and

volume of P&O fig.1 and an undamped natural frequency of 0.01 Hertz yields

M =1014.5 x 0.442

102.97 x (2 x 0.01)2 = 487

lbf

in2in4

in3sec2

rad2

Convert to more meaningful units:

1lbf sec2

in = 386.4 lbm

M = 1.88 x 105lbm.

Thus the mass required to yield an undamped natural frequency that low is unbelievably large84 tons!

However, the mass required to yield an undamped natural frequency of 0.1 Hz is a little morereasonable1,880 pounds.

To choose an initial velocity I assumed undamped oscillations and used

vm=V

Am

V = V sin nt

vm(0) =Vn

Am

Simulations

Nonlinear simulations of P&O figs. 1, 3 and 4 are as follows.

-0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5P&O fig. 1, nonlinear

normalized volume

no

rmalizedpressure



29/289

-0.05 0 0.05410

420

430

440

450

460

470P&O fig. 3, nonlinear

normalized volume

pressure(psi)

-0.1 -0.05 0 0.05 0.1380

400

420

440

460

480

500P&O fig. 4, nonlinear

normalized volume

pressure(psi)

Note that in all cases these simulations are quite close to P&Os even though those authors useda much more elaborate model of the gas. For 10% volume changes, the ideal gas model isessentially identical to their data. For 5% volume changes the ideal gas model is essentiallylinear.



30/289

Transient responses

Simulations of the transient responses for the large mass assuming thermally damped, adiabaticand isothermal conditions are as follows.

0 50 100 150 200 250 300 350 400-4

-2

0

2

4

Transient response of complete nonlinear system

time (seconds)

velocity(inch/sec)

-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25-0.4

-0.2

0

0.2

0.4

normalized volume

normalizedpressure

The plot shows the velocity of the piston mass. Note that the oscillation is steadily losingamplitude as time increases. Also note that no mechanical losses such as friction have beenincluded in our model. This loss in energy is due to the irreversible entropy production due toconduction in the gas-charged accumulator. Assuming only a reversible work storage in the gasaccumulator would not predict this behavior. Just like a bicycle pump, compression of the gasgenerates heat that flows through the walls of the chamber. Not all of this heat energy isrecovered when the gas expands, so energy is lost and we have thermal damping.

The adiabatic case was modeled by adding an ideal flow source on the thermal side of the two-port capacitor. Setting this flow source to zero sets the entropy flow to zero, i.e. zero entropyflows which in this case means no heat flow. The result of this simulation shows oscillations asin the previous case. However, the amplitude of the oscillations does not decrease. We haveessentially removed the heat lost through the wall leaving only the reversible work storage incompressing the gas. The system without thermal damping behaves like an undamped spring-mass system.



31/289

0 50 100 150 200 250 300 350 400-4

-2

0

2

4Adiabatic transient response

time (seconds)

velocity(inch/sec)

-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25-0.4

-0.2

0

0.2

0.4

normalized volume

normalizedpressure

The isothermal case was modeled by setting dT/dt = 0 and setting the initial gas temperatureequal to Tw. This is equivalent to assuming no thermal resistance across the wall of the

accumulator. Hence there is no entropy production and again, no thermal damping.

0 50 100 150 200 250 300 350 400-4

-2

0

2

4Isothermal transient response

time (seconds)

velocity(inch/sec)

-0.3 -0.2 -0.1 0 0.1 0.2 0.3-0.4

-0.2

0

0.2

0.4

normalized volume

norm

alizedpressure

The isothermal and adiabatic cases are undamped, as expected. Note that the isothermalfrequency of oscillation is lower than the damped case which in turn is lower than the adiabaticcase. This is also as expected.



32/289

Finally, a simulation of the transient response for the smaller (more reasonable) mass is asfollows.

0 20 40 60 80 100 120 140 160 180 200-40

-20

0

20

40Transient response for reduced piston mass

time (seconds)

velocity(inch/sec)

-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25-0.4

-0.2

0

0.2

0.4

normalized volume

normalizedpressure

Note that even though the normalized pressure-volume plot shows very little evidence of energydissipation (i.e., the area inside the loop is small) the effect on the transient response issubstantial. I conclude that thermal damping is likely to be a real phenomenon in practicaldynamic systems.



33/289

Not used:

U = mcvTo

VVo

Rcvexp

S Somcv

1 + Uo

Choose an operating point at thermal equilibrium with the environment. Assume no forcing, i.e.,Qa= 0

To= Tw

For P&O fig. 4:

Po= 69.95 bar = 1014.5 psia (operating pressure in the accumulator)

Vo= 1687cc = 103 cu. in. (operating volume of the accumulator)

Gas properties of nitrogen @ 300K (room temperature):

cp= 1041 Joules/kg K (specific heat at constant pressure)

= 1.399 (polytropic index)R = 297 Joules/kg K (gas constant)

Remember: cv=cp

= cp- R

cp= 2560 in-lbf/lbm R

= 2560/1900 = 1.35



34/289

MEMORANDUM

To: 2.141 classFrom: Professor Neville HoganDate: November 26, 2002Re: Assignment #3

All of you did well on this assignment. Maybe Im making them too easy...

One common difficulty was to conclude that entropy production is an essentially nonlinearphenomenon. This is a subtle problem: if you did the obvious (as I didsee my solution) andlinearized the two-port resistor about conditions with the same (ambient) temperature on bothsides, you linearized about equilibrium. That corresponds to assuming a reversible process andhence the linearized model will not describe entropy generation. However, if you linearize aboutnon-equilibrium conditions (i.e., different temperatures on the two ports of the resistor,corresponding to a non-zero heat flux) the resulting linearized model will describe entropy

production.

Class statistics on this part: mean 93%; standard deviation 4%.

28


35/289



Assignment #4 Out: 10/24/02Due: 11/7/02

Control Through Singularities

Motivation: One common and productive use of modeling and simulation is to study andunderstand proposed engineering designs. In this assignment you will test the behavior of aproposed robot controller.

A common form of robot motion control specifies a workspace position or trajectory (e.g., adesired time-course of tool position in Cartesian coordinates) and transforms that specification toa corresponding configuration-space position or trajectory (e.g., a time-course of joint angles).

However, most robot mechanisms have kinematic singularities, configurations at which therelation between workspace and configuration-space becomes ill-defined. As a result, most robotmotions controllers do not operate at or near these singular points.

However, an energy-based analysis of mechanics shows that the transformation of positions andvelocities is well-defined in one direction while the transformation of efforts and momenta iswell-defined in the other. This implies that a controller that takes advantage of these facts shouldbe able to operate at and close to mechanism singularities. A simple impedance controllerattempts to impose the workspace behavior of a damped spring connected to a movable virtualposition.

JxBLxKJ oot

where is a vector of generalized coordinates, a vector of (conjugate) generalized efforts, xand xoare vectors of actual and virtual Cartesian tip coordinates, L() and J() are linkagekinematic equations and Jacobian respectively, and Kand Bare stiffness and dampingrespectively. Note that this controller requires neither the inverse of the kinematic equations northe inverse of the Jacobian.

In this assignment you are to test whether this simple impedance controller can operate at andclose to mechanism singularities.

Tasks:

Assume a planar mechanism open-chain mechanism with two links of equal length L = 0.5 m,operating in the horizontal plane (i.e., ignore gravity) and driven by ideal controllable-torqueactuators, one driving the inner (shoulder) link relative to ground, the other driving the outer(elbow) link relative to the inner link. Sensors mounted co-axially with the actuators providemeasurements of joint angle and angular velocity.



36/289

1. Write kinematic equations relating generalized coordinates to the position coordinates of thetip, expressed in a Cartesian coordinate frame with its origin at the axis connecting the innerlink to ground.

2. Find the corresponding Jacobian (to relate generalized velocities to tip Cartesian velocities).

3. Identify the set of singular configurations for this linkage (at which the relation between tipCartesian coordinates and joint angles becomes ill-defined) and show that they include thecenter of the workspace as well as at the limits of reach.

4. Formulate a dynamicmodel of the mechanism relating input actuator torques to outputmotion of the tip in Cartesian coordinates. Assume the links are rods of uniform cross sectionand mass m = 0.5 kg and that the joints are frictionless. (Hint: be careful in your choice ofgeneralized coordinates.)

5. Simulate the behavior of this mechanism under the action of a simple impedance controller.Assume uniform tip stiffness and damping (i.e., the stiffness and damping matrices have the

form: and where kandbare constants).

10kK 01 01

10bB

5.a Choose the stiffness and damping matrices so that when the mechanism is making smallmotions about a configuration with the inner link aligned along the Cartesian x-axis andthe outer link aligned parallel to the Cartesian y-axis the highest-bandwidth transferfunction between virtual and actual position has critically-damped poles with anundamped natural frequency of 2 Hz. (Hint: it may be easiest to first transform thestiffness and damping to generalized coordinates.)

5.b Simulate the response to the following virtual trajectories and plot at least the path of the

tip in Cartesian coordinates:

1) Starting from rest at {x = L, y = 0} ending at rest at {x = 2L, y = 0}; trapezoidal speedprofile with acceleration to peak speed in 250 ms, constant speed for 1.5 sec anddeceleration to rest in 250 ms.

2) Starting from rest at {x = -L, y = 0} ending at rest at {x = L, y = 0}; trapezoidal speedprofile with acceleration to peak speed in 250 ms, constant speed for 1.5 sec anddeceleration to rest in 250 ms.

3) Starting from rest at {x = -L, y = L/20} ending at rest at {x = L, y = L/10}; trapezoidalspeed profile with acceleration to peak speed in 250 ms, constant speed for 1.5 sec and

deceleration to rest in 250 ms.

6. Repeat the simulation of 5.b.3 with kand bchosen to yield critically-damped poles with anundamped natural frequency of 20 Hz.

7. Repeat the simulation of 5.b.3 with the outer link 10% shorter than the inner link. How doesthis change the set of singular configurations?



37/289

8. Comment briefly on (a) whether and (b) how well this controller operates nearmechanism singularities.

Tent-Pole Instability

The complex behavior of multi-body mechanical systems stems from kinematically modulatedtransformation of energy. It can give rise to some highly counter-intuitive phenomena, one ofwhich you are to explore in this assignment.

A tent-pole is to be supported in the upright position by guy-ropes. However, the support is notcompletely rigid as the guy ropes inevitably have some elasticity. It may seem natural to increasethe tension in the ropes to stiffen the support but in fact, this can have the opposite effect. Togain insight you are to develop a model and numerical simulation to explore this phenomenon.

Assume the tent pole is a rigid rod of height, h = 2 m, and that it is mounted on a hinge at its baseso that it can only move in a plane. Assume that two elastic cords are attached to its top and tothe ground in the plane of motion at a horizontal distance, w = 1 m, on either side. Assume the

cords have negligible mass, linear elasticity (i.e., obeying Hooke's law) with stiffness, k, andpretension, Fp.

1. Develop a model to describe the mechanical system comprised of the pole and the twocords.

To understand this system, simulate the following cases:

2. Assume the cords have pretension but zero stiffness; assume the pole starts in the uprightposition; and assume no gravity (for now). Show thatthis mechanical system is unstablefor any value of Fpeven in the absence of gravity.

3. Assume the cords have non-zero stiffness but no pretension; assume the pole starts in theupright position; and assume no gravity. Test whetherthis system is stable for (a) smallperturbations (e.g., initial conditions 1 from upright) and (b) large perturbations (e.g.,initial conditions 45 from upright).

4. Now that you understand what is going on, include both stiffness and pretension in thecords and include gravity and any other physical behavior relevant to system stability.Using the numerical simulation,show thatfor a given cord stiffness there is a maximumpretension for stable behavior and find how it varies with stiffness (a few representativevalues will suffice).

This system is simple enough to be amenable to symbolic (rather than numerical) analysis.

5. Derive an algebraic expression for the relation between the maximum pretension forstable upright behavior (i.e., corresponding to 4. above).How well do your numericalsimulation results agree with your analytical results?



38/289

Thomas A. Bowers2.141 Fall 2002

Assignment 4: Kinematics

Two-Link Planar Robotic Mechanism

1. The configuration being analyzed in this problem is shown below in Figure 1.

2

1

L1

L2

(x,y)

(xc1,yc1)

(xc2,yc2)

Figure 1: Cartesian and Generalized Coordinates of 2-Link Robotic Mechanism

The equations relating the Cartesian coordinates to the generalized coordinates, 1and 2,are as follows:

xL1 cos1L2 cos2yL sin L2 sin21 1

L1c1 cos1

2

L1y sin12

c1

L1 cos L2 cos2

2xc2 1

y L1 sin L2 sin22

c2 1

The kinematic equations relating the generalized and Cartesian coordinates can be foundfrom derivation.

x L sin L2 sin1 1 1 2 2 yL cos L cos1 1 1 2 2 2

32


39/289

L11x c1 sin12L11y cos1c1

2L22x

c2

L1

sin sin21 1 2

2y L1 cos

L2 cos2c2 1 12

2. The Jacobian matrices for the tip and the centers of mass of the mechanism can befound directly from the kinematic equations:

L1 sin L2 sin2 Jtip L1 cos

1

L2 cos2

1

L

2

1

sin1 0 L1 cos1 0

2 0 J 1

L2 cm L1 sin sin2 1 2 L1 cos1

L2 cos2

2 0 1

3. To find the singular values, the determinant of the Jacobian for the tip of the

mechanism is set to zero. For these angles the inverse of the Jacobian is not defined sothere is no transformation from Cartesian to generalized coordinates.

detJ LL 2sin cos2 sin 2 cos1 sin 2 01 1 1 ,0....k1 2

This corresponds to all configurations where the links are parallel or anti-parallel.

4. To formulate the dynamic model of the system the inertances of the links must bedetermined. The links each have a translational and rotational inertia:

I m

mL2J

12

The mass matrix for the system is then:

33


40/289

2

m1 0 0 0 0 0

0 m1 0 0 0 02Lm 10 0 1 0 0 0

12

M0 0 0 m2 0 0

0 0 0 0 m2 0Lm 220 0 0 0 0 2

12

This can be converted to the generalized inertia of the system.

TMJJI

T

L1 L1 sin 1 m1 0 0 0 0 0 sin 1 0 0

2

2

2 2

0 1 0 0 0 0L1m L1

21

1 0

cos 10 0cos 2Lm 110 0 0 0 0

1 0120 0 0 m 0 02

0 0 0 0 m2 0

1

L2

21

L2

2sin 2L1 L1 sin sin sin

L LLm 222

2

2

2

21

0 1

1

0 1

cos 2L1 L1cos cos cos0 0 0 0 0 12

2 2 L1 Lm 11 LL 21 sin 1sin

2 2 sin 2 2 21 1LL1 2 sin 1

1 m1 sin 2 1

Lm 22

cosLm 12

m1 cos cos cos24 12 2 2L2

4sin 2 22

LL1

cos 1

2 2

LL1

cos 2sin m cos m2 22 12

2 2

Lm 11 Lm 11 1

2 22m cos 2m cos 2 2 Lm 12Lm 12

2 23 2 3 2

2222LL1 Lm2

3

LL1 Lm2

22

3 1 2

2 cosm22m cos2

2

This expression of the generalized inertance is used to find the co-energy of the system:

1 T* )( E Ik2

2Lm 1 LL1 1Lm

2 2m cos 2

2

1

2

1

2

2 1 2

1 3 2

2

2LL1 Lm2

22

3

2 cosm2

2 2

1

2 2

1 Lm13

m2

LL 21

2

LL 21 Lm2

3

1

2

2

1

21 Lm 12 cos 2 cos 2

2

m 22

2 2

2

1

2

2

2 Lm1

3

Lm2

3

1 LLm 22 1

21 Lm 12

2 cos 1 22

2

2

34


41/289


42/289

T T FJ xKJ

L1 sin1 L1 cos1k

L2 cos 0 L1 L2 0 L

2

21

22

2 L sin2 L2 cos2 L1 sin

L2 0

k

L11

2

kL1 0 2

2 1

2 L2K kL

0

1

2 0

Similarly, the damping matrix is transformed to generalized coordinates using the

Jacobian.

T T T L1 sin1 L1 cos1L1 sin L2 sin2

2 0 FJ xBJ J BJ b

L sin2 L2 cos2 L1 cos

1

bL1

2L2 cos2 2 1 0 L2 2 0

B bL1

2 0 L2

The desired natural frequency for the system is 2 Hz. This allows the stiffness constant, k,to be evaluated:

1

n K I1

2 2 L2k I

L

0

1

2 0 n

Lm 12

m 2L1 m 2LL 2 1 2 1

I cos2 6

0 1

3

LL 22

2 1

m 2 cos2m 2L 2 0 1

2 3 24

1 2 0 3 2k I

02

L22

1

16 6

0 0 4 164 2

L1 0 01

4 0

2

1 0

24 6 8 2k max3

If the stiffness coefficient is larger than kmax then the system will oscillate faster than 2Hzin the horizontal direction, which is the lower impedance direction for this configuration.

The damping coefficient must be found to provide critical damping to the system in thisconfiguration. The critical damping coefficient is found from the following expression:

36


43/289

B 2 K I

1 2 0 L22 6

0 2b

L1 02

2

4 8

L1

2 0 0 0 L2 3

1 24

11 1 1

1b 2

322 0 460 16

0 322 0 6 1 1 3 2 3

100 0 0

4 24 16 3

b4

critical3

5b. The system is driven by a simple impedance controller with the following

characteristic:

T 0 J XK L VB J0

These torques are the torques corresponding to changes in the angles 1and 2, which arenot the same as the motor torques on the links. If the motor torques are needed in order tocontrol actual hardware, the following relationships are used. The equivalent torque

acting on each of the links is depicted below in Figure 2.

1-21

L1 L2

1

2

2

Figure 2: Uncoupled 2-Link Mechanism Indicating Torque Input

From this figure it is evident that the torque corresponding to 1 is the difference between

the motor torques: 2 . Also, the torque corresponding to 2is the torque1corresponding to 1minus the torque corresponding to2, which results in 1.

1) The first simulation describes the motion of the two link mechanism from theCartesian position (L, 0) to (2L, 0) with a trapezoidal speed command. Figure 3 shows

the position of the mechanism at 50ms intervals during its translation.

37


44/289

0 0.2 0.4 0.6 0.8 1-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

x position (m)

yposition(m)

Time Plot of Mechanism Position for First S imulation

Figure 3: Time Plot of Mechanism for Virtual Trajectory from (L,0) to (2L,0)

2) The second simulation describes the motion of the mechanism as it passes through the

singular point (0, 0). The virtual trajectory of the robot is from (-L, 0) to (L, 0). Figure 4demonstrates the behavior of the mechanism for this input.

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1

0

0.1

0.2

0.3

0.4

0.5

x position (m)

yposition(m)

Time Plot of Mechanism Position for Second Simulation

Figure 4: Time Plot of Mechanism for Virtual Trajectory from (-L,0) to (L,0)

38


45/289

It is clear from this plot that the compliance in the system allows the outer link to rotatearound its pivot since there is less inertia in this direction than in the direction of the

virtual trajectory. This is seen as the mechanism begins to move and again after it passesthe singular configuration. There is some overshoot at the end of the movement due to the

inertance of the mechanism.

3) The third simulation attempts to pass the tip of the mechanism near the singularconfiguration at the origin. The virtual trajectory is from (-L, L/20) to (L, L/10). This

results in the following behavior for the mechanism.

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

x position (m)

yposition(m)

Time Plot of Mechanism Position for Third Simulation

Figure 5: Time Plot of Mechanism for Virtual Trajectory from (-L,L/20) to (L,L/10)

Figure 5 demonstrates that the mechanism passes through the singular configurationinstead of following the virtual trajectory. In order for the mechanism to pass near the

singular configuration without actually passing through it, the links of the mechanismwould need to quickly rotate through 180

oas the mechanism approached the singular

configuration. This requires very large torques on the mechanism as it passes near thesingular point.

6. The natural frequency of the controller is now adjusted to 20Hz to determine theimpact this has on the system. The corresponding kand bare found as follows:

2

max

2

3

800

06

13

20

1600

k

k

39


46/289

340

03

26

10

3

3200

16

10

016

1

24

10

06

1

04

14

10

3

3200 2

1

22

criticalb

b

Incorporating these into the model of the controller, yields the following output for the

conditions specified in the third simulation.

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

x position (m)

yposition(m)

Time Plot of Mechanism Position for Third Simulation with 20Hz Natural Frequency

Figure 6: Time Plot for Trajectory from (-L,L/20) to (L,L/10) with 20Hz Critically Damped Poles

Figure 6 demonstrates the need for the mechanism to rotate 180oas it passes near the

singular configuration if it is stiffly linked to the virtual trajectory.

7. Finally, simulation three is repeated with the outer link 10% shorter than the inner link

(.45m instead of .5m). With this discrepancy in the link lengths the mechanism can nolonger reach points within .05 meters of the origin. Additionally, it cannot reach beyond a

circle of radius .95 meters. The resulting behavior of the mechanism is shown below in

Figure 7.

40


47/289

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

x position (m)

yposition(m)

Time Plot of Mechanism Position for Third Simulation with 20Hz Natural Frequency

Figure 7: Time Plot for Trajectory from (-L,L/20) to (L,L/10) with Uneven Link Lengths

This figure shows that the low rotational inertia of the outer link again causes it to rotate

away from the virtual trajectory. As the mechanism approaches the origin in thissimulation it is unable to pass within .05 meters. This causes it to rotate away from the

origin and the prescribed virtual trajectory. After passing through the singularconfiguration at (0,.05) it converges back onto the virtual trajectory and overshoots its

final destination again due to the inertance of the rotating links.

8. Despite the fact that the Jacobian of the mechanism is undefined at its singularconfigurations, the simple impedance controller allows the mechanism to operate at its

singular configurations. In the first simulation, the final desired position of themechanism was the singular position corresponding to both links fully extended to the

maximum attainable radius. With the impedance controller the mechanism was able toreach this configuration. However, due to the low translational inertance that corresponds

to this position and relatively high rotational inertance the mechanism rotated beyond thedesired position pulling the tip of the mechanism back towards the origin. If the time

scale is extended, it is seen that the mechanism continues to oscillate around this positionwith decreasing frequency. At 200 seconds the mechanism still has 1

ooscillations with a

period of 36 seconds. The actual trajectory and virtual trajectory of the mechanism areshown below in Figure 8, which includes only the 2 second command interval.

41


48/289

Actual Trajectory and Desired Trajectory for First Simulation0.1

0.08

0.06

0.04

VerticalPosition(m)

0.02

0

-0.02

-0.04

-0.06

-0.08

-0.1 0.5 0.6 0.7 0.8 0.9 1

Horizontal Posit ion (m)

Figure 8: The Actual and Virtual Trajectory for Desired Movement from (L,0) to (2L,0)

For the second simulation, the mechanism was commanded to pass directly through the

singular position at the origin. Because the impedance controller does not require theinverse of the Jacobian, the mechanism was able to pass through this point. Indeed the

mechanism prefers to pass through this point, as shown below in Figure 9, since it is thelowest energy path through the origin.

Actual Trajectory and Desired Trajectory for Second Simulation

0

i

l

i

(m)

-0.1

-0.08

-0.06

-0.04

-0.02

0.02

0.04

0.06

0.08

0.1

Vertca

Postion

-0.4 -0.2 0 0.2 0.4 0.6


Figure 9: The Actual and Virtual Trajectory for Desired Movement from (-L,0) to (L,0)

42


49/289

This is also seen on the third simulation, which attempts to pass the mechanism near theorigin. Rather than follow the virtual trajectory, which passes just above the origin, the

mechanism deviates from the virtual trajectory to pass through the lowest energy path:directly through the origin. This is clearly evident in Figure 10, which shows the virtual

trajectory passing through (0, .0375) and the actual trajectory passing through (0, 0).

0

l i i i l ion

i

l

iti

(m)

-0.1

-0.08

-0.06

-0.04

-0.02

0.02

0.04

0.06

0.08

0.1Actua Trajectory and Des red Trajectory for Th rd S mu at

VertcaPos

on

-0.4 -0.2 0 0.2 0.4 0.6

Horizontal Position (m)

Figure 10: The Actual and Virtual Trajectory for Desired Movement from (-L,L/20) to (L,L/10)

When the stiffness of the virtual linkage was increased, the mechanism was constrained

more closely to the virtual trajectory. This is seen clearly in Figure 11.

Actual Trajectory and Desired Trajectory for High Stiffness Simulation

0

i

l

iti

(m)

-0.1

-0.08

-0.06

-0.04

-0.02

0.02

0.04

0.06

0.08

0.1

VertcaPos

on

-0.4 -0.2 0 0.2 0.4 0.6

Horizontal Position (m)

Figure 11: Actual & Virtual Trajectory from (-L,L/20) to (L,L/10) with Increased Stiffness

43


50/289

While the increased stiffness resulted in better performance away from the singularposition, near the singular position the mechanism required quick rotations and huge

torques in order to maintain the desired trajectory. This was seen previously in Figure 6.The mechanism performs well, but requires much more energy to reach the same desired

end state.

VerticalPosition(m)

The final simulation demonstrated how a larger singular region around the origin affected

the behavior of the robot. With unequal link lengths the robot could not attain anyposition within .05 meters of the origin. However, the desired trajectory of the robot was

through the point (0, .0375). Because this configuration could not be attained, themechanism passed through the closest point with the least resistance, (0, .05). This is seen

in Figure 12, which shows the virtual trajectory and actual trajectory of the mechanism.This figure also demonstrates that when the singular configuration is closer to the desired

trajectory, the mechanism can more accurately follow the desired path. This is especiallyapparent when comparing Figure 12 to Figure 10.

Actual Trajectory and Desired Trajectory for Uneven Link Simulation0.1

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08

-0.1-0.4 -0.2 0 0.2 0.4 0.6


Figure 12: Actual & Virtual Trajectory from (-L,L/20) to (L,L/10) with Uneven Link Lengths

44


51/289

Tent Pole Instability

1. The mechanical system of a tent pole and guy-ropes is shown below in Figure 13.

1 m 1 m

k kl1 l2

2 m

Figure 13: Tent-Pole & Guy-Rope System

The relationship between the variables l1, l2and can be determined from the Law of

Cosines:

l1 5 cos4

l2 5 cos4

The Jacobian for the transformation between coordinates is found by differentiating each

of these terms with respect to .

sin2 l1 5 cos4 sin2

l2

5 cos4

sin2

5 cos4 J

sin2

5 cos4

The inertia of the pole is simply mL

2

/3. The co-energy, therefore, is dependent only on .This results in the following relationship between torque and angular acceleration:

32mL

The torque on the pole is proportional to the forces from each of the guy-ropes, according

to the following equation:

45


52/289

FJT

The forces from the guy-ropes are found from the next equation, where kis the elasticity

of the rope and F0 is the pretension.

0FlkF

There is also a torque component due to gravity, which results in the final expression fortorque:

cos2

mgLFJT

2. The first simulation is intended to show the mechanism is unstable for any pretension

if there is no elasticity in the guy-ropes. Gravity is also not to be considered. With these

assumptions and a pretension of only 10N the following behavior is exhibited.

-1 -0.5 0 0.5 1 1.5 2 2.5-1

-0.5

0

0.5

1

1.5

2

2.5


V

erticalPosition(m)

Tent-Pole Simulation 2

Figure 14: Time Plot of Tent-Pole with 10N Pretension and Zero Elasticity

As the pretension increases in the guy-ropes the system becomes more unstable due to thehigher torques generated.

3. The next simulation removed the pretension in the guy-ropes and added in theelasticity. Gravity effects were again omitted in order to emphasize the effect of the

elasticity on the system. Initially, the system was only perturbed 1oto show that it was

stable. This produced the following result where the pole oscillated between 1o.

46


53/289

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1

1.5

2

2.5


VerticalPosition(m)

Tent-Pole Simulation 3a

Figure 15: Time Plot of Tent-Pole with Elasticity and No Pretension for 1oPerturbation

The next task was to see if the tent-pole was still stable for perturbations up to 45o. Figure

16 demonstrates the stability of the pole with an initial perturbation of 45o.

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1

1.5

2

2.5


VerticalPosition(m)

Tent-Pole Simulation 3b

Figure 16: Time Plot of Tent-Pole with Elasticity and No Pretension for 45oPerturbation

47


54/289

4. With stiffness, pretension, and gravity all considered in the model of the system, thefollowing behavior is witnessed. Figure 17 shows the case where the system is stable, and

Figure 18 demonstrates the unstable case.

Tent-Pole Simulation 4 with F02.23607k

-1

0

1

2

i

l

i

(m)

-0.5

0.5

1.5

2.5

Vertca

Postion

-1 -0.5 0 0.5 1 1.5 2 2.5


Figure 18: Time Plot of Complete Tent-Pole System with Unstable Behavior

48


55/289

5. To analytically solve for the stability of the tent-pole system the equivalent stiffness ofthe joint is found. The expression for the joint stiffness is:

K JT

F kJTJ

The system becomes unstable at the point where there is no restoring torque on the

system for a small displacement . For this analysis the mass of the system is neglected.This yields the following equation for the torque on the tent pole.

TJ kl1

JTF 0

l2

sin2 sin2 sin2 5 cos4

sin2 sin2 F0 5 cos4 5 cos4

k

sin2 5 cos4 5 cos4 F0

5 cos4

4k

sin4 2 sin4 2 F0

sin2 sin2

5 cos4 5 cos4 5 cos4

5 cos4

For stability about the vertical position, = /2, where there are small displacements, ,this expression becomes:

2 2

4kcos cos

F0

cos2 cos2

5 4 5 4 5 4 5 4

10

2F0 cos

5 4 5 44kcos 2 2 2

25 16 25 16

F0 20cos 2k 25 16 5 4 5 4

Using LHopitals Rule:

cos20 20sin 20lim 5

x 0 2 2 4 16 2 5 4 5 4 25 16

5 4

5 4 5 2 5

Therefore, the maximum value for the pretension is k 5 or 2.23607k. This result can becompared to the ratios found through simulation, which are shown in Table 1:

25 16

Table 1: Ratio of F0to k for MATLAB Simulation of Tent-Pole System without Gravity

F0 k F0/k % error

50 N 22.36 N/m 2.23614 -0.003

100 N 44.72 N/m 2.23614 -0.003

150 N 67.08 N/m 2.23614 -0.003

2000 N 894.43 N/m 2.23606 -0.0003

49


56/289

It is clear from the simulation data, that MATLAB accurately predicts the stability pointfor the system. The precision on the ratio becomes greater as the value of the pretension

is increased. When the mass is considered as well, the stability point decreases as theangle from vertical increases. The mass has a larger effect on the stability of the system

for lower values of pretension.

50


57/289


58/289




Assignment #5 Out: 11/7/02

Due: 11/21/02

Simple convection and throttling processes

The main point of this assignment is to give you experience with modeling transportprocesses in a simple thermofluid system. A secondary purpose is to show you that evensimple systems involving matter transport can exhibit interesting behavior.

Two air cylinders are connected by a valve, which is initially closed. One cylindercontains air at 20 psi, the other, air at ambient pressure, 14.7 psi. Both cylinders are

initially at ambient temperature, 70F. You are to model and simulate the transient thatfollows when the valve is abruptly opened.

70F,20 psi

70F,14.7 psi

Assume air may adequately be described as an ideal gas. Assume each cylinder is 10inches in diameter and 30 inches long. Assume the valve orifice has diameter 0.1 in.Model the heat transfer between the cylinders and their surroundings, but you mayneglect any heat transfer across the valve or between the valve and its surroundings. Varythe heat transfer coefficient over a range of values to represent adiabatic conditions,

isothermal conditions and at least one value in between. I suggest K (BTU/secR) = 0,0.0001, 0.001 and (or, more realistically, 1).

1a. Draw a bond graph of your system model, clearly identifying key quantities andpower fluxes.

1b. Choose state variables and develop (nonlinear) state equations.

2a. Simulate the transient changes in (a) pressure (b) temperature and (c) mass flowrate into the right cylinder.

2b. An interesting aspect of this problem is that since thermal systems are composed

only of capacitors and resistors, and we might intuitively expect their transientresponses to exhibit no overshoot. You should be able to show that this is notalways true. Comment and explain.

2c. Check the veracity of your simulation by computing the equilibrium conditions ineach tank after the transients have expired. (Calculating equilibrium states beforeand after is standard engineering thermodynamics.)

52


59/289


3. A much more common situation (as any SCUBA diver knows) is to fill a cylinderfrom a (nominally) constant-pressure supply. One way to model this situation is toassume the left-hand cylinder in the diagram above is extremely large (so thatwithdrawing air from it changes its pressure very little). Use the model youdeveloped to simulate this situation. What is the minimum set of properties needed

to specify the air supply?

4. Modify your model to include heat transfer across the valve due to any temperaturedifference between the two cylinders. (Just show me a bond graph. Equations andsimulation are optional.)

5. The pressure difference above was chosen to keep valve flow in the non-chokedflow regime. Of course, realistic pressure differences are much higher. (A typicalSCUBA tank is routinely filled to 3,000 psi.) Revise your model and simulation tocover this situation. In this case, how valid is the assumption of no heat transfer atthe valve?

53


60/289

2.141, Assignment#5 21 November, 2002

Matter-Transport and Convection

Will Booth

54


61/289

W. Booth

08/18/04

1

1 Nomenclature

Arabic

A1 [in2] Surface area of tank 1

A2 [in2] Surface area of tank 2

At [in2] Valve (throat) cross-sectional area

Cd none Coefficient of discharge for air expansion on valve exit

Cv_air [ft-lbf/lb-mol-R] Specific heat of air at constant volumeD1 [in] Diameter of tank 1

D2 [in] Diameter of tank 2

Dt [in] Diameter of valve (throat)

h [ft-lbf/sec-in2-R] Heat transfer coefficient (tankenvironment)

L1 [in] Length of tank 1

L2 [in] Length of tank 2

Mair [1/mol] Molecular weight of air

N1 [lb-mol] Mass of air in tank 1

N2 [lb-mol] Mass of air in tank 2

P1 [psi] Pressure in tank 1

P2 [psi] Pressure in tank 2

Ru [ft-lbf/lb-mol-R] Universal gas constant

S1 [ft-lbf/R] Total entropy of tank 1S2 [ft-lbf/R] Total entropy of tank 2

dSenv1/dt [ft-lbf/sec-R] Entropy flux to the environment from tank 1

dSenv2/dt [ft-lbf/sec-R] Entropy flux to the environment from tank 2

dSvlv/dt [ft-lbf/sec-R] Entropy flux across valve

T1 [R] Temperature of tank 1

T2 [R] Temperature of tank 2

Tamb [R] Ambient temperature

V1 [in3] Volume of tank 1

V2 [in3] Volume of tank 2

Greek

none ratio of specific heats (cp/ cv)

1 [lb-mol/in3] density in tank 12 [lb-mol/in

3] density in tank 2

55


62/289

W. Booth

08/18/04

2

2 Results

Part 1 Bond graph and state equation derivation

The bond graph for the two-tank / valve system is shown below. State variables S1, N1, S2, N2are chosen.

C R

0

0

0

0

C

R R

0

Se

su

.sd

.s2

.s1

.

senv2

.senv1.

N1

.N2

.Nu

.Nd

.

T1 T2

Tamb

::

1

2

: :

: hh :

Sf: 0Sf: 0P1 P2

Figure 2-1: Bond graph for 2-tank system with throttling through a valve and heat transfer to environment.

State equations are derived as follows:

Looking at the throttling process with respect to the throat, relative to the upstream

conditions:

Following the logic outlined in the notes, the non-bulk flow terms can be neglected, because

we are satisfying continuity with:

tm

in

d

d tm

throat

d

d

i.e. no leakage. Therefore, the remaining terms in a power balance are the "flow work rate",

and the "kinetic energy transport rate", to use the suggested terminology.

Thus, a power balance from input to throat is:

Pu

Mair u

vu2

2

tmu

d

d

Pt

Mair u

vt2

2

tmt

d

d

56


63/289

W. Booth

08/18/04

3

where the units of are [lb-mol / in3] and the units of M airare [1 / mol]. Subscript ' u' denotesupstream parameters, and subscript 't' denotes parameters at the throat.

Cancelling mass flow rates, and assuming there is negligible upstream velocity, re-arranging,

gives:

vt2

Mair

Pu

u

Pt

u

i.e. in terms of molar mass, the mass flow-rate at the throat is:

tmt

d

d tMairN

d

dMair t At

2

Mair

Pu

u

Pt

u

and making the suggested assumption that all the flow work goes into speeding up the flow,

and not compressing the air:

where the units of N are [lb-mol] and the units

of P are [psi]

t

Nd

d

At

2 u

Mair

Pu Pt

A problem arises, since from the ideal gas law:

Pu

Pt

Tu

Tt

Therefore, to get mass flow, we require Pu> Pt, which implies Tu> Tt, and therefore, uu>

ut, and hu> htwhich disagrees with the common assumption that throttling is isenthalpic.

To reconcile this inconsistency, it is recognised that we measure the flow, pressure and

temperature downstream of the throttle, therefore mixing will occur between throat-flow and

the downstream gas. If this mixing is assumed to occur at constant pressure, and continues

until Tu= T

dthen, the process will be isenthalpic, as commonly assumed.

To determine equations for entropy flux, per the definition of specific entropy:

sS

Ntherefore

The entropy flux in a particular tank is, given the sign conventions in the system bond graph

tS

d

d

S

N tN

d

d

tSenv

d

d (in terms of total entropy)

i.e. the entropy flux is equal to the specific entropy multiplied by the mass flux minus the

change in entropy due to heat transfer to the environment.

Specifically:

tS1

d

d

S1

N1 tN1

d

d

tSenv1

d

d

and

tS2

d

d

S2

N2 tN2

d

d

tSenv2

d

d

57


64/289

W. Booth

08/18/04

4

PN

VRu T

and since the volume of each tank is fixed:

(where Ruis the universal gas constant)P Ru T

The pressure in the chamber can be calculated from the ideal gas law, since we are

assuming that we can model air as an ideal gas, therefore:

(where Tiis the initial temperature, and S iis the reference

entropy)

T

Ti

expS Si

M N cv

Therefore:

M N cv dT T dS

So:

(where M is the molar weight, and N has units of lb-mol)U M N cv T

From the definition of internal energy:

(since the tank is of fixed volume)dU T dS P dV T dS

Therefore, in terms of total quantities:

(specific quantities)du T ds Pdv

Recalling the expression of the 1st law from assignment 3:

But, to determine the heat lost to the environment and the pressure in each tank, we need an

expression for the temperature of each tank, which can be derived as follows:

since all the mass which leaves tank 1 arrives in tank 2 (assumption of no leakage)

tN1

d

d tN2

d

d

Choosing state variables, S1, N1, S2, and N2, the above equations are 2 of the state

equations, and the other 2 are the equations for mass flowrate through the throttle above,

where

An attempt at linearization of the equations for the 3-port capacitor with the mechanical work port closed, is shown

in Appendix A. This system could then be coupled with the linearized four port resistor model shown in the class

notes, to provide some insight to the system. I could not successfully linearize the capacitor model, however,because the cross-partial terms representing the couple between domains werent equal, therefore not satisfying

Maxwells reciprocity conditions.

Part 2a Simulation results

Matlab m-files for all the simulations are contained in Appendix B. Results for simulations with h=0, 0.0001, 0.001,

0.01 and 1 ft-lbf / sec-in2-R are shown in Figure 2-2. I used values for h suggested in the problem statement

(directly in ft-lbf/sec-in2-R as opposed to BTU/sec-R), but when multiplied by the surface area of the tanks, theresult is approximately equivalent hence the results show the range of responses intended by the suggested values.

58


65/289

W. Booth

08/18/04

5

Figure 2-2: Simulation results for part 2a with various values of h

One aspect of note, regarding the simulation, came about in discussion of this problem set with Tom Bowers. He

had found that the solution of the problem was more efficient and less noisy when using a stiff ODE solver, likeode15s, rather than a non-stiff solver like ode45. Matlab does not offer a definition of stiffness, so I went looking

on the web for a definition of a stiff vs. non-stiff system. One reference indicates there is no exact definition of

stiffness, but that when using Runge-Kutta solvers of order 4 or 5, symptoms of stiffness are non-efficient numericalsolution (small time steps) and oscillations in the response. In addition, if I had been able to linearize my system

equations, I could have plotted the eigenvalues for the initial time, and then at some later time, and looked to see that

they move significantly, indicating the system is stiff.

When I solve my system equations with no heat transfer (h = 0) and using ode45, and vary the relative tolerance I

get the following results:

RelTol time (sec) successful steps failed attempts

1.e-3 1.1 57 1

1.e-5 8.74 565 71.e-7 93.21 5587 8

whereas with ode15s solver:

RelTol time (sec) successful steps failed attempts

1.e-3 0.44 23 5

1.e-5 3.57 245 152

1.e-7 1.54 110 55

59


66/289

W. Booth

08/18/04

6

These results, suggest symptoms of a stiff problem, and the plot of ode45 results for the first two values of relativetolerance shows that when the tolerance is large, the response is oscillatory, whereas if the tolerance is much

reduced, the solver is less efficient, but produces a smooth response.

Figure 2-3: Simulation results for ode45 solver with different relative tolerances

The initial conditions of the system are shown in the Table 2-1, with corresponding units given in section 1.

N1(init) T1(init) P1(init) S1(init) N2(init) T2(init) P2(init) S2(init)

0.0048 529.67 20 174.5244 0.0035 529.67 14.7 129.9494

Table 2-1: Table of initial conditions

For the adiabatic process (h=0), by definition there is no heat loss to the environment, therefore, the initial pressuredifference between the two tanks drives mass flow from left to right, until the pressures are in equilibrium, at which

point there is no further changes. The mass of air in tank 1 remains greater than the mass in tank 2, whereas the

temperature in tank 2 is greater than the temperature in tank 1, i.e., N1> N2, and T1< T2. What tank 1 loses inpressure, it must also lose in temperature, from the ideal gas law. Comparing the equilibrium conditions in Table

2-2 with the initial conditions in Table 2-1, the total entropy decrease in tank 1, is equal to the total entropy increasein tank 2. As expected, the ideal system conserves entropy.

Part 2b & 2c Transient Overshoot and Tank Equilibrium Conditions

Figure 2-2 shows that as the heat transfer between the tanks and the environment increases, the following occurs:

1. Settling time of pressure transient increases

2. Temperatures settle to ambient (if h 0) and settle faster as h increases.3. Total entropy of the tanks return to their initial values (while the entropy of the environment increases)4. Mass of air in the two tanks equalizes, as the tank temperatures reduce to ambient

60


67/289

W. Booth

08/18/04

7

I had not expected results 1,3 or 4. I expected the pressure transients to settle independent of the heat transfercoefficient. The reason this is not the case, is because of the coupling between the pressure, temperature and mass

flowrate, i.e. the fact that this is a 3-port capacitor. The pressure difference between the tanks dominates the initialresponse, but thereafter, it is the heat transfer to the environment, which governs the settling time of the system.

Some of the initial entropy produced is lost to the environment, therefore the temperature of tank 2 does not initially

increase as rapidly as in the adiabatic case, nor does tank 1s temperature decrease as rapidly. Therefore, thepressure difference between the two tanks is maintained for longer, until the response is governed primarily by the

heat transfer to the environment, and thus, the pressure difference can only decay as quickly as the pressure

difference (see settling times for h=0.001 and h=0.01 in Table 2-3.

The pressure difference between the two tanks is plotted in Figure 2-4.

Figure 2-4: Pressure difference between tanks 1 and 2

Given that the R elements all have temperature in, entropy flow rate out (i.e. conductance) causality on both

ports, and through-power flow, these elements guarantee satisfaction of the 2ndlaw of thermodynamics. It is not theabsolute value of entropy which can never decrease, it is the rate of entropy production that must never decrease,

therefore my initial surprise that the entropy of tank 1 initially decreases and subsequently returns to its initial value

is expected. This is due to the fact that when T1> Tambthen the entropy of tank 1 will decrease (disregarding mass

transfer for the time being), but if T1< Tambthe entropy of tank 1 will increase.

An interesting result occurs when the heat transfer coefficient is low, i.e. h=0.0001 ft-lbf/sec-in2-R. The pressuredifference decays to the adiabatic equilibrium condition of ~17.5psi almost as quickly as the adiabatic case itself, i.e.

the settling time for the initial temperature/entropy and pressure/mass transients is 22sec. After which, the heatdrain to the environment causes the pressure in both tanks to decay to the isothermal equilibrium conditions (for

h=1) over a much longer timer (>600sec). At first glance, there is no pressure difference to cause this final mass

Documents

2 141 Fall 2002 Modeling and Simulation of Dynamic Systems Volume III