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

2.141 — Fall 2002 Modeling and Simulation of Dynamic Systems Volume III Neville Hogan 2.141 — Fall 2002 Modeling and Simulation of Dynamic Systems Neville Hogan MIT OpenCourseWare Cambridge, MA http://ocw.mit.edu/ © 2006 Massachusetts Institute of Technology The material in this book is provided under a Creative Commons license that grants you certain privileges to use, copy, or adapt the contents for non-commercial educational purposes as long as you give credit to MIT OpenCourseWare and to the faculty author, and you make any derivative works freely and openly available to others under the same terms as our license. Please refer to the full text of the license and other notices at the back of this book. Printed and bound in the United States of America. MIT OpenCourseWare Building 9-213 77 Massachusetts Avenue Cambridge, MA 02139-4307 USA ISBN <<ENTER ISBN>> 2006XXXXXX 10 9 8 7 6 5 4 3 2 1 What is MIT OpenCourseWare? 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Contents Highlights of this Course ...................................................................... 1 Course Description.............................................................................. 1 Syllabus ............................................................................................ 2 Calendar............................................................................................ 4 Lecture Notes ..................................................................................... 8 Assignments .................................................................................... 11 Assignment 3: Gas-charged Accumulator, Solution and Commentary Assignment 4: Control through Singularities, Solution and Commentary Assignment 5: Simple Convection and Throttling, Solution and Commentary Projects ........................................................................................... 75 Suggested Term Project Topics Sample Student Projects Study Materials................................................................................135 Introductory Modeling Notes Old Quiz for Study Highlights of this Course This course features an extensive collection of lecture notes plus background study materials on modeling. Course Description This course deals with modeling multi-domain engineering systems at a 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 and Hamiltonian forms and control-relevant properties. Application examples may include electro-mechanical transducers, mechanisms, electronics, fluid and thermal systems, compressible flow, chemical processes, diffusion, and wave transmission. Course Meeting Times Lectures: Two sessions / week 1.5 hours / session Level Graduate This bond graph models the free-flight and contact behaviors of a ball bouncing off of a another ball. (Image courtesy of Prof. Neville Hogan.) 1 Syllabus Prerequisites 2.151 or equivalent exposure to physical system modeling -- see me to clarify. Textbook Required Brown, Forbes T. Engineering System Dynamics. New York: Marcel Dekker, Inc., 2001. ISBN: 0824706161. Course Description This course is about Modeling multi-domain engineering systems at a level of detail suitable for design and control system implementation. It also describes Network representation, state-space models, Multiport energy storage and dissipation, Legendre transforms, Nonlinear mechanics, transformation theory, Lagrangian and Hamiltonian forms, Control-relevant properties. The application examples may include electro-mechanical transducers, mechanisms, electronics, fluid and thermal systems, compressible flow, chemical processes, diffusion, and wave transmission. 2 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 brief interim progress report will be required by lecture 16. Term projects must be completed by the last week of the term, at which time a final report will be due. A brief oral presentation of each term project will also be required. ACTIVITIES PERCENTAGES Homework 60% Term Project 40% Ethics Policy Collaboration Collaboration on and discussion of homework assignments is encouraged but each student must submit an individual solution. Collaboration on term projects is encouraged provided some means for clearly identifying individual contributions is proposed and approved by the instructor. Use of Material from Previous Years Use of material from prior offerings of this subject in preparing homework assignments defeats the purpose of the assignments and is forbidden. 3 Calendar LEC # TOPICS KEY DATES Week 1 1 Introduction Multi-domain Modeling Syllabus, Policies and Expectations Assignment 1 out Week 2 2 Review: Network Models of Physical System Dynamics Bond-graph Notation 3 Review (cont.): Equivalent Behavior in Different Domains Block Diagrams, Bond Graphs, Causality Week 3 4 Thévenin and Norton Equivalent Networks Impedance Control and Applications 5 Energy-storing Coupling between Domains Multi-port Capacitor Maxwell's Reciprocity Assignment 1 due Assignment 2 out Week 4 6 Energy, Co-energy, Legendre Transformation, Causal Assignment Intrinsic stability 4 7 Review Revisited: Magnetism and Electro- magnetism Week 5 8 Electro-magnetic-mechanical Transduction Use of Co-energy Functions Term project proposal due 9 Linearized Energy-storing Transducer Models Assignment 2 due Assignmnet 3 out Week 6 10 Cycle Processes Work-to-heat Transduction Thermodynamics of Simple Substances 11 Causal Assignments and Co-energy Functions Second Law for Heat Transfer Multi-port Resistors Week 7 12 Nonlinear Mechanical Systems Modulated Transformers and Gyrators Assignment 3 due Week 8 13 Lagrangian Mechanics Coordinates, State Variables and Independent Energy Storage Variables 14 Nonlinear Mechanical Transformations and Assignment 4 5 Impedance Control out Week 9 15 Hamiltonian Mechanics Stable Interaction Control Canonical Transformation Theory 16 Term Project Progress Report Discussion Term project progress report due Week 10 17 Identification of Physical and Behavioral Parameters Model structure 18 (Internally) Modulated Sources Non-equilibrium Multi-port Resistors Nodicity Assignment 4 due Assignment 5 out Week 11 19 Amplifying Processes Small-signal and Large-signal Models 20 Thermodynamics of Open Systems Convection and Matter Transport Lagrangian vs. Eulerian Frames Week 12 21 Power Conjugates for Matter Transport Second Law for Non-heat-transfer Processes 6 Throttling and Mixing 22 Bernoulli's Incompressible Equation The "Bernoulli Resistor" and "Pseudo-bond- graphs" Assignment 5 due Week 13 23 Chemical Reaction and Diffusion Systems Gibbs-Duhem equation Week 14 24 Control-relevant Properties of Physical System Models Causal Analysis, Relative Degree, Passivity and Interaction Stability 25 Transmission Line Models Term Project Presentations Week 15 26 Wrap-up Discussion Term Project Presentations (cont.) 7 Lecture Notes and Topics Lecture # Topic Lecture 1 Introduction; Multi-domain Modeling Lecture 2 Review: Network Models of Physical System Dynamics; Bond Graph Notation, Block Diagrams, Causality Lecture 3 Review (cont.): Equivalent Behavior in Different Domains Lecture 4 Thevenin and Nortan Equivalent Networks; Impedance Control and Applications Lecture 5 Energy-storing Coupling between Domains; Multi-port Capacitor; Maxwell’s Reciprocity Lecture 6 Energy, Co-energy, Legendre Transformation, Causal Assignment; Intrinsic stability Lecture 7 Review Revisited: Magnetism and Electro-magnetism Lecture 8 Electro-magnetic-mechanical Transduction; Use of Co-energy Functions Lecture 9 Linearized Energy-storing Transducer Models Lecture 10 Cycle Processes; Work-to-heat Transduction; Thermodynamics of Simple Substances 8 Lecture 11 Causal Assignments and Co-energy Functions; Second Law for Heat Transfer; Multi-port Resistors Lecture 12 Nonlinear Mechanical Systems; Modulated Transformers and Gyrators Lecture 13 Lagrangian Mechanics; Coordinates, State Variables and Independent Energy Storage Variables Lecture 14 Nonlinear Mechanical Transformations and Impedance Control Lecture 15 Hamiltonian Mechanics; Stable Interaction Control; Canonical Transformation Theory Lecture 16 Term Project Progress Report Discussion Lecture 17 Identification of Physical and Behavioral Parameters; Model Structure Lecture 18 (Internally) Modulated Sources; Non-equilibrium Multi-port Resistors; Nodicity Lecture 19 Amplifying Processes; Small-signal and Large-signal Models Lecture 20 Thermodynamics of Open Systems; Convection and Matter Transport; Lagrangian vs. Eulerian Frames Lecture 21 Power Conjugates for Matter Transport; Second Law for Non- heat-transfer Processes; Throttling and Mixing 9 Lecture 22 Bernoulli’s Incompressible Equation; The “Bernoulli Resistor” and “Pseudo-bond-graphs” Lecture 23 Chemical Reaction and Diffusion Systems; Gibbs-Duhem equation Lecture 24 Control-relevant Properties of Physical System Models; Causal Analysis, Relative Degree, Passivity and Interaction Stability Lecture 25 Transmission Line Models; Term Project Presentations Lecture 26 Wrap-up Discussion; Term Project Discussions (cont.) 10 Massachusetts Institute of Technology Department of Mechanical Engineering 2.141 Modeling and Simulation of Dynamic Systems Assignment #3 Out: 10/3/02 Due: 10/17/02 Gas-charged accumulator Gas-charged accumulators are commonly used in hydraulic systems, primarily to reduce the magnitude of the pressure transients resulting from abrupt changes in flow rate. The attached paper 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 (PV n = 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 to support 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 the Dynamic 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 and represent it by a bond graph. Assuming the flow rate Q a as 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 of the "anelastic model" of P. & O. fig. 2. 2b. Develop a bond graph corresponding to your linearized model using single-port storage and dissipative elements and idealized transducers (e.g. gyrators, transformers) and show that it yields the same linearized equations. 2c. Does your linearized model describe entropy production? Is entropy production an essentially 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 plots corresponding to P. & O. '84 figs. 1, 3 & 4. 3b. Comment on the degree of (dis)agreement between these models and P. & O.'s experiments and simulations; that is, how plausible is the ideal gas assumption in this case? (Bear in mind that your nonlinear model is different from the one P. & O. used -- see their reference #3.) 2.141 2002 page 1 assignment #3 11 4. Develop a model of the complete hydraulic system of P. & O. '84, fig. 5 and represent it by a bond graph. Assuming the flow rate Q s and the force F as inputs and that nitrogen may be modeled as an ideal gas, derive nonlinear dynamic equations suitable for simulating the transient response of the system. Hint: Keep in mind that your choice of state variables can influence the complexity of your equations. 5. Assuming that both inputs are zero (Q s = 0, F = P o A m ) simulate the pressure in the accumulator in response to an initial velocity of the mass in the direction which would compress 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. Choose parameters 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 linear actuator 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 times higher (i.e., 0.1Hz) and repeat 5c. above. Based on this analysis, what would you conclude about the importance of the "thermal damping" phenomenon? 2.141 2002 page 2 assignment #3 12 Massachusetts Institute of Technology Department of Mechanical Engineering 2.141 Modeling and Simulation of Dynamic Systems 2.141 Assignment #3: GAS-CHARGED ACCUMULATOR The figure below (after Pourmovahed & Otis, 1984) is a schematic diagram of the hydraulic and mechanical system. gas oil motor m F y piston or diaphragm Q m Q s A m P Q a P P, v accumulator Our first goal is to model the thermofluid subsystem assuming the flow rate Q a as an input. The following bond graph is the simplest representation of the accumulator subsystem. R T w dS w /dt C P –dV/dt dS/dt S f S e Q a (t): hydraulic domain thermal domain T :T w The two-port capacitor on the left represents the reversible energy storage and work-to-heat transduction in the ideal gas in the accumulator. The two-port resistor on the right represents irreversible heat conduction through the wall of the accumulator with the concomitant entropy production. The outside of the wall is assumed to remain at a constant temperature (i.e., ambient temperature). The direction of the power bonds has been chosen to follow the usual convention that power is positive into the storage element. However, the sign convention for the two-port resistor follows a more intuitive “power in = power out” convention but assumes for convenience that positive heat flow is from the wall to the gas. Nonlinear equations Causal assignment indicates that both ports of the capacitor may be given the preferred integral causal form. Note that the two-port resistor assumes its preferred causal form with temperatures as 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 equally simple. Gas-charged accumulator solution page 1 Neville Hogan 13 Hydraulic side: We could choose either pressure, P, or volume, V, as state variable but volume seems to be the most sensible choice as the corresponding state equation is simply ˙ V = –Q a (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. We could also use total internal energy, U, as a state variable. To choose between these, consider the available information. Two-port capacitor constitutive equations are derived assuming nitrogen may be modeled as an ideal 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 c v dT where c v is the specific heat at constant volume. Integrating: U - U o = m c v (T – T o ) where subscript o denotes reference state. The two-port resistor equations assume Fourier’s-Law (i.e., linear) heat conduction. ˙ Q = h A w (T w – T) where h is a heat transfer coefficient and A w and T w are the area and temperature of the wall. Consider using S and V as state variables. T ˙ S = ˙ Q = h A w (T w – T) hence ˙ S = h A w ( T w T – 1) However, I need T = T(S,V) and P = P(S,V) to form state equations. From class notes, T = T o \ | . | | V V o – R c v exp \ | . | | S – S o mc v P = P o \ | . | | V V o – \ | . | | R c v + 1 exp \ | . | | S – S o mc v These are output equations. State equations are Gas-charged accumulator solution page 2 Neville Hogan 14 ˙ S = h A w \ | . | | | T w T o \ | . | | V V o R c v exp \ | . | | S o – S mc v – 1 ˙ V = –Q a These are highly nonlinear, coupled differential equations. Note that the “thermal time-constant” is far from obvious. Consider using U and V as state variables. Pressure and temperature are determined as follows by rearranging the expression for internal energy. T = U m c v + T o P = m RT V = m R V \ | . | | U m c v + T o These are output equations. State equations may be found from an energy balance equation (the first law): ˙ U = ˙ Q – P ˙ V ˙ V = –Q a ˙ U = –h A w \ | . | | U m c v + T o – T w + m R V \ | . | | U m c v + T o Q a These equations are also nonlinear and coupled. However, the “thermal time constant” 1 t = h A w m c v is clearly identified as may be seen by re-arranging. ˙ U = – \ | . | | h A w m c v U + h A w (T w – T o ) + m R V \ | . | | U m c v + T o Q a Note that choosing absolute zero for the reference state would simplify this equation to ˙ U = – \ | . | | h A w m c v U + h A w T w + U V \ | . | | R c v Q a Finally consider using T and V as state variables. (Note that this is loosely analogous to choosing displacement and velocity — the Lagrangian choice — as the state variables to describe a mechanism.) ˙ U = ˙ Q – P ˙ V ˙ U = m c v ˙ T = h A w (T w – T) + m RT V Q a Gas-charged accumulator solution page 3 Neville Hogan 15 ˙ T = – h A w m c v T + h A w m c v T w + R c v T V Q a ˙ V = –Q a 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 less important. 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 state variable. That shouldn’t be surprising given the simple relation between temperature and internal energy for and ideal gas. The point to remember is that the integral causal form does not constrain your choice of state variables. State variables should be chosen for convenience and to maximize insight. I will use temperature and volume as state variables. Linearized equations The second task is to linearize the model and compare to P&O’s linearized model. Linearizing the state equations is always helpful. In this case it can provide insight into the physical system and 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 form—any of the four causal forms will serve. Use the causal form with temperature and volume as (integrated) inputs, i.e. C P –dV/dt dS/dt T P = m RT V Denote the operating point by subscript o and linearize. P – P o - m R V o (T – T o ) – m RT o V o 2 (V – V o ) Write AV = V – V o , AT = T – T o and AP = P – P o A - m R V o AT + m RT o V o 2 (–AV) Remember that the positive work compresses the gas, hence –AV is the appropriate displacement variable on the hydraulic side. This indicates that the hydraulic side of the linearized model may be represented by a transformer between thermal and hydraulic domains connected to a fluid capacitor by a 1- junction (common flow). transformer modulus: m R V o Gas-charged accumulator solution page 4 Neville Hogan 16 (inverse) fluid capacitance: m RT o V o 2 The second constitutive equation is in the form S = S(T,V) It may be derived from the internal energy equation and the first law. dU = m c v dT = T dS – P dV dS = m c v T dT + m R V dV ˙ S = m c v T o ˙ T – m R V o (– ˙ V) Again, remember that – ˙ V is the appropriate flow variable of the hydraulic side given the power sign we have assumed. This indicates that the thermal side of the linearized model may be represented by a thermal capacitor connected to a transformer between thermal and hydraulic domains by a 0-junction (common effort). transformer modulus: m R V o (as before) thermal capacitance: m c v V o A bond graph of the linearized two-port capacitor is as follows. TF –dV/dt T 1 0 C C : m c v /T o m R/V o : V o 2 /m R T o Note that this linearized model indicates that the strength of the coupling between the thermal and mechanical domains is proportional to the mass of gas and inversely proportional to the volume it occupies—a useful insight. The two-port resistor equations describe entropy flow as a function of temperature. For the gas ˙ S = ˙ Q T = h A w \ | . | | T w T – 1 Linearizing ˙ S – ˙ S o - – \ | . | | h A w T w T o 2 (T o – T) ˙ S - h A w \ | . | | T w T o – 1 – \ | . | | h A w T w T o 2 (T o – T) Gas-charged accumulator solution page 5 Neville Hogan 17 Assume the operating point (and also the reference thermal state) is at equilibrium with the wall temperature. T o = T w ˙ S - – \ | . | | h A w T o (T o – T) Under the same operating conditions, the entropy flow rate from the wall is identical ˙ S w = ˙ Q T w ˙ S w - – \ | . | | h A w T o (T o – T) A bond graph of this linearization of the two-port resistor is as follows. S e dS/dt 1 R : T o /h A w :T w Note that this linearized model does NOT reflect the second law of thermodynamics— linearization has eliminated entropy production due to heat transfer. This is due to the particular choice of operating point. By choosing T o = T w we make the entropy flow from the wall identical to the entropy flow to the gas. ˙ S net = ˙ S – ˙ S w = 0 However, entropy production is not an essentially nonlinear phenomenon—it’s the choice of operating point that matters. Linearization about equilibrium eliminates entropy production; a model linearized about a non-equilibrium operating point will describe entropy production. A bond graph of the complete linearized system is as follows. S e dS/dt 1 R : T o /h A w :T w TF –dV/dt T 1 0 C C m c v /T o : m R/V o V o 2 /m R T o : S f Q a : Using the causal assignment shown, linearized equations may be read from the graph. A ˙ T = T o m c v \ | . | | A ˙ S – m R V o A ˙ V Gas-charged accumulator solution page 6 Neville Hogan 18 A ˙ S = h A w T o AT State equations A ˙ V = –Q a A ˙ T = – h A w m c v AT + R c v T o V o Q a Output equation AP = m R V o AT – m RT o V o 2 AV To compare with P&O’s linearized equations, take the Laplace transform. AT \ | . | | s + 1 t = R c v T o V o Q a AV = – Q a s AP = m R T o V o 2 \ | . | | | R c v + s + 1 t (s + 1 t s) Q a Substitute m R T o V o 2 = P o V o R c v + 1 = R + c v c v = ¸ P Q a = P o V o ¸ t s + 1 s (t 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 transformer linking 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. Gas-charged accumulator solution page 7 Neville Hogan 19 S e dS/dt R : –F :T w TF P T 0 1 I C : 1/M A m S f Q s : S e –dV/dt v m The translational inertia is an ideal linear element. It makes little difference whether momentum or velocity is used as its state variable. For clarity I will use velocity with temperature and volume as the gas state variables. State and output equations for the complete system are as follows. ˙ V = – A m v m – Q s ˙ T = h A w m c v ( ) T w – T + R c v T V ( ) A m v m + Q s ˙ v m = 1 M \ | . | | m R A m T V – F P = m RT V where A m is the piston area, v m is the velocity of the inertia and M its mass. Gas-charged accumulator solution page 8 Neville Hogan 20 Parameters for simulation For P&O fig. 1: P o = 1014.5 psia (operating pressure in the accumulator) V o = 103 cu. in. (operating volume of the accumulator) m = 0.293 lbm (mass of gas) AV/V o = ±0.25 For P&O fig. 3: P o = 438.5 psia V o = 122 cu. in. m = 0.15 lbm AV/V o = ±0.05 For P&O fig. 4: P o = 436.7 psia V o = 122 cu. in. m = 0.149 lbm AV/V o = ±0.01 In all cases: T w = 540°R (room temperature) R = 660 in-lbf/lbm °R (gas constant) c v = 1900 in-lbf/lbm °R t = m c v h A w = 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 domain from participating in the dynamics. The undamped natural frequency is given by e n = P o A m 2 M V o hence the estimated mass is M = P o A m 2 V o e n 2 Gas-charged accumulator solution page 9 Neville Hogan 21 Assuming a piston diameter of 0.75 inches, A m is 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 0.44 2 102.97 x (2 t x 0.01) 2 = 487 lbf in 2 in 4 in 3 sec 2 rad 2 Convert to more meaningful units: 1 lbf sec 2 in = 386.4 lbm M = 1.88 x 10 5 lbm. Thus the mass required to yield an undamped natural frequency that low is unbelievably large— 84 tons! However, the mass required to yield an undamped natural frequency of 0.1 Hz is a little more reasonable—1,880 pounds. To choose an initial velocity I assumed undamped oscillations and used v m = ˙ V A m V = AV sin e n t v m (0) = AV e n A m 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 0.5 P&O fig. 1, nonlinear normalized volume n o r m a l i z e d p r e s s u r e Gas-charged accumulator solution page 10 Neville Hogan 22 -0.05 0 0.05 410 420 430 440 450 460 470 P&O fig. 3, nonlinear normalized volume p r e s s u r e ( p s i ) -0.1 -0.05 0 0.05 0.1 380 400 420 440 460 480 500 P&O fig. 4, nonlinear normalized volume p r e s s u r e ( p s i ) Note that in all cases these simulations are quite close to P&O’s even though those authors used a much more elaborate model of the gas. For 10% volume changes, the ideal gas model is essentially identical to their data. For 5% volume changes the ideal gas model is essentially linear. Gas-charged accumulator solution page 11 Neville Hogan 23 Transient responses Simulations of the transient responses for the large mass assuming thermally damped, adiabatic and 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) v e l o c i t y ( i n c h / s e c ) -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 n o r m a l i z e d p r e s s u r e The plot shows the velocity of the piston mass. Note that the oscillation is steadily losing amplitude as time increases. Also note that no mechanical losses such as friction have been included in our model. This loss in energy is due to the irreversible entropy production due to conduction in the gas-charged accumulator. Assuming only a reversible work storage in the gas accumulator would not predict this behavior. Just like a bicycle pump, compression of the gas generates heat that flows through the walls of the chamber. Not all of this heat energy is recovered 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 entropy flows which in this case means no heat flow. The result of this simulation shows oscillations as in the previous case. However, the amplitude of the oscillations does not decrease. We have essentially removed the heat lost through the wall leaving only the reversible work storage in compressing the gas. The system without thermal damping behaves like an undamped spring- mass system. Gas-charged accumulator solution page 12 Neville Hogan 24 0 50 100 150 200 250 300 350 400 -4 -2 0 2 4 Adiabatic transient response time (seconds) v e l o c i t y ( i n c h / s e c ) -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 n o r m a l i z e d p r e s s u r e The isothermal case was modeled by setting dT/dt = 0 and setting the initial gas temperature equal to T w . 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 4 Isothermal transient response time (seconds) v e l o c i t y ( i n c h / s e c ) -0.3 -0.2 -0.1 0 0.1 0.2 0.3 -0.4 -0.2 0 0.2 0.4 normalized volume n o r m a l i z e d p r e s s u r e The isothermal and adiabatic cases are undamped, as expected. Note that the isothermal frequency of oscillation is lower than the damped case which in turn is lower than the adiabatic case. This is also as expected. Gas-charged accumulator solution page 13 Neville Hogan 25 Finally, a simulation of the transient response for the smaller (more reasonable) mass is as follows. 0 20 40 60 80 100 120 140 160 180 200 -40 -20 0 20 40 Transient response for reduced piston mass time (seconds) v e l o c i t y ( i n c h / s e c ) -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 n o r m a l i z e d p r e s s u r e Note that even though the normalized pressure-volume plot shows very little evidence of energy dissipation (i.e., the area inside the “loop” is small) the effect on the transient response is substantial. I conclude that “thermal damping” is likely to be a real phenomenon in practical dynamic systems. Gas-charged accumulator solution page 14 Neville Hogan 26 Not used: U = mc v T o ¸ ¸ ( ( ( \ | . | | V V o – R c v exp \ | . | | S – S o mc v – 1 + U o Choose an operating point at thermal equilibrium with the environment. Assume no forcing, i.e., Q a = 0 T o = T w For P&O fig. 4: P o = 69.95 bar = 1014.5 psia (operating pressure in the accumulator) V o = 1687cc = 103 cu. in. (operating volume of the accumulator) Gas properties of nitrogen @ 300°K (room temperature): c p = 1041 Joules/kg °K (specific heat at constant pressure) ¸ = 1.399 (polytropic index) R = 297 Joules/kg °K (gas constant) Remember: c v = c p ¸ = c p - R c p = 2560 in-lbf/lbm °R ¸ = 2560/1900 = 1.35 Gas-charged accumulator solution page 15 Neville Hogan 27 MEMORANDUM To: 2.141 class From: Professor Neville Hogan Date: November 26, 2002 Re: Assignment #3 All of you did well on this assignment. Maybe I’m making them too easy... One common difficulty was to conclude that entropy production is an essentially nonlinear phenomenon. This is a subtle problem: if you “did the obvious” (as I did—see my solution) and linearized the two-port resistor about conditions with the same (ambient) temperature on both sides, you linearized about equilibrium. That corresponds to assuming a reversible process and hence the linearized model will not describe entropy generation. However, if you linearize about non-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 Massachusetts Institute of Technology Department of Mechanical Engineering 2.141 Modeling and Simulation of Dynamic Systems Assignment #4 Out: 10/24/02 Due: 11/7/02 Control Through Singularities Motivation: One common and productive use of modeling and simulation is to study and understand proposed engineering designs. In this assignment you will test the behavior of a proposed robot controller. A common form of robot motion control specifies a workspace position or trajectory (e.g., a desired time-course of tool position in Cartesian coordinates) and transforms that specification to a corresponding configuration-space position or trajectory (e.g., a time-course of joint angles). However, most robot mechanisms have kinematic singularities, configurations at which the relation between workspace and configuration-space becomes ill-defined. As a result, most robot motions controllers do not operate at or near these singular points. However, an energy-based analysis of mechanics shows that the transformation of positions and velocities is well-defined in one direction while the transformation of efforts and momenta is well-defined in the other. This implies that a controller that takes advantage of these facts should be able to operate at and close to mechanism singularities. A “simple” impedance controller attempts to impose the workspace behavior of a damped spring connected to a movable “virtual position”. ( ) ( ) ( ) ( ) ( ) { } J x B L x K J & & ÷ + ÷ = o o t where u is a vector of generalized coordinates, t a vector of (conjugate) generalized efforts, x and x o are vectors of actual and virtual Cartesian tip coordinates, L() and J() are linkage kinematic equations and Jacobian respectively, and K and B are stiffness and damping respectively. Note that this controller requires neither the inverse of the kinematic equations nor the inverse of the Jacobian. In this assignment you are to test whether this simple impedance controller can operate at and close 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-torque actuators, 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 provide measurements of joint angle and angular velocity. 2.141 2002 page 1 assignment #4 29 1. Write kinematic equations relating generalized coordinates to the position coordinates of the tip, expressed in a Cartesian coordinate frame with its origin at the axis connecting the inner link 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 tip Cartesian coordinates and joint angles becomes ill-defined) and show that they include the center of the workspace as well as at the limits of reach. 4. Formulate a dynamic model of the mechanism relating input actuator torques to output motion of the tip in Cartesian coordinates. Assume the links are rods of uniform cross section and mass m = 0.5 kg and that the joints are frictionless. (Hint: be careful in your choice of generalized 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 k and b are constants). ( ¸ ( ¸ = 1 0 k K 0 1 0 1 ( ¸ ( ¸ = 1 0 b B 5.a Choose the stiffness and damping matrices so that when the mechanism is making small motions about a configuration with the inner link aligned along the Cartesian x-axis and the outer link aligned parallel to the Cartesian y-axis the highest-bandwidth transfer function between virtual and actual position has critically-damped poles with an undamped natural frequency of 2 Hz. (Hint: it may be easiest to first transform the stiffness 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 speed profile with acceleration to peak speed in 250 ms, constant speed for 1.5 sec and deceleration to rest in 250 ms. 2) Starting from rest at {x = -L, y = 0} ending at rest at {x = L, y = 0}; trapezoidal speed profile with acceleration to peak speed in 250 ms, constant speed for 1.5 sec and deceleration to rest in 250 ms. 3) Starting from rest at {x = -L, y = L/20} ending at rest at {x = L, y = L/10}; trapezoidal speed 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 k and b chosen to yield critically-damped poles with an undamped natural frequency of 20 Hz. 7. Repeat the simulation of 5.b.3 with the outer link 10% shorter than the inner link. How does this change the set of singular configurations? 2.141 2002 page 2 assignment #4 30 8. Comment briefly on (a) whether and (b) how well this controller operates near mechanism singularities. “Tent-Pole” Instability The complex behavior of multi-body mechanical systems stems from kinematically modulated transformation of energy. It can give rise to some highly counter-intuitive phenomena, one of which 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 not completely rigid as the guy ropes inevitably have some elasticity. It may seem natural to increase the tension in the ropes to stiffen the support but in fact, this can have the opposite effect. To gain 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 base so that it can only move in a plane. Assume that two elastic cords are attached to its top and to the 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, and pretension, F p . 1. Develop a model to describe the mechanical system comprised of the pole and the two cords. To understand this system, simulate the following cases: 2. Assume the cords have pretension but zero stiffness; assume the pole starts in the upright position; and assume no gravity (for now). Show that this mechanical system is unstable for any value of F p even in the absence of gravity. 3. Assume the cords have non-zero stiffness but no pretension; assume the pole starts in the upright position; and assume no gravity. Test whether this system is stable for (a) small perturbations (e.g., initial conditions 1º from upright) and (b) large perturbations (e.g., initial conditions 45º from upright). 4. Now that you understand what is going on, include both stiffness and pretension in the cords and include gravity and any other physical behavior relevant to system stability. Using the numerical simulation, show that for a given cord stiffness there is a maximum pretension for stable behavior and find how it varies with stiffness (a few representative values 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 for stable upright behavior (i.e., corresponding to 4. above). How well do your numerical simulation results agree with your analytical results? 2.141 2002 page 3 assignment #4 31 Thomas A. Bowers 2.141 Fall 2002 Assignment 4: Kinematics Two-Link Planar Robotic Mechanism 1. The configuration being analyzed in this problem is shown below in Figure 1. u 2 u 1 L 1 L 2 (x,y) (x c1 , y c1 ) (x c2 , y c2 ) Figure 1: Cartesian and Generalized Coordinates of 2-Link Robotic Mechanism The equations relating the Cartesian coordinates to the generalized coordinates, u 1 and u 2 , are as follows: x = L 1 cosu 1 + L 2 cosu 2 y = L sinu + L 2 sinu 2 1 1 L 1 c1 = cosu 1 2 L 1 y = sinu 1 2 c1 = L 1 cosu + L 2 cosu 2 2 x c2 1 y = L 1 sinu + L 2 sinu 2 2 c2 1 The kinematic equations relating the generalized and Cartesian coordinates can be found from derivation. x ÷ = L u sinu ÷ L 2 u sinu 1 1 1 2 2 y = L u cosu + L u cosu 1 1 1 2 2 2 32 ÷ L 1 u 1 x c1 = sinu 1 2 L 1 u 1 y = cosu 1 c1 2 L 2 u 2 x c2 ÷ = L 1 u sinu ÷ sinu 2 1 1 2 2 y = L 1 u cosu + L 2 u cosu 2 c2 1 1 2 2. The Jacobian matrices for the tip and the centers of mass of the mechanism can be found directly from the kinematic equations: ÷ L 1 sinu ÷ L 2 sinu 2 ( = J tip ¸ L 1 cosu 1 L 2 cosu 2 ¸ ( 1 ( ÷ L 2 1 sinu 1 0 ( L 1 cosu 1 0 ( ( 2 ( 0 ( J = 1 L 2 ( cm ÷ L 1 sinu ÷ sinu 2 ( 1 2 L 1 cosu 1 L 2 cosu 2 ( ( 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 so there is no transformation from Cartesian to generalized coordinates. det J ÷ = L L 2 (sinu cosu 2 ÷sinu 2 cosu 1 ) = sin(u ÷u 2 ) = 0 1 1 1 u ÷u = , 0 t....kt 1 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 be determined. The links each have a translational and rotational inertia: I = m mL 2 J = 12 The mass matrix for the system is then: 33 2 ( m 1 0 0 0 0 0 0 m 1 0 0 0 0 2 L m 1 0 0 1 0 0 0 12 ( ( ( ( ( ( ( ( ( ¸ M = 0 0 0 m 2 0 0 0 0 0 0 m 2 0 L m 2 2 0 0 0 0 0 2 12 ¸ This can be converted to the generalized inertia of the system. T MJ J I = T L 1 L 1 ( ( sin 1 u m 1 0 0 0 0 0 u sin 1 ÷ 0 ÷ 0 ( ¸ 2 ( ( ( ( ( ( ( ( ( ( ( ¸ u 2 u ¸ ( ( ( ( ( ( ( ( ( ( ( ¸ 2 2 ( ( ( ( ( ( ( ( ( 0 1 0 0 0 0 L 1 m L 1 2 1 1 0 u u cos 1 0 0 cos 2 L m 1 1 0 0 0 0 0 1 0 12 0 0 0 m 0 0 2 0 0 0 0 m 2 0 = 1 ÷ u L 2 2 ÷ u 1 L 2 2 u sin 2 L 1 ÷ L 1 ÷ sin sin sin L L L m 2 2 2 2 2 2 2 1 0 1 u u 1 0 1 u cos 2 L 1 L 1 cos cos cos 0 0 0 0 0 ¸ 12 ¸ 2 2 L 1 L m 1 1 u ( L L 2 1 sin 1 sin 2 ( ) ( 2 sin ) 2 2 2 1 1 L L 1 2 u sin 1 u ( u u ) 1 u m 1 u sin 2 u 1 L m 2 2 u cos L m 1 2 + + + + + ¸ m 1 cos cos cos 2 4 12 2 = 2 L 2 4 (sin 2 ) 2 u 2 L L 1 u cos 1 u ( 2 u 2 L L 1 u cos 2 sin + + + m cos m 2 2 2 12 2 2 ( ( ( ( ( ¸ L m 1 1 L m 1 1 u ( 1 ¸ = ( ( ( ( ¸ ) 2 2 2 m cos µ 2 m cos 2 u ÷ 2 L m 1 2 L m 1 2 + + ¸ 2 2 3 2 3 2 = 2 2 2 2 L L 1 L m 2 3 L L 1 µ L m 2 2 2 3 u ( 1 u ÷ 2 ) 2 cos m 2 2 m cos 2 2 This expression of the generalized inertance is used to find the co-energy of the system: e 1 T * ) ( e u E I = k 2 2 L m 1 L L 1 1 L m ( 2 µ 2 m cos 2 + ¸ | u 2 u u ( ( ( ( ¸ 1 2 ( ( ¸ 1 2 2 1 2 u | 1 3 2 ¸ = 2 2 L L 1 µ L m 2 2 2 3 2 cos m 2 ( 2 2 | \ ¸ u | | | 1 | u | | | 2 2 u 1 ( L m 1 3 | \ m 2 L L 2 1 2 L L 2 1 L m 2 3 1 2 u | | . 2 | \ u | | . 1 + 2 1 L m 1 2 µ cos 2 µ cos 2 + + u ¸ ( ( ¸ 2 ( ¸ m = 2 2 . \ . | | 2 2 | u | | | 2 1 | u | | | 2 2 2 L m 1 3 L m 2 3 1 ( L L m 2 2 1 ) u u \ \ 2 1 L m 1 2 \ µ . 2 + + + cos = 1 2 2 . ( ( ( ( ( ¸ ) | | . 2 2 34 2 1 2 1 1 The angular momentum of the system is related to the torque applied and the kinetic co- energy as follows: d I d dt dt ( q ( )u u ) E * k c = ÷ t = u c 2 L m 1 L L 1 1 ( 2 2 m cos 2 µ m L 1 + ¸ 1 2 u u ( ( ( ( ¸ ( ( ¸ 2 2 3 2 = q ¸ 2 L L 1 L m 2 cos m 2 2 2 2 2 3 L m 1 2 L L 2 1 1 µ ( 2 cos 2 µ L 1 + ¸ 1 2 u u ( ( ( ( ¸ m m ( ( ¸ ÷ | \ u 2 ( ( ¸ L L 2 1 2 d dt q | | . µ 2 2 2 u 1 ( ) 3 2 u sin 2 ÷ ¸ u ¸ 1 = m 2 2 L L 1 L 2 m 2 cos m 2 2 2 2 3 µ | | | | 2 2 | | | u | | | 2 2 2 L m 1 3 L 2 3 1 m \ 2 m 1 2 ) µ ( u | | . * 2 1 m \ u u E 2 L 1 L L 1 + + + = cos k 2 2 1 2 2 . \ . m 2 L L 2 1 ( | \ u u | | . 2 u u | | . µ ÷ sin ¸ ( ( ( ( ¸ * 1 2 E u c c 2 k = m 2 L L 2 1 2 | \ sin 2 µ 1 2 To express the flow variables in terms of the torques the momentum equation must be inverted. At this point it is easier to substitute values for the variables. ÷ ( 1 | µ 2 L L 2 1 ( L L 2 1 | \ m | | . u u | \ | \ ¸ + ( ( ( ( ¸ | | . 2 L m 1 L L 1 1 ( u u u ÷ 2 2 | | . µ 2 u u µ sin 2 ÷ sin 2 2 m cos L 1 m ¸ + ¸ ( ( ( ( ¸ m t 1 2 1 2 2 1 2 u ( ( 2 2 2 2 3 2 ÷ u 2 ¸ = ( ¸ t ¸ 2 ( ¸ 2 L L 2 1 L L 2 1 L L 1 L 2 | \ | | . u ( 2 1 m 2 m cos µ 2 2 2 3 µ sin 2 µ sin 2 u u 2 1 m m 2 1 2 2 2 2 2 \ ÷1 ( 1 8 1 16 + | \ | | . u u \ | \ ÷ | u | | . 2 2 1 1 ( t t ¸ 1 2 µ sin 2 µ sin 2 µ cos 2 ( ( ( ( ¸ u = ( ( ¸ ( ( ( 1 2 6 16 u ¸ 1 1 1 1 16 | ÷ t \ t ¸ ( ( ( ( ¸ 1 2 | | . u u u | | . 2 1 µ 16 2 24 µ sin 2 µ sin 2 + cos 1 2 ¸ ¸ 8 µ µ 2 cos 2 ÷ cos 144 96 ( ( + | \ 1 16 | | . ( 2 u u 2 1 ( ( ( ( ¸ ) ) u ÷ 2 2 µ sin 2 ¸ u ( 2 2 µ µ µ 2 2 cos 2 16 ÷ 16 ÷ cos u ¸ = ( ¸ ÷ cos 144 384 \ 1 ÷ | | | . ( 2 u u 2 1 u ÷ 2 1 µ sin 2 2 2 µ cos 2 16 16 ÷ 16 ÷ The motion of the tip can be found from the kinematic relationship between the generalized coordinates and the tip Cartesian coordinates. 5a. To simulate the behavior of the system under the action of the impedance controller, the tip stiffness and damping matrices must be transformed to generalized coordinates. For small motions about a configuration where the inner link is aligned with the x-axis and the outer link is parallel to the y-axis, sinAu 1 = Au 1 and cos(t/2 + Au 2 ) = –Au 2 . ( ( ( ( ¸ ) ) (| | | | | | ÷ . 35 T T A t = F J = x K J t |( ÷ L 1 sinu 1 L 1 cosu 1 ( ( k L 2 cos | A + u | ( 0 L 1 ( ÷ L 2 Au ( 0 L 2 2 (Au 1 ( t = \ 2 2 2 ( ¸ ÷ L sinu 2 L 2 cosu 2 ¸ L 1 sin Au . ( ¸ = ¸ ÷ L 2 0 ( ¸ k ¸ L 1 Au 1 2 ( ¸ = k ¸ L 1 0 ¸¸ Au 2 ¸ ( 2 ¸ 1 2 ( L 2 K( ) = k ¸ L 0 1 2 0 ¸ ( u Similarly, the damping matrix is transformed to generalized coordinates using the Jacobian. T T T ÷ L 1 sinu 1 L 1 cosu 1 (÷ L 1 sinu ÷ L 2 sinu 2 ( 2 0 ( t = F J = x B J = J BJ u = b ¸ ÷ L sinu 2 L 2 cosu 2 ¸ ( ¸ L 1 cosu 1 ( u = b L 1 2 ( u L 2 cosu 2 ¸ 2 1 ¸ 0 L 2 ¸ 2 0 ( B( ) = b L 1 2 ( u ¸ 0 L 2 ¸ The desired natural frequency for the system is 2 Hz. This allows the stiffness constant, k, to be evaluated: u ÷1 e n = K( ) I (u ) ÷1 2 ( 2 L 2 u k = (e )I ( ) ¸ L 0 1 2 0 ¸ ( n L m 1 2 + m 2 L 1 m 2 L L 2 ( 1 ( 2 1 I ( ) = cos µ 2 ( 6 0 ( 1 3 ( = u L L 2 2 2 1 m 2 cos µ 2 m 2 L 2 ( 0 1 ( ( ¸ 2 3 ¸ ( ¸ 24 ¸ 1 ( 2 ( 0 3 ( 2 k = ( ) I (u ) 0 2 L 2 2 ( ( ÷1 = 16t 6 0 (0 4( = 16t 4t 2 ¸ L 1 0 ¸ 0 1 ( ( ¸ 4 0 ¸ ( 2 1 0 ( ( ¸ 24 ¸ ¸ 6 ¸ 8 2 k = t max 3 If the stiffness coefficient is larger than k max then the system will oscillate faster than 2Hz in 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 this configuration. The critical damping coefficient is found from the following expression: 36 B( ) = 2 K( ) ( ) u u u I 1 ( 2 0 L 2 2 ( 6 0 ( 2 b | L 1 0 2 ( ( | | | 2 = 4 8 t ¸ L 1 2 0 ¸ 0 \ ¸ 0 L 2 ¸ . 3 ( 1 ( ( ¸ 24¸ 1(1 ( 1 ( ÷1 1( b 2 32t 2 0 4 ( 6 0 ( 16 0 ( 32t 2 0 6 ( = = ( 1 ( 1 ( 3 2 ( 3 1 0(0 ( 0 ( 0( ¸4 ¸¸ 24¸¸ 16¸ ¸3 ¸ b 4 critical = 3 t 5b. The system is driven by a simple impedance controller with the following characteristic: u T ( u ( 0 t = J( ) ( X K ÷ L( )) + V B ÷ J(u )u)) 0 These torques are the torques corresponding to changes in the angles u 1 and u 2 , which are not the same as the motor torques on the links. If the motor torques are needed in order to control actual hardware, the following relationships are used. The equivalent torque acting on each of the links is depicted below in Figure 2. u 1 -u 2 u 1 L 1 L 2 t 1 t 2 t 2 Figure 2: Uncoupled 2-Link Mechanism Indicating Torque Input From this figure it is evident that the torque corresponding to u 1 is the difference between the motor torques: t ÷t 2 . Also, the torque corresponding to u 2 is the torque 1 corresponding to u 1 minus the torque corresponding to µ 2 , which results in t 1 . 1) The first simulation describes the motion of the two link mechanism from the Cartesian 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 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) y p o s i t i o n ( m ) Time Plot of Mechanism Position for First Simulation 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 4 demonstrates 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) y p o s i t i o n ( 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 It is clear from this plot that the compliance in the system allows the outer link to rotate around 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 passes the 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 singular configuration 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) y p o s i t i o n ( 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 configuration instead 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 mechanism would need to quickly rotate through 180 o as the mechanism approached the singular configuration. This requires very large torques on the mechanism as it passes near the singular point. 6. The natural frequency of the controller is now adjusted to 20Hz to determine the impact this has on the system. The corresponding k and b are found as follows: 2 max 2 3 800 0 6 1 3 2 0 1600 t t = ( ( ( ¸ ( ¸ = k k 39 t t t 3 40 0 3 2 6 1 0 3 3200 16 1 0 0 16 1 24 1 0 0 6 1 0 4 1 4 1 0 3 3200 2 1 2 2 = ( ( ( ¸ ( ¸ = ( ( ( ¸ ( ¸ ( ( ( ¸ ( ¸ ( ( ( ¸ ( ¸ = ÷ critical b 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) y p o s i t i o n ( 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 180 o as 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 no longer 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 -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) y p o s i t i o n ( 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 this simulation 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 singular configuration 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 singular configurations, the simple impedance controller allows the mechanism to operate at its singular configurations. In the first simulation, the final desired position of the mechanism was the singular position corresponding to both links fully extended to the maximum attainable radius. With the impedance controller the mechanism was able to reach this configuration. However, due to the low translational inertance that corresponds to this position and relatively high rotational inertance the mechanism rotated beyond the desired 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 position with decreasing frequency. At 200 seconds the mechanism still has 1 o oscillations with a period of 36 seconds. The actual trajectory and virtual trajectory of the mechanism are shown below in Figure 8, which includes only the 2 second command interval. 41 Actual Trajectory and Desired Trajectory for First Simulation 0.1 0.08 0.06 0.04 V e r t i c a l P o s i t i o n ( 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 Position (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 the inverse 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 the lowest 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 V e r t c a P o s t i o n -0.4 -0.2 0 0.2 0.4 0.6 Horizontal Position (m) Figure 9: The Actual and Virtual Trajectory for Desired Movement from (-L,0) to (L,0) 42 This is also seen on the third simulation, which attempts to pass the mechanism near the origin. 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 i t i ( m ) -0.1 -0.08 -0.06 -0.04 -0.02 0.02 0.04 0.06 0.08 0.1 Actua Trajectory and Des red Trajectory for Th rd S mu at V e r t c a P o s o n -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 i t i ( m ) -0.1 -0.08 -0.06 -0.04 -0.02 0.02 0.04 0.06 0.08 0.1 V e r t c a P o s o n -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 While the increased stiffness resulted in better performance away from the singular position, 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. V e r t i c a l P o s i t i o n ( 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 any position 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, the mechanism 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 especially apparent when comparing Figure 12 to Figure 10. Actual Trajectory and Desired Trajectory for Uneven Link Simulation 0.1 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 Horizontal Position (m) Figure 12: Actual & Virtual Trajectory from (-L,L/20) to (L,L/10) with Uneven Link Lengths 44 Tent Pole Instability 1. The mechanical system of a tent pole and guy-ropes is shown below in Figure 13. 1 m 1 m u k k l 1 l 2 2 m Figure 13: Tent-Pole & Guy-Rope System The relationship between the variables l 1 , l 2 and u can be determined from the Law of Cosines: l 1 = 5÷ cos 4 u ÷ l 2 = 5÷ cos 4 ( u t ) The Jacobian for the transformation between coordinates is found by differentiating each of these terms with respect to u. sin 2 u l 1 = u 5÷ cos 4 u ÷ sin 2 ( u t ) u ÷ = ÷ l 2 5÷ cos 4 ( u t ) sin 2 u ( 5÷ cos 4 u ( J = ÷ sin 2 ( u t ) ( ÷ ( 5÷ cos 4 ( u t ) ( ¸ ¸ ÷ The inertia of the pole is simply mL 2 /3. The co-energy, therefore, is dependent only on e. This results in the following relationship between torque and angular acceleration: 3t u = 2 mL The torque on the pole is proportional to the forces from each of the guy-ropes, according to the following equation: 45 F J T = t The forces from the guy-ropes are found from the next equation, where k is the elasticity of the rope and F 0 is the pretension. 0 F l k F + A = There is also a torque component due to gravity, which results in the final expression for torque: u t cos 2 mgL F J T + = 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 Horizontal Position (m) V e r t i c a l P o s i t i o n ( 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 the higher torques generated. 3. The next simulation removed the pretension in the guy-ropes and added in the elasticity. Gravity effects were again omitted in order to emphasize the effect of the elasticity on the system. Initially, the system was only perturbed 1 o to show that it was stable. This produced the following result where the pole oscillated between ±1 o . 46 -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 Horizontal Position (m) V e r t i c a l P o s i t i o n ( m ) Tent-Pole Simulation 3a Figure 15: Time Plot of Tent-Pole with Elasticity and No Pretension for 1 o Perturbation The next task was to see if the tent-pole was still stable for perturbations up to 45 o . Figure 16 demonstrates the stability of the pole with an initial perturbation of 45 o . -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 Horizontal Position (m) V e r t i c a l P o s i t i o n ( m ) Tent-Pole Simulation 3b Figure 16: Time Plot of Tent-Pole with Elasticity and No Pretension for 45 o Perturbation 47 4. With stiffness, pretension, and gravity all considered in the model of the system, the following 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 F0<2.23607k -1 0 1 2 i l i ( m ) -0.5 0.5 1.5 2.5 V e r t c a P o s t i o n -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Horizontal Position (m) Figure 17: Time Plot of Complete Tent-Pole System with Stable Behavior Tent-Pole Simulation 4 with F0>2.23607k -1 0 1 2 i l i ( m ) -0.5 0.5 1.5 2.5 V e r t c a P o s t i o n -1 -0.5 0 0.5 1 1.5 2 2.5 Horizontal Position (m) Figure 18: Time Plot of Complete Tent-Pole System with Unstable Behavior 48 5. To analytically solve for the stability of the tent-pole system the equivalent stiffness of the joint is found. The expression for the joint stiffness is: K = c ÷ J T F + kJ T J cu The system becomes unstable at the point where there is no restoring torque on the system for a small displacement Au. For this analysis the mass of the system is neglected. This yields the following equation for the torque on the tent pole. T t = J k Al 1 ( ( ÷ J T F = 0 ¸ Al 2 ¸ sin 2 u ( ( sin 2 u ÷ sin 2 (t ÷u ) ( 5 ÷ cos 4 u (Au = sin 2 u ÷ sin 2 (t ÷u ) ( F 0 ( ( 5 ÷ cos 4 u 5 ÷ cos 4 (t ÷u ) ( ( ¸ k ÷ sin 2 (t ÷u ) ( 5 ÷ cos 4 u 5 ÷ cos 4 (t ÷u ) ( ¸ ¸ F 0 ¸ ( ¸ ¸ 5 ÷ cos 4 (t ÷u ) ( ¸ ¸ | 4k | sin 4 2 u sin 4 2 (t ÷u ) | | |Au = F 0 sin 2 u sin 2 (t ÷u ) | + \ 5 ÷ cos 4 u 5 ÷ cos 4 (t ÷u ) . \ 5 ÷ cos 4 u ÷ 5 ÷ cos 4 (t ÷u ) . | | For stability about the vertical position, u = t/2, where there are small displacements, Au, this expression becomes: 2 2 4k | cos Au cos Au | | | Au = F 0 | cos 2 Au cos 2 Au | + ÷ | | \ 5÷ 4Au 5+ 4Au . \ 5÷ 4Au 5+ 4Au . 10 | | Au = 2F 0 cos Au | 5+ 4Au ÷ 5÷ 4Au | | 4k cos 2 Au | 2 2 \ 25 ÷16Au . \ 25 ÷16Au . | F 0 = 20Au cos Au 2 k 25 ÷16Au ( 5+ 4Au ÷ 5÷ 4Au ) Using L’Hopital’s Rule: cos 20 Au ÷ 20Au sin Au 20 lim = = = 5 x 0 | 2 2 | | 4 | ÷16Au 2 ( 5+ 4Au ÷ 5÷ 4Au )+ 25÷16Au \ 5+ 4Au + 5÷ 4Au . | | 5 | | 2 \ 5 . Therefore, the maximum value for the pretension is k 5 or 2.23607k. This result can be compared to the ratios found through simulation, which are shown in Table 1: 25÷16Au Table 1: Ratio of F 0 to k for MATLAB Simulation of Tent-Pole System without Gravity F 0 k F 0 /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 It is clear from the simulation data, that MATLAB accurately predicts the stability point for 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 the angle from vertical increases. The mass has a larger effect on the stability of the system for lower values of pretension. 50 MEMORANDUM To: 2.141 class From: Professor Neville Hogan Date: December 9, 2002 Re: Assignment #4 Performance on assignment #4 was generally excellent. In the problem on control in and near kinematic singularities, the main challenge was to deal with the nonlinear kinematics. It is substantially easier to use configuration variables (angles) measured with respect to an inertial reference frame to formulate the kinetic energy and derive state equations. The sensor measures relative angles and the motors generate relative torques, but the transformation between relative angles and inertial angles is simple as is the transformation of relative torques to generalized torques. To evaluate the controller parameters you could either transform the inertia tensor from configuration space to end-point coordinates or transform the controller stiffness and damping from end-point coordinates to configuration space. In the configuration specified (links at right angles) the Jacobian is particularly simple. The natural frequency and damping ratio for small motions is the same in either frame. In the tent-pole instability problem, “pencil-and-paper” analysis up front can do a lot to guide numerical simulation. Some of you need to look more closely at your numbers; the simulations were generally correct, but non-negligible errors can and did occur. In this problem analysis shows that the relation between stiffness and preload should be a straight line. That’s a good way to check your simulations. 51 2.141 2002 page 1 assignment #5 Massachusetts Institute of Technology Department of Mechanical Engineering 2.141 Modeling and Simulation of Dynamic Systems 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 transport processes in a simple thermofluid system. A secondary purpose is to show you that even simple systems involving matter transport can exhibit interesting behavior. Two air cylinders are connected by a valve, which is initially closed. One cylinder contains air at 20 psi, the other, air at ambient pressure, 14.7 psi. Both cylinders are initially at ambient temperature, 70°F. You are to model and simulate the transient that follows when the valve is abruptly opened. 70°F, 20 psi 70°F, 14.7 psi Assume air may adequately be described as an ideal gas. Assume each cylinder is 10 inches 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 may neglect any heat transfer across the valve or between the valve and its surroundings. Vary the 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/sec˚R) = 0, 0.0001, 0.001 and · (or, more realistically, 1). 1a. Draw a bond graph of your system model, clearly identifying key quantities and power fluxes. 1b. Choose state variables and develop (nonlinear) state equations. 2a. Simulate the transient changes in (a) pressure (b) temperature and (c) mass flow rate 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 transient responses to exhibit no overshoot. You should be able to show that this is not always true. Comment and explain. 2c. Check the veracity of your simulation by computing the equilibrium conditions in each tank after the transients have expired. (Calculating equilibrium states before and after is standard engineering thermodynamics.) 52 2.141 2002 page 2 assignment #5 3. A much more common situation (as any SCUBA diver knows) is to fill a cylinder from a (nominally) constant-pressure supply. One way to model this situation is to assume the left-hand cylinder in the diagram above is extremely large (so that withdrawing air from it changes its pressure very little). Use the model you developed 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 temperature difference between the two cylinders. (Just show me a bond graph. Equations and simulation are optional.) 5. The pressure difference above was chosen to keep valve flow in the non-choked flow regime. Of course, realistic pressure differences are much higher. (A typical SCUBA tank is routinely filled to 3,000 psi.) Revise your model and simulation to cover this situation. In this case, how valid is the assumption of no heat transfer at the valve? 53 2.141, Assignment#5 – 21 November, 2002 Matter-Transport and Convection Will Booth 54 W. Booth 08/18/04 1 1 Nomenclature Arabic A 1 [in 2 ] Surface area of tank 1 A 2 [in 2 ] Surface area of tank 2 A t [in 2 ] Valve (throat) cross-sectional area C d none Coefficient of discharge for air expansion on valve exit C v_air [ft-lbf/lb-mol-R] Specific heat of air at constant volume D 1 [in] Diameter of tank 1 D 2 [in] Diameter of tank 2 D t [in] Diameter of valve (throat) h [ft-lbf/sec-in 2 -R] Heat transfer coefficient (tank environment) L 1 [in] Length of tank 1 L 2 [in] Length of tank 2 M air [1/mol] Molecular weight of air N 1 [lb-mol] Mass of air in tank 1 N 2 [lb-mol] Mass of air in tank 2 P 1 [psi] Pressure in tank 1 P 2 [psi] Pressure in tank 2 R u [ft-lbf/lb-mol-R] Universal gas constant S 1 [ft-lbf/R] Total entropy of tank 1 S 2 [ft-lbf/R] Total entropy of tank 2 dS env1 /dt [ft-lbf/sec-R] Entropy flux to the environment from tank 1 dS env2 /dt [ft-lbf/sec-R] Entropy flux to the environment from tank 2 dS vlv /dt [ft-lbf/sec-R] Entropy flux across valve T 1 [R] Temperature of tank 1 T 2 [R] Temperature of tank 2 T amb [R] Ambient temperature V 1 [in 3 ] Volume of tank 1 V 2 [in 3 ] Volume of tank 2 Greek ¸ none ratio of specific heats (c p / c v ) µ 1 [lb-mol/in 3 ] density in tank 1 µ 2 [lb-mol/in 3 ] density in tank 2 55 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 S 1 , N 1 , S 2 , N 2 are chosen. C R 0 0 0 0 C R R 0 S e s u . s d . s 2 . s 1 . s env2 . s env1 . N 1 . N 2 . N u . N d . T 1 T 2 T amb : : µ 1 µ 2 :: : h h : S f : 0 S f : 0 P 1 P 2 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: t m in d d t m 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: P u M air µ u v u 2 2 + | \ | . t m u d d P t M air µ u v t 2 2 + | \ | . t m t d d 56 W. Booth 08/18/04 3 where the units of µ are [lb-mol / in 3 ] and the units of M air are [1 / mol]. Subscript ' u ' denotes upstream parameters, and subscript ' t ' denotes parameters at the throat. Cancelling mass flow rates, and assuming there is negligible upstream velocity, re-arranging, gives: v t 2 M air P u µ u P t µ u ÷ | \ | . i.e. in terms of molar mass, the mass flow-rate at the throat is: t m t d d t M air N ( ) d d M air µ t A t 2 M air P u µ u P t µ 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 N d d A t 2 µ u M air P u P t ÷ ( ) A problem arises, since from the ideal gas law: P u P t T u T t Therefore, to get mass flow, we require P u > P t , which implies T u > T t , and therefore, u u > u t , and h u > h t which 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 T u = T d then, the process will be isenthalpic, as commonly assumed. To determine equations for entropy flux, per the definition of specific entropy: s S N therefore The entropy flux in a particular tank is, given the sign conventions in the system bond graph t S d d S N t N d d t S env 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: t S 1 d d S 1 N 1 t N 1 d d t S env1 d d ÷ and t S 2 d d S 2 N 2 t N 2 d d t S env2 d d ÷ 57 W. Booth 08/18/04 4 P N V R u T and since the volume of each tank is fixed: (where R u is the universal gas constant) P µ R u 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 T i is the initial temperature, and S i is the reference entropy) T T i exp S S i ÷ M N c v | \ | . Therefore: M N c v dT T dS So: (where M is the molar weight, and N has units of lb-mol) U M N c v 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) t N 1 d d t N 2 d d ÷ Choosing state variables, S 1 , N 1 , S 2 , and N 2 , 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 weren’t equal, therefore not satisfying Maxwell’s 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-in 2 -R are shown in Figure 2-2. I used values for h suggested in the problem statement (directly in ft-lbf/sec-in 2 -R as opposed to BTU/sec-R), but when multiplied by the surface area of the tanks, the result is approximately equivalent hence the results show the range of responses intended by the suggested values. 58 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, like ode15s, 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 numerical solution (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 7 1.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 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 relative tolerance 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. N 1 (init) T 1 (init) P 1 (init) S 1 (init) N 2 (init) T 2 (init) P 2 (init) S 2 (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 pressure difference 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., N 1 > N 2 , and T 1 < T 2 . What tank 1 loses in pressure, 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 increase in 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 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 transfer coefficient. 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 initial response, 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 1’s temperature decrease as rapidly. Therefore, the pressure 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 2 nd law of thermodynamics. It is not the absolute 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 T 1 > T amb then the entropy of tank 1 will decrease (disregarding mass transfer for the time being), but if T 1 < T amb the 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-in 2 -R. The pressure difference 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 heat drain 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 flow to equilibrium. However, the gradual entropy generation as the temperature reduces to ambient, decreases the pressure more in tank 2 than in tank 1 (since T 2 > T 1 ) which results in mass flow between tank 1 and tank 2, tending to equalize the masses of air in both tanks. This is as expected, since at equilibrium, even though not shown in Figure 2-2, both tanks will be at the same temperature and pressure. h increasing 61 W. Booth 08/18/04 8 This case (h=0.0001) is the perfect example of the 2-stage process alluded to earlier, since the stages are distinct. The first stage is mass flow due to significant pressure differences between the tanks. The second stage is gradual change of tank temperature to ambient, and the equalization of air masses in each tank, due to heat loss to the environment. The time constant of the second stage is much longer than the time constant of the first stage, hence the prolonged pressure settling times as the heat transfer coefficient increases (see Table 2-3). The temperature and entropy overshoot is a result of the trade-off between the time constants of the 2 processes, the initial mass flow due to significant pressure differences between the tanks and the subsequent change in tank temperatures due to heat transfer to the environment. In the extreme cases for h=0 (adiabatic) or h=1 (isothermal), there is no overshoot, since there is only a single process, due to mass flow as a result of AP. As the heat transfer coefficient increases to isothermal conditions (h=1) the temperature and entropy transients will tend to settle faster, and in the isothermal limit, there is no transient since temperature and entropy in each tank are always equal to their initial conditions. The following shows the equilibrium conditions for the mass, temperature, pressure and entropy in each tank at the end of the transients, for the various heat transfer coefficients (with units given in section 1). h N 1 T 1 P 1 S 1 N 2 T 2 P 2 S 2 P diff 0 0.0047 470.30 17.54 172.35 0.0036 621.03 17.53 132.15 0.0088 0.0001 0.0046 480.30 17.29 172.79 0.0037 585.87 17.29 131.41 0.0087 0.001 0.0042 525.99 17.34 174.41 0.0041 532.52 17.33 130.04 0.0042 0.01 0.0042 529.67 17.35 174.52 0.0042 529.67 17.35 129.95 0.0014 1 0.0042 529.67 17.35 174.52 0.0042 529.67 17.35 129.95 -0.0011 Table 2-2: Table of equilibrium conditions h Temperature / Entropy settling time [seconds] Pressure / Mass settling time [seconds] 0 20 20 0.0001 22, > 600 22, > 600 0.001 600 400 0.01 200 230 1 0 190 Table 2-3: Table of approximate settling times ????????????DO STUDY WITH T1 = T2 < TAMB ????? ????????????OR T1 < T2 = TAMB = 100°F ???????? Part 3 – Tank 1 = Constant Pressure Supply Tank 1 is now given dimensions of 40” diameter and 60” length, to simulate a constant pressure supply. The results are shown in Figure 2-5. The pressure of tank 1 is more or less constant, as desired, and its temperature does not change from its initial condition. The entropy and mass of the air in tank 1 are significantly greater than that in tank 2 therefore, only results for tank 2 are plotted. The results show that the qualitative response of tank 2 is identical to that shown in Figure 2-2, except that the tank equilibrium pressure is now = to the supply pressure (20psia) since the supply pressure does not change. The state equations assume knowledge of the initial entropy of the supply air, as well as its initial pressure, so both these quantities must be specified in order to determine the transient response of the system. 62 W. Booth 08/18/04 9 Figure 2-5: Simulations for Tank1 = constant-pressure supply Part 4 – Model heat transfer across the valve Adding heat transfer across the valve, introduces an additional R element into the bond graph, shown in Figure 2-6. My initial calculations for a heat transfer coefficient, kA/L for a 0.005” coating of cadmium on a one meter long interconnecting tube, set kA/L = 2.062E-5 ft-lbf/sec-R. Initial simulations with this value of kA/L gave no change in the transient response, because the entropy change due to heat transfer across the valve was minimal, even without heat transfer to the environment. Therefore, to show the effects of heat transfer across the valve, kA/L was artificially inflated by a factor of 10 4 , and the simulation results are shown in Figure 2-7. As expected, even with no heat transfer to the environment, the temperatures in the two tanks will equilibrate over time. The pressure transient shows an interesting response, due to the slow changing temperature and tank air mass responses. 63 W. Booth 08/18/04 10 C R 0 0 0 0 C R R 0 S e s u . s d . s 2 . s 1 . s env2 . s env1 . N 1 . N 2 . N u . N d . T 1 T 2 T amb : : : µ 1 µ 2 :: : h h kA/L : S f : 0 S f : 0 P 1 P 2 s vlv1 s vlv2 . . R Figure 2-6: System bond graph with added model of heat transfer across the valve 64 W. Booth 08/18/04 11 Figure 2-7: Simulations with heat transfer across valve (kA/L = 2.062E-1 ft-lbf/sec-R), and h=0 (no heat transfer to environment) Part 5 – Choked flow Using the following more accurate equations for subsonic flow: t N d d C d A t 2 ¸ ¸ 1 ÷ µ u P u P d P u | \ | . 2 ¸ P d P u | \ | . ¸ 1 + ¸ ÷ ¸ ( ( ( ( ¸ and the following equation for choked flow: t N d d C d A t 2 ¸ 1 + | \ | . 1 ¸ 1 ÷ µ u P u ¸ 2 ¸ 1 + The system equations are modified, and the results, for P 1 (init) = 200psi are shown in Figure 2-8. 200psi was chosen as representative of choked flow, since P d / P 1 < 0.528. The results are not very satisfactory, because: 1) The ode15s solver no longer converged, so I had to resort to ode45, which gives oscillatory response at equilbrium 2) More importantly, the temperature of tank 2 can get incredibly high, if there is no heat transfer to the environment, therefore there will be a large temperature differential across the valve due to the high pressures involved, even without considering the effects of the shock. Thus I conclude that I need a better model of the heat transfer across the valve, because the one for the cadmium-coated interconnecting pipe did not give me an accurate coefficient of heat transfer. 65 W. Booth 08/18/04 12 Figure 2-8: Simulations for choked flow, with P 1 (init) = 200psia References: Hogan, N. 2.141 Lecture Notes MIT. Fall, 2002. Massey, B.S. Mechanics of Fluids 6 th ed. London: Chapman & Hall, 1992. Rogers, G.F.C and Mayhew, Y.R. Thermodynamic and Transport Properties of Fluids 5 th ed. Oxford: Blackwell, 1995. http://thermal.sdsu.edu/testcenter/indexjavaapplets.html 66 W. Booth 08/18/04 13 APPENDIX A: Linearization of multi-port capacitor equations Given the 3-port capacitor, with the mechanical port closed, we must choose a causal-assignment to determine the form of the linearized equations, hence, choose form as above, with S and N as inputs, and T and G as outputs on each port, respectively. Hence, the form we are looking for is the following: oT oG | \ | . S T d d S G d d N T d d N G d d | \ | | | . oS oN | \ | . (evaluated at S o , N o operating point) The expression for T is as used in the analysis. The expression for G is as follows: G U T S ÷ P V + where T(S) and P(S): Evaluating the matrix of partials: S T d d T i M c v N o exp S o S i ÷ M c v N o | \ | . N T d d S o S i ÷ M c v | \ | . ÷ T i N o 2 exp S o S i ÷ M c v N o | \ | . S G d d M N o c v S T d d | \ | . T i exp S o S i ÷ M c v N o | \ | . ÷ S o S T d d | \ | . ÷ V S P d d + substituting for dT/dS and evaluating dP/dS and factoring common factor, gives: S G d d T i exp S o S i ÷ M c v N o | \ | . R c v S o M c v N o ÷ | \ | . N G d d M c v T S ( ) M c v N o N T d d | \ | . + S o N T d d | \ | . ÷ V N P d d | \ | . + substituting for dT/dN and evaluating dP/dN and factoring common factor, gives: N G d d T i exp S o S i ÷ M c v N o | \ | . M c p S o M c v N o ÷ ( ) S o S i ÷ M c v | \ | . 1 N o 2 + ¸ ( ( ( ¸ Therefore, it appears that my linear model does not satisfy Maxwell’s reciprocity conditions, because dG/dS does not equal dT/dN. I thought I had kept track of all my variables in terms of the state variables, S, and N, but I still do not see where I have made my mistake? 67 W. Booth 08/18/04 14 APPENDIX B: Matlab M-Files %Assignment #5 -- 2.141 W.Booth %due: 21 Nov 02 %Assumptions: % 1) ... % 2) ... global T1_init N1_init S1_init M_air Cv_air V_tank1 V_tank2 R_universal T2_init N2_init S2_init At h Asurf1 Asurf2 Tamb Question gamma_air Cd %Change as required, to simulate various parts of problem. Question = 5 %given D_tank2 = 10; %in L_tank2 = 30; %in if Question == 3 D_tank1 = 40; %in L_tank1 = 60; %in else D_tank1 = 10; %in L_tank1 = 30; %in end if Question ~= 5 P1_init = 20; %lbf/in^2 else P1_init = 200; %lbf/in^2 end P2_init = 14.7; %lbf/in^2 T1_init = 70 + 459.67; %R -- need calc. in absolute temperature T2_init = 70 + 459.67; Tamb = 70 + 459.67; Dt = 0.1; %in %known R_universal = 1545; %ft-lbf/lb-mol-R -- p.26 of Rogers' and Mayhew's "Thermodynamic & Transport Prop. of Fluids" M_air = 28.94; %1/mol -- http://www.rwc.uc.edu/koehler/biophys/8a.html gamma_air = 1.4; %p.16 of Rogers and Mayhew Cd = 0.5 ; %discharge coefficient %calculations V_tank1 = (1/4)*pi*(D_tank1^2)*L_tank1; %in^3 V_tank2 = (1/4)*pi*(D_tank2^2)*L_tank2; %in^3 At = (1/4)*pi*(Dt^2); %in^2 -- throttle area Asurf1 = pi*D_tank1*L_tank1; %in^2 -- surface area of tank Asurf2 = pi*D_tank2*L_tank2; %in^2 -- surface area of tank Cv_air = R_universal/(gamma_air-1); %ft-lbf/lb-mol-R %initial conditions N1_init = P1_init *V_tank1/ ( 12 * R_universal *T1_init); %lb-mol = lbf/(in^2)*(in^3)/ ((in/ft)*(ft-lbf/lb-mol-R)*R ) 68 W. Booth 08/18/04 15 N2_init = P2_init*V_tank2/(12*R_universal*T2_init); %lb-mol %Found values of specific entropy for initial conditions at % http://thermal.sdsu.edu/testcenter/indexjavaapplets.html %to calc. total entropy must multiply by initial mass in tank 1 (m1 = M_air * N1), hence: S1_init = 1256.7 * M_air * N1_init; %ft-lbf/R = %ft-lbf/lb-R * (1/mol)* (lb-mol) S2_init = 1273.1*M_air*N2_init; %matrix of initial conditions: init = [N1_init T1_init P1_init S1_init N2_init T2_init P2_init S2_init]; if Question == 5 tMAX=150; %run for less time b/c numerical problems with divide by zero ~200sec ?? else tMAX=600; %seconds end colour1=['b' 'g' 'r' 'c' 'k']'; hv=[0 0.0001 0.001 0.01 1]'; %heat transfer coefficients %tab = []; %matrix of equilibrium conditions equil = []; %Run model for various heat transfer coefficients for i=1:length(hv), %Run model with various ODESET parameters (see CODE REFERENCE [1] below) %rtol=[ 1.e-3 1.e-5 1.e-7 ]; %for i=1:length(rtol), %op=odeset('RelTol',rtol(i),'Stats','on'); %legend('ode45, RelTol=1.e-3 (tank1)','ode45, RelTol=1.e-3 (tank2)',... % 'ode45, RelTol=1.e-5 (tank1)','ode45, RelTol=1.e-5 (tank2)') h = hv(i); %h = hv(1) clear N1 S1 N2 S2 T1 P1 T2 P2 choked %execute model: %tic; if Question ~= 5 [t,X]=ode15s(@throttling,[0 tMAX],[N1_init S1_init N2_init S2_init]);%,op); else %ode15s could not converge with choked flow, therefore use ode45 [t,X]=ode45(@throttling,[0 tMAX],[N1_init S1_init N2_init S2_init]);%,op); end %tt = toc; %tab = [ tab ; rtol(i) tt ] N1=X(:,1); %molar mass of air in tank 1 S1=X(:,2); %entropy of tank 1 N2=X(:,3); %molar mass of air in tank 2 S2=X(:,4); %entropy of tank 2 69 W. Booth 08/18/04 16 T1=T1_init.*exp(((S1-S1_init).*M_air)./(M_air.*N1.*Cv_air)); P1 = 12*N1.*R_universal.*T1./V_tank1; T2=T2_init.*exp(((S2-S2_init).*M_air)./(M_air.*N2.*Cv_air)); P2 = 12*N2.*R_universal.*T2./V_tank2; Pdiff = P1-P2; %equilibrium conditions, assuming tMAX > settling time equil = [equil ; h N1(length(t)) T1(length(t)) P1(length(t)) S1(length(t)) N2(length(t)) T2(length(t)) P2(length(t)) S2(length(t)) Pdiff(length(t))]; %1=choked, 0=not choked choked=zeros(length(P1)); choked=(P2<=0.528*P1); if (Question == 5) num_sub_plots = 5; else num_sub_plots = 4; end figure(1) subplot(num_sub_plots,1,1),plot(t,P1,colour1(i),t,P2,strcat(colour1(i),':')),ylabel('Pressure [psi]'),axis([0 t(length(t)) P2_init P1_init]),hold on subplot(num_sub_plots,1,2),plot(t,T1,colour1(i),t,T2,strcat(colour1(i),':')),ylabel('Temp [degR]'),hold on subplot(num_sub_plots,1,3),plot(t,S1,colour1(i),t,S2,strcat(colour1(i),':')),ylabel('Entropy [ft-lbf/R]'),hold on if (Question == 3); axis([0 t(length(t)) 120 140]), ylabel('Entropy of tank2 [ft-lbf/R]'); end subplot(num_sub_plots,1,4),plot(t,N1,colour1(i),t,N2,strcat(colour1(i),':')),ylabel('Mass [lb-mol]'),xlabel('Time [sec]'),hold on if (Question == 3); axis([0 t(length(t)) 3.5e-3 5e-3 ]),ylabel('Mass of tank 2 [lb-mol]'); end if (Question == 5) subplot(num_sub_plots,1,4),xlabel('') subplot(num_sub_plots,1,5),plot(t,choked,colour1(i)),ylabel('Choked [1=YES / 0=NO]'),xlabel('Time [sec]'),axis([0 t(length(t)) -1 2]),hold on end figure(2) plot(t,Pdiff,colour1(i)),ylabel('Delta P = P1-P2 [psi]'),xlabel('Time [sec]'),hold on,grid end %of heat transfer coeff. for-loop figure(1) legend(strcat('h = ',num2str(hv(1)),' -- tank1'),... strcat('h = ',num2str(hv(1)),' -- tank2'),... strcat('h = ',num2str(hv(2)),' -- tank1'),... strcat('h = ',num2str(hv(2)),' -- tank2'),... strcat('h = ',num2str(hv(3)),' -- tank1'),... strcat('h = ',num2str(hv(3)),' -- tank2'),... strcat('h = ',num2str(hv(4)),' -- tank1'),... strcat('h = ',num2str(hv(4)),' -- tank2'),... strcat('h = ',num2str(hv(5)),' -- tank1'),... strcat('h = ',num2str(hv(5)),' -- tank2')) if Question == 5 legend off %legend doesn't print properly ? end 70 W. Booth 08/18/04 17 figure(2) legend(strcat('h = ',num2str(hv(1))),... strcat('h = ',num2str(hv(2))),... strcat('h = ',num2str(hv(3))),... strcat('h = ',num2str(hv(4))),... strcat('h = ',num2str(hv(5)))) %CODE REFERENCE [1] %CODE TO VARY ODESET PARAMETERS AND LOOK AT STIFFNESS OF DIFF'L EQUATIONS %FROM PAPER ON STIFF DIFF'L EQUATIONS FROM http://www.maths.uq.edu.au/~gac/math3201/mn_ode2.pdf %tab = [ ] ; %for atol=[ 1.e-5 1.e-7 1.e-9 ] %op=odeset('reltol',1.e-15,'abstol',atol,'stats','on'); %tic ; %[t,Y]=ode23('rob',[0,1],[1;0;0],op); %tt = toc ; %tab = [ tab ; abstol tt ] %end 71 W. Booth 08/18/04 18 %throttling.m -- for use with ode45 function function xdot=throttling(t,x) global T1_init N1_init S1_init M_air Cv_air V_tank1 V_tank2 R_universal T2_init N2_init S2_init At h Asurf1 Asurf2 Tamb Question gamma_air Cd %x = [N1 S1 N2 S2]' xdot=zeros(4,1); %column vector of derivatives N1=x(1); S1=x(2); N2=x(3); S2=x(4); %"1" stands for left tank, initially at higher pressure, "2" stands for right tank %calculating temp from M.Cv.N.dT = T.dS (similar to assign #3) %Units on S-S0 is ft-lbf/R (total entropy), so need to multiply by M_air (1/mol) so (S-S0)/N*Cv expression is unitless T1=T1_init*exp(((S1-S1_init)*M_air)/(M_air*N1*Cv_air)); %calculating pressure from ideal gas law: PV = N*Ru*T P1 = 12*N1*R_universal*T1/V_tank1; %calculating density from definition: ro = N/V ro1 = N1/V_tank1; %same for tank 2 T2=T2_init*exp(((S2-S2_init)*M_air)/(M_air*N2*Cv_air)); P2 = 12*N2*R_universal*T2/V_tank2; ro2 = N2/V_tank2; if (Question ~=5) if (P1 >= P2) %If P1 > P2, then flow is passing from left to right, and equations are as derived in notes: N1dot = -At * sqrt(2*ro1*(P1-P2)/M_air); %mass flow leaving tank 1 N2dot = -N1dot; %same mass flow arrives at tank 2 else %P2 > P1 and flow is reversed, hence, N1dot is +ve and N2dot is -ve N1dot = At * sqrt(2*ro1*(P2-P1)/M_air); %mass flow arriving at tank 1 N2dot = -N1dot; %same mass flow leaves tank 2 end else %QUESTION 5 if (P1 >=P2) %If P1 > P2, then flow is passing from left to right if (P2 <= 0.528*P1) %Subsonic flow -- better model N1dot = -Cd * At * sqrt((2*gamma_air/(gamma_air-1))*ro1*(1/M_air)*P1*(((P2/P1)^(2/gamma_air))- ((P2/P1)^((gamma_air+1)/gamma_air)))); N2dot = -N1dot; else %CHOKED FLOW 72 W. Booth 08/18/04 19 N1dot = -Cd * At * ((2/(gamma_air+1))^(1/(gamma_air-1))) * sqrt((2*gamma_air/(gamma_air+1))*ro1*(1/M_air)*P1); N2dot = -N1dot; end else %P2 > P1 and flow is reversed, hence N1dot is +ve and N2dot is -ve if (P1 <= 0.528*P2) %Subsonic flow -- better model N1dot = Cd * At * sqrt((2*gamma_air/(gamma_air-1))*ro2*(1/M_air)*P2*(((P1/P2)^(2/gamma_air))- ((P1/P2)^((gamma_air+1)/gamma_air)))); N2dot = -N1dot; else %CHOKED FLOW N1dot = Cd * At * ((2/(gamma_air+1))^(1/(gamma_air-1))) * sqrt((2*gamma_air/(gamma_air+1))*ro2*(1/M_air)*P2); N2dot = -N1dot; end end end %entropy flux to environment due to convective heat transfer: Sdot_env1 = h * Asurf1 * (T1-Tamb)/T1 ; Sdot_env2 = h * Asurf2 * (T2-Tamb)/T2 ; %entropy flux between tanks (across valve): Sdot_vlv1 = 2.062E-5*(T1-T2)/T1; %assuming 0.005" covering of Cadmium on 1m long interconnecting tube %conductivity of cadmium = 96.9J/m-sec-K Sdot_vlv2 = 2.062E-5*(T1-T2)/T2; %Sdot_vlv1 and Sdot_vlv2 will both be >0 if T1>T2, but Sdot_vlv1 > Sdot_vlv2 %i.e. entropy lost at valve by tank 1 is > entropy gained at valve by tank 2 %irrespective of flow direction, the entropy fluxes can be defined: if (Question ~= 4) S1dot = (S1/N1)*N1dot-Sdot_env1; S2dot = (S2/N2)*N2dot-Sdot_env2; elseif (Question == 4) S1dot = (S1/N1)*N1dot-Sdot_env1-Sdot_vlv1; S2dot = (S2/N2)*N2dot-Sdot_env2+Sdot_vlv2; end %assigning state variable derivates to xdot-vector: xdot(1)=N1dot; xdot(2)=S1dot; xdot(3)=N2dot; xdot(4)=S2dot; 73 MEMORANDUM To: 2.141 class From: Professor Neville Hogan Date: December 9, 2002 Re: Assignment #5 Performance on assignment #5 was mixed. The effort put in was excellent but there were some difficulties. State variables are associated with energy-storage elements. In the two-tank throttling problem, as the tank volumes are fixed, you need at most two state variables for each tank, four in all. However, the system doesn’t leak so the total mass in the system is constant, hence only three of the four state variables are independent. In principle you could work with “excess” state variables but it’s inefficient and inelegant (if you care about that sort of thing) and you run a risk of generating numerical problems, depending on how you formulate your state equations—think of computing the rate of change of a constant to be numerically indistinguishable from zero instead of recognizing that it is identically zero. Most of you exhibited significant numerical artifacts in your simulations (some more than others). You should be very suspicious about abrupt rates of change in a physical system model unless you are quite certain you deliberately introduced them. If you aren’t sure, try changing parameters of either the numerical integration routine or the model. You should be able to change the integrator parameters (especially in the direction of greater precision) with no significant quantitative effect on the results. Because physical system behavior usually varies smoothly with model parameters, small changes of model parameters should result in no significant qualitative changes, only small quantitative changes. Another potential cause of difficulty is your choice of state variables. Most of you used total entropy and total number of moles in each chamber but the exponentials in the resulting state equations can cause numerical difficulties. Alternative choices (e.g., temperature instead of entropy) can make things simpler. 74 2.141 Term Project Pump Fault Detection and Diagnosis (FDD) Based on Electrical Startup Transient 2.141\termprj\MPL5rpt.doc 2002.12.12 The repeatability of start transients for a typical 3-phase, single speed HVAC pump is illustrated in Figure 1. The plot shows six smoothed 1 time histories of phase-to-neutral power observed on the A-phase. We wish to model the system (or as much of it as necessary) to explore the possibility of using this relatively easily observed signal [Laughman] as a basis for fault detection and diagnosis. Figure 1. 50-hp chilled water pump startup, A-phase power, 120Hz sampling, 6 repetitions A common approach to fault detection and diagnosis is to model the target system, then simulate its behavior under various conditions. After the correctly functioning system has been characterized, various faults can be modeled and the simulations repeated. At this point there are two possible paths. One is to find distinctive features of the responses that can be associated with each fault. This may be difficult. The other approach is to use the simulation results to see if system parameters, changes in which might be associated with various faults, can be estimated from the responses. This is also difficult. In either case, the first step is to model the system in order to better understand it, to find out how it responds under various conditions—special excitations as well as typical—and to learn if the responses contain information that could be used to identify faults or nascent failure by one of the two diagnostic paths. At some point, real data is needed to test detection and diagnostic procedures. Since we already have measured responses of a real system, the main goals of this term project have been to understand the system and develop a verifiable model, i.e. a model that reproduces the measured responses. System Overview. The main elements of the system to be modeled are the power source, motor, pump, and hydraulic load. Inertances and resistances are likely to be important determinants of startup behavior. For example, on the electrical side, inrush current is governed by source [Shaw, 2000] and stator resistances and inductances. The inrush current time constant is likely to be small compared to other system time constants. The flow transient is governed, in the fluid 1 by a “spectral envelope” filter (Appendix A)—a low-pass filter found [Leeb] useful in categorizing the start transient signatures of individual electric loads of buildings and other energy-using facilities. Children's Court 17 June P4, 6 reps 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 22000 1 13 25 37 49 61 73 85 97 109 121 133 145 157 169 181 193 Time (1s/tick) s i g n a l ( A - p h a s e A ) 1 2 3 4 5 6 75 Armstrong 2/29 domain, by inertances and resistances in the piping circuit; it may, depending on coupling and relative time scales, be significantly affected by motor and pump rotor inertances. The non-linear natures of motor, pump, and load static performance are, of course, very basic to the process because the entire range of operation is traversed during startup. Ringing, evident on the tail in Figure 1 2 , could stem from compliances in the motor field, in mechanical coupling of motor to pump, or in the hydraulic circuit 3 . In the latter case, compliances could be near to or remote from the pump—or distributed over the circuit. Model 1. We can represent the motor, pump, and load by their non-linear static behaviors. The source/nature of ringing can then be explored by inserting plausible compliances, inertances, and damping. Typical motor and pump static performance curves, taken from 9.3.3-4 of [Brown], are shown in Figures 2 and 3. Figure 2. Typical induction motor torque-speed curve. The pump characteristics are given in terms of non-dimensional shaft torque, static pressure, and flow rate. The general impeller pump similitude relations [Sabersky] are: | | . | \ | = | | . | \ | = L Q L Q g L P L Q L Q f L T µ e µ e µ µ e µ e µ , ' , ' 3 2 2 3 2 5 where T = impeller shaft torque, P = static pressure between the pump’s discharge and suction ports, Q = volumetric flow rate, µ = fluid viscosity, = fluid density (constant for incompressible flow), L = impeller outside diameter 4 , and = shaft speed. 2 Very similar ringing profiles have been observed consistently in a wide range (5-60 hp) of 3-phase induction motor driven loads including belt-driven fans as well as direct-coupled pumps 3 It could also result from electric side inertance which transforms, via motor gyrator action, to mechanical- side compliance; our initial sense is that the transformed inductances typically encountered would be too small to account for the rather low resonant frequency observed. 4 In the flow coefficient, Q/L 3 , L 3 may by replaced by WL 2 where W is the impeller discharge width. 0 20 40 60 80 100 120 140 160 180 200 0 5 10 15 f t - l b torque (ft-lb) shaft speed (rad/sec) 76 Armstrong 3/29 Figure 3. Radial-flow pump efficiency, torque, pressure versus dimensionless flow Q/L 3 e. The effect of Reynolds number, Re = Q/µL, is customarily neglected even though viscous effects tend to be significant at low Re (low Q). Our justification for using the simpler form (Eulerian similitude) is that torque and speed are dominated by inertial effects early in the start transient when Q is low. Thus we have: | | . | \ | = = | | . | \ | = = e µ e µ e µ e µ 3 2 2 3 2 5 Pr L Q g L P essure nd L Q f L T ndTorque The system can be readily modeled with a shaft coupling compliance and separate motor and pump rotor inertias. Including the hydraulic inertia is also straightforward. The bond graph representation is shown in Figure 4 and the governing equations are: 0 2 4 6 8 10 12 14 0 0.5 1 pumpEfficiency=eta 0 2 4 6 8 10 12 14 0.5 1 1.5 2 ndTorque=f'(flowCoeff) 0 2 4 6 8 10 12 14 0 0.1 0.2 ndPressure=g'(flowCoeff) 77 Armstrong 4/29 Figure 4. Bond graph of Model 1 with a shaft coupling compliance. ) ( ) , ( ) , ( ) ( Q P Q P I Q Q T C I C T I L p p p Q p p p p p m m m m p m ÷ = ÷ = ÷ = ÷ = e e e e e e e e e where m , p I m , I p are angular velocities and inertances of the motor and pump rotors, , C are the angular deflection and compliance of the motor-pump coupling, Q, I Q are hydraulic flow rate and inertance, T p (),P p () are pump’s static head and torque characteristics, T m () is the motor’s torque-speed curve, and P L () is the hydraulic load’s pressure-flow rate curve. Note that the motor is treated as a black box in this initial model. Back emf and, to the extent that they affect developed steady-state torque, the nonlinear self- and mutual- inductance relations involving rotor position wrt the rotating fields 5 are embedded in this black-box. But the torque-speed curve does not explicitly say anything about load current. We hope, at most, to see if the ringing apparent in the electrical response might plausibly be associated with mechanical and hydraulic inertances coupled by a compliance. Model 1 Simulation. The equations governing Model 1 are coded for Matlab simulation in Appendix D. The simulated system responses during a startup are shown in Figure 5a with I m =0.008, I p =0.042 ft-lb/s 2 . The effect of removing the compliance is estimated by the same model with the coupling made very stiff (low compliance). Results of one such simulation are shown in Figure 5d. Note that the panel labeled “motor torque” in all of the Figure 5 cases should be interpreted as approximate electromagnetic torque, rather than torque available at the shaft, since rotor inertia is modeled downstream of the static motor performance block. 5 note that the magnetic storage field is not modeled and hence the (modulated) effective mechanical compliance, observed at the shaft of a real induction motor, does not appear in this model of the system. T(e m ) 1 0 C c I m 1 I p T p (e p ,Q) P p (e p ,Q) 1 I fluid P L (Q) e m e p o Q T(e m ) 1 0 C c I m 1 I p T p (e p ,Q) P p (e p ,Q) 1 I fluid P L (Q) e m e p o Q 78 Armstrong 5/29 Figure 5. Start transient of the Model1 with I m =0.008, I p =0.042 (ft-lb/s 2 ), C=. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 100 200 pump speed [1/s] 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 100 200 motor speed [1/s] 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 0.2 0.4 flow rate [cfs] time (s) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 2 4 6 x 10 -3 h y d . l o a d ( p s i ) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 5 10 15 m o . t o r q u e ( f t - l b ) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 100 200 c p l g . d e f l ( r a d i a n s ) 79 Armstrong 6/29 Figure 5d. Start transient with I m =0.0001, I p =0.0499 (ft-lb/s 2 ). 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 2 4 6 x 10 -3 hyd.load (psi) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 5 10 15 mo.torque (ft-lb) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 5 10 15 cplg.defl (radians) time (s) 80 Armstrong 7/29 Model 2. The ringing obtained in the simulation of Figure 5, which involved shaft compliance, is similar to the observed ringing depicted in Figure 1. However, it is not quite plausible that—even during the brief period of startup when the motor develops peak torque—a shaft coupler would deflect 10’s or 100’s of radians! We expect that similar responses might be obtained by inserting a suitable compliance at any one of several other points in the system. In fact, we know that change in flux linkage with shaft position appears as a mechanical compliance and a given compliance inserted at this bond-graph location will produce the slowest and largest oscillations because it puts all of the significant (mechanical and hydraulic) inertances downstream. To model this variant of the system we must invert the torque characteristic. With the shaft coupling made rigid, motor and pump rotor inertances are combined. The bond graph representation is shown in Figure 6 and the governing equations are: ) ( ) , ( ) , ( ) ( ) ( Q P Q P I Q Q T C I I C L p p p Q p p p p m p p m ÷ = ÷ = + ÷ = e e e e e e e where I m + I p is the inertance of the motor and pump rotors combined, m (T) is the motor’s inverted torque-speed curve, and , C are now the angular deflection and compliance of the stator-rotor “coupling.” Figure 6. Bond graph representation of the system with a motor field compliance. The inverted torque function is two-valued except at the torque peak. For the purpose of simulating the start transient, this difficulty might be overcome by taking the value of shaft speed that is higher than the value used in the previous time step (the initial value is given as zero) or by taking the value closest to the pump shaft speed. More likely, the causality problem is simply confirmation that an effort source cannot be easily tricked into exhibiting compliance. In any event, other difficulties remain: 1) the electromagnetic compliance is, in reality, far from linear and 2) the responses of main interest—phase currents and total motor power—have not been addressed. e m (T) 0 C em 1 I m+p T p (e p ,Q) P p (e p ,Q) 1 I fluid P L (Q) e p o Q e m (T) 0 C em 1 I m+p T p (e p ,Q) P p (e p ,Q) 1 I fluid P L (Q) e p o Q 81 Armstrong 8/29 Induction Motor Model. Since model 2 faces a number of difficulties, the approach of using a linear mechanical compliance to approximate the air-gap flux to rotor perturbation relation was abandoned in favor of using a more realistic and informative motor model. Standard 3-phase induction motor designs have multiple stator polls (most commonly 2 or 4 poles per phase) and a smooth rotor with a distributed “winding” such that the current in each rotor bar is distinct. The apparent system order is depressingly large. If one considers air gap flux the situation is somewhat better. However, a transformation developed in the 1920’s [Park] that allows the electrical elements to be modeled in terms of one complex stator flow-effort pair (i qds ,v qds ) and one complex rotor flow-effort pair (i qdr ,v qdr ) has been used ever since, with variations [Stanley, Kron, Krause, Karnopp, Breedveld], to simulate dynamic performance of induction and synchronous machines. The assumptions commonly made in the analysis of 3-phase, line-powered machines are 1) balanced construction and excitation, 2) linear magnetics (to be relaxed later), and 3) sinusoidal distribution of air-gap flux. In the qd domain the very simple equivalent circuit model shown in Figure 7 is used to evaluate input and output power when the standard assumptions hold. Although not competent to model the significant transients expected on startup, an even simpler model (essentially half of Figure 7; see Appendix C) applies under steady state conditions. These circuits’ similarity to the usual transformer equivalent circuit can help one understand most of the important behaviors, such as locked rotor currents and how rotor resistance shapes the steady-state torque-speed curve. Note that the two circuits of Figure 7 contain just four independent energy storage elements, e.g. the direct and quadrature components of stator and rotor leakage flux. (The primes indicate rotor parameters referred to the stator side by the stator-rotor turns ratio). Also note that there is no difficulty adding source impedance elements to the left of the stator RL elements. Figure 7. Magnetically superposed equivalent circuits in qd coordinates rotating at e. Since machine behavior with rotor and air-gap flux transients is non-linear, we must numerically integrate the state equations. Park’s transformation, besides reducing the number of electrical energy storage state variables, makes the treatment of the two independent angular velocities manageable. The transformation is applied as shown in Figure 8 [after Krause, slightly modified]. Each of the transformation matrices, K s and 82 Armstrong 9/29 K r , is a function of e and analogous to kinematic’s position modulated transformer. For induction motors, wherin v qdr =0, synchronous speed (e=e e ) is the most convenient choice for generating state equations and results in v qds being constant rather than periodic. Figure 8. Simulation block showing conjugate power variables and their causality If the input voltages are given as boundary conditions (e.g., v abcr = [0 0 0] T and v abcr = 1.414sin(t)[277 277 277] T where [a b c] T is a column vector) the transformed voltages may also be treated as given. Thus a simulation can be run with just the two center blocks. The i abcs and i abcr currents, if needed, can be computed by a post processor. The usual constitutive equation for inductive circuit elements, = Li = Xi, must be expressed in matrix form when multiple windings share various portions of the magnetic circuit. As it happens, the matrix resulting from the topology of Figure 7 has four diagonal sub-elements such that it can be inverted by inspection, thus: K s K s -1 K r K r -1 Flux Linkages Currents Torque v qds i qds i qdr v qdr v abcs i abcs i adcr v abcr Rotor Speed T L T e = e r r m lr rr m ls ss m rr ss dr qr ds qs ss M ss M M rr M rr dr qr ds qs dr qr ds qs rr M rr M M ss M ss dr qr ds qs X X X and X X X X X X D where X X X X X X X X D i i i i i i i i X X X X X X X X + ' = ' + = ÷ ' = ( ( ( ( ¸ ( ¸ ' ' ( ( ( ( ¸ ( ¸ ÷ ÷ ÷ ' ÷ ' = ( ( ( ( ¸ ( ¸ ' ' ( ( ( ( ¸ ( ¸ ' ' ( ( ( ( ¸ ( ¸ ' ' = ( ( ( ( ¸ ( ¸ ' ' , , 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 2 ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ 83 Armstrong 10/29 Model 3. The topology of Figure 8 maps rather directly to the bond graph depicted in Figure 9. We choose state equations that use the foregoing linear transformation, X, of the currents, i, to flux linkage rates, , because use of flux states is both more convenient (one state variable derivative in each equation) and (by using the net magnetizing flux instead of separate rotor and stator contributions) more intuitive. The resulting state equations are: Mechanical load can be modeled as a simple linear load by adding to the rotor moment of inertia (increasing J) and letting T Load = Br, or the mechanical and hydraulic load static coupler equations from model 1 or 2 can be added instead to model non-linear pump and load characteristics. Figure 9. Bond graph of Figure 8 with mechanical/hydraulic load elements added. poles of number P and X X X Y where T X Y P J X X Y r v X X Y r v X X Y r v X X Y r v m rr ss Load e qr ds dr qs m r qr r ds m dr ss r dr e dr dr r qs m qr ss r qr e qr qs dr m ds rr s ds e ds ds qr m qs rr s qs e qs = ÷ ' = ÷ ' ÷ ' = ' ÷ ÷ ÷ ' ' ÷ ' = ' ' ÷ ÷ ÷ ' ' ÷ ' = ' ÷ ' ÷ ' ÷ = ÷ ' ÷ ' ÷ = ÷ 2 / 1 2 2 2 2 2 2 ) ( / ) ( ) 4 / 3 ( ) ( )) ( ( ) ( )) ( ( )) ( ( )) ( ( e ¢ ¢ ¢ ¢ e ¢ e e ¢ ¢ e ¢ ¢ e e ¢ ¢ e ¢ e¢ ¢ ¢ e ¢ e¢ ¢ ¢ e ¢ S e 1 I m +I p T p (e p ,Q) P p (e p ,Q) 1 I fluid P L (Q) e p = e r Q IC MTF MTF S e 3 2 2 3 e 84 Armstrong 11/29 Model 3 Simulation Results. The no-load start responses of a 50hp motor using parameters suggested by Krause (from Cathay; reproduced in Appendix B) are shown in Figures 10a-d. Figure 10a shows the 3-phase electrical power trajectory of interest. The origin of the early “chop” observed in Figure 1 is now clearly evident. It is essentially the load of a transformer with its secondary shorted and its mutual inductance modulated by rotor slip. The response of rotor speed, Figure 10b, shows that air-gap torque is similarly modulated even though only a small fraction of input power is being converted to mechanical power in this early part of the motor startup. Note that 3-phase motor power can be evaluated two ways 1) in machine coordinates: the sum of three phase loads, each the product of phase-to-neutral voltage and the in-phase component of current, and 2) in qd coordinates: the product of direct stator current and voltage. Figure 10c shows the three phase currents in panel 1 and the stator direct and quadrature currents in panel 2. The equivalent direct stator current (note scale change from 10a) is also plotted—and seen to deviate very little from the direct component of qd current. Figure 11 shows the static torque-speed curve (red) and the torque-speed state trajectory obtained in the simulation. Figures 12 and 13 show the effect of reducing rotor resistance. The rotor resistance used in the previous simulation (Figures 10-11) is typical of a 50hp motor produced before 1973. The trend since 1973 has been to market “energy-efficient” motors with lower stator and rotor resistances for a given nominal power rating. Stator resistance affects motor efficiency. Rotor resistance also affects motor efficiency but, in addition, has a profound effect on the shape of the static torque-speed curve (Appendix C). In Figure 12 the rotor resistance is halved and in Figure 13 the rotor resistance is halved again. The increased steepness of the torque speed curve above the torque peak is apparently equivalent to an increased spring constant in the mechanical domain. This implies that slip (rotor deviation from synchronous speed normalized by synchronous speed) is the displacement variable associated with air gap flux storage. Note that induction and synchronous motors are very different in this respect: torque angle (deviation from synchronous position) is the equivalent of mechanical displacement in the latter. For a given peak torque, an induction motor is much more compliant than the corresponding synchronous motor. The results of Figures 10-13 show that the ringing observed in the measured start transients of Figure 1 (and all of the twenty-some other delta-connected pumps and fans tested to date 6 ) can be reproduced by adjusting the motor resistances and a single parameter representing the system aggregate mechanical inertance. 6 7.5 to 50 hp for heating and cooling water pumps, supply, return, and cooling tower fans, in four buildings. 85 Armstrong 12/29 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 50 100 150 200 250 300 350 400 Time (s) R o t o r S p e e d ( r a d / s ) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 -200 0 200 400 600 s t a t o r ( a , b , c ) c u r r e n t s ( A ) -1000 -500 0 500 1000 S t a t o r L o a d ( A ) 3-Phase Load: ids(red) iqs(blue) P/Vrms(green) |ids+jiqs|(blk) total: black re(ia): red im(ia): blue 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 100 200 300 400 500 600 n r m l z d 3 - p h s l o a d ( P a + P b + P c ) / V r m s ( A ) Figure 10. Simulated startup of a 50hp induction motor with no load. 86 Armstrong 13/29 0 50 100 150 200 250 300 350 -1000 -500 0 500 1000 1500 2000 Rotor Speed (rad/s) T o r q u e ( N - m ) Figure 11. Torque-speed curve (red) and corresponding state path during simulated startup 0 0.2 0.4 0.6 0.8 1 1.2 1.4 -100 0 100 200 300 400 500 600 n r m l z d 3 - p h s l o a d ( P a + P b + P c ) / V r m s ( A ) p.rs = 0.087; p.rr = 0.114; p.J = 1.662; p.Bl = 0; Figure 12. Simulated startup of the 50hp induction motor with r' r = 0.114O and no load. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 -100 0 100 200 300 400 500 600 700 n r m l z d 3 - p h s l o a d ( P a + P b + P c ) / V r m s ( A ) p.rs = 0.087; p.rr = 0.057; p.J = 1.662; p.Bl = 0; Figure 13. Simulated startup of the 50hp induction motor with r' r = 0.057O and no load. 87 Armstrong 14/29 0 50 100 150 200 250 300 350 -600 -400 -200 0 200 400 600 800 1000 Rotor Speed (rad/s) T o r q u e ( N - m ) Figure 14. Transient and nominal torque-speed curves, 50hp motor, r' r = 0.057O, no load. Thermal Model. The first order model considers i 2 R heating of the balanced stator windings as a single lumped thermal capacitance with conductance of the resulting heat to ambient. The rotor is the same but has its own parameters and separately tracked temperature state. This two-node model can be tweaked to work well early in startup before penetration of the heat waves into the stator and rotor core masses becomes significant. With a thoughtfully elaborated conductance network, it can also give good agreement with steady state operating temperatures. In reality, portions of the rotor and stator windings do exhibit approximate first order behavior. Parts that lose most of their heat via the iron core, however, should be modeled by a distributed parameter diffusion model or some multi-node approximation thereof. The bond graph is shown in Figure 15. The possibility of using distributed resistance-capacitance (RC) elements to model the motor’s iron core elements is indicated by dashed lines surrounding the iron elements currently represented as lumped RC elements. Notation is defined below: u s i 2 r s = heat generated in fraction, u s , of stator winding that is embedded in the stator core, u r i 2 r r = heat generated in fraction, u r , of rotor winding that is embedded in the rotor core, (1-u s )i 2 r s = heat generated in fraction of stator winding that is exposed to cooling air, (1-u r )i 2 r r = heat generated in fraction of rotor winding that is exposed to cooling air, R(e) = one of several 7 thermal conductances to the cooling air, functions of rotor speed, R = one of several 7 thermal resistances in the winding and core materials, C cs = thermal capacitance of portion of the stator conductors embedded in the stator core, C csair = thermal capacitance of portion of the stator conductors exposed to cooling air, C isid = thermal capacitance of the inner (air gap-cooled) portions of the stator iron, C isod = thermal capacitance of the outer portions of the stator iron and motor frame, C cr = thermal capacitance of portion of the rotor conductors embedded in the rotor core, C crair = thermal capacitance of portion of the rotor conductors exposed to cooling air, C ir = thermal capacitance of the rotor iron, and C agap = thermal capacitance of the cooling air (mainly to eliminate an algebraic equation). 7 thermal resistance subscripts have been omitted to reduce clutter in the figure; the function of a given resistance is indicated by the subscripts on the capacitances associated with the zero junctions connected by the resistance in question. 88 Armstrong 15/29 Figure 15. Thermal model involving four stator and three rotor capacitances; alternatively, consider this one of many cross-coupled sections arrayed along shaft with R agap –T amb ap- pearing only in the first slice and R–C c*air –R only in the first and last (half to each) slices. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 50 100 150 200 250 300 T s - b l u e , T r - r e d ( C ) -100 0 100 200 300 400 500 600 700 ( P a + P b + P c ) / V r m s ( A ) Figure 16. Response of the 50hp motor start using lumped rotor and stator thermal masses. 0 C isid R(e) 0 0 C ir R(e) 0 C cs R(e) 0 C cr R(e) R R R 0 C csair R 0 0 C crair R RR (1-u s )i 2 r s (1-u r )i 2 r r u r i 2 r r u s i 2 r s R R T amb 0 C isod C agap 0 C isid R(e) 0 0 C ir R(e) 0 C cs R(e) 0 C cr R(e) R R R 0 C csair R 0 0 C crair R RR (1-u s )i 2 r s (1-u r )i 2 r r u r i 2 r r u s i 2 r s R R T amb 0 C isod C agap 0 C isid R(e) 0 0 C ir R(e) 0 C cs R(e) 0 C cr R(e) R R R 0 C csair R 0 0 C crair R RR (1-u s )i 2 r s (1-u r )i 2 r r u r i 2 r r u s i 2 r s R RR T amb 0 C isod C agap 89 Armstrong 16/29 A first order version of this model has been implemented. Results, for the 50hp motor with reduced (r r = 0.057) rotor and default (r s = 0.087) stator resistances, are shown in Figures 16 and 17. It is important to implement a more realistic model 8 of at least the rotor because its electrical resistance has such a profound effect on the torque-speed curve (compare Figure 17 to the corresponding result with no thermal model, Figure 14). 0 50 100 150 200 250 300 350 -600 -400 -200 0 200 400 600 800 1000 Rotor Speed (rad/s) T o r q u e ( N - m ) Figure 17. Deviation of transient from nominal torque stems largely from rotor heating. Saturation Model. The simplest reasonably competent saturation model characterizes a flux-current relation for the entire balanced magnetic circuit using the magnetizing current in qd coordinates [Corzine]. The magnetizing reactance, X m in Model 3, can no longer be treated as a constant, rather it is a function of magnetizing flux. The state equations of model 3 become: This introduces an algebraic loop because X = is a function of the direct component of the stator-rotor common flux, md , which, in turn, is given in terms of X = : 8 Locked-rotor tests can be performed to estimate the thermal parameters if thermocouples are staked or drilled into a rotor bar and external airflow is provided to simulate the rotor fan airflow developed at rated speed. Responses to various faults that keep the rotor stalled might also be observed with the fan off. 1 1 1 1 ) ( )) ) / / ( )( / ( ( ) ( )) ) / / ( )( / ( ( )) ) / / ( )( / ( ( )) ) / / ( )( / ( ( ÷ = = = = = | | . | \ | + ' + = ' ÷ + ' ÷ ' ' ÷ ' ' + ' = ' ' ÷ ÷ ' ÷ ' ' ÷ ' ' + ' = ' + ÷ ' ' ÷ + = ÷ ÷ ' ' ÷ + = m lr ls qr r dr lr dr ls ds lr r dr e dr dr r qr lr qr ls qs lr r qr e qr qs ds lr dr ls ds ls s ds e ds ds qs lr qr ls qs ls s qs e qs X X X X where X X X X r v X X X X r v X X X X r v X X X X r v ¢ e e ¢ ¢ ¢ e ¢ ¢ e e ¢ ¢ ¢ e ¢ e¢ ¢ ¢ ¢ e ¢ e¢ ¢ ¢ ¢ e ¢ ) ( ( qr qs ls md X Y ¢ o ¢ ¢ ' ÷ = = 90 Armstrong 17/29 The non-linear reactance, X m ( md ), could be implemented by table lookup but a safer approach is to use the arctan function recommended by Corzine to avoid bad derivative behavior. This simplest of saturation models ignores the fact that leakage inductances become variable, increasing significantly at the onset of saturation. Next Steps. Effects of saturation and thermal transient parameters need to be explored. A test motor with known magnetic and thermal parameters is needed. A number of other tasks should also be pursued. For verification, the simulation results for A-phase direct and quadrature current and power must be passed through the appropriate spectral envelope filter before comparing to responses of the type depicted in Figure 1. We need to simulate a known 9 system and, if possible, one in which at least the piping circuit inertance and static characteristics can be varied, e.g. by two shunt valves, one close to and one remote from the pump. Measurements of rotor speed and pump pressure during startup would be very useful. Extension 10 of the system model to include resistance and inertance in the power supply as shown in Figure 18, should be implemented. Source impedance can move downstream of the MTF but dissipation from stator resistance must be separate from source dissipation. Adequacy of the Eulerian similitude model [Sabersky] needs to be checked for operation at low speeds and when the impeller is subjected to the much higher than normal torques that are developed briefly during startup. Faults, such as broken impeller vane, broken rotor bar, and nicked stator conductor, should be introduced into a test pump-motor system to establish the extent to which such parameter changes can be identified by parameter estimation based on electric-side start transient measurements. (review bkn rotor bar literature: start excitation? thermal model?} Figure 18. Bond graph of motor-pump-load system with source impedance added. Discussion. A 6 th order model, with rotor, impeller, and hydraulic inertances included, was developed and shown to reproduce the main features of measured start transients. The thermal behavior of the stator, that might raise stator resistance fast enough to explain the observed diminishing inrush current while the motor is still acting like a transformer 9 for which the start transient has been accurately and repeatably measured and for which all the of the system parameters referenced in the model have been measured or can be reasonably estimated. 10 This extension is not needed if the FDD system measures voltage waveform, as well as the phase of its fundamental and current, at the motor terminals. S e I m +I p T p (e p ,Q) P p (e p ,Q) 1 I fluid P L (Q) e p = e r Q IC MTF MTF S e 3 2 2 3 e 1 1 3 I src R src 91 Armstrong 18/29 with its secondary shorted, has not been modeled yet. The state equations that account for saturation have been developed but not yet coded in Matlab. The proposed FDD system measures all three phase currents and voltages, making detection of electrical imbalances in both the power supply and load very straightforward. This study therefore focused on mechanical and hydraulic aspects of the system to explore the extent to which they might be characterized and monitored by the electric-side measurements. Two derived responses, total real and total reactive motor power, were found, as expected based on the qd (Park’s) transformation, to be useful, given balanced electric-side conditions, for these purposes. The effects of parameter changes were explored. The following parameters had effects that were found to be important: stator and rotor winding resistances lumped mechanical/hydraulic inertance dominant compliance (function of motor parameters) stator and rotor thermal capacitances When a compliant coupling (belt or flexible shaft coupling) is added between the motor and fan or pump, the rotor inertance is decoupled from the downstream inertances by the compliance. Preliminary simulations indicate that the additional poles are too highly damped or too far from the real axis to be detected reliably from a typical start transient. Future work should explore whether such poles might be detected from response to step changes or impulses in excitation during normal run conditions, e.g. brief power interruption with stator leads open or shorted. There may be cases where compliances in the hydraulic system are significant to the observed responses. Further field testing is needed to explore this hypothesis. The most practical approach to FDD appears to involve a priori data. Blocked-rotor and no-load run-up responses are easily obtained by 1) making modest extensions of the FDD monitoring system’s software and 2) calling for such tests as part of building commissioning. The static torque curve is important, particularly the slope around nominal steady-state load. It is not clear if the curves given by manufacturers are sufficiently reliable and, if not, whether reliable electrical parameters (mutual inductance, rotor and stator resistance and leakage inductance) can be obtained in the field. The possibility of obtaining a reasonable static torque-speed curve during commissioning seems tenuous but may be worth exploring. Note that there are two important reasons to integrate the FDD system with the control system. First, it can simplify building commissioning and, second, it enables an active testing component of FDD—i.e., extends passive-monitoring-based FDD by enabling automated active testing processes. There is a third reason that does not necessarily apply to pumps but would certainly apply to other pieces of the plant, e.g., chillers: model-based FDD and model-based control can share the same model. To aid further development and refinement of on-line-model based FDD, it may be important to have the 3-phase, high bandwidth monitoring subsystem (under development) output the unfiltered (or lightly filtered) 3-phase power signal, sampled at a fairly high frequency (e.g. 900-1200 Hz), in addition to the spectral envelopes output by the preprocessor stage in the current single-phase implementation. 92 Armstrong 19/29 The problem of tracking shaft speed during transients is a difficult one [Velez-Reyes, Shaw 1999] that has not been addressed here. We have so far focused on FDD measurements, models and approaches that do not require a measurement or estimate of shaft speed. Belt driven shaft speeds, if needed, can be estimated in steady operation from the spectral envelope data [5/01 CEC progress report] and might also be reliably estimated for direct coupled pumps were the 3-phase power measurement, discussed above, made available. It would be reasonable to measure transient shaft speeds during commissioning if there is sufficient value in doing so. These are additional possibilities to be explored in the future. References. Breedveld, P.C., 2001. A generic dynamic model of multiphase electromechanical transduction in rotating machinery, Proc. WESIC. Brown, F.T. 2001. Engineering System Dynamics, Marcel Dekker, New York.. Cathay, J.J., et al, 1973. Transient load model of an induction machine, IEEE Trans. PAS 92(1399-1406). Corzine, K.A., et al, 1998. An improved method for incorporating magnetic saturation in the q-d synchronous machine model, IEEE Trans. Energy Conversion, 13(3) pp.270-275. Karnopp, D. 1991. State functions and bond graph dynamic models for rotary, multi- winding electrical machines, J. Franklin Institute, 328(1) pp.45-54. Krause, P.C., et al, 2002. Analysis of Electric Machinery and Drive Systems, 2 nd Edition, IEEE Press, Piscataway, NJ. Kron, G., 1951. Equivalent Circuits of Electric Machinery, J. Wiley and Sons, NY. Laughman, C.R., et al, 2002. Advanced non-intrusive monitoring of electric loads, submitted October 2002 to IEEE CAP. Leeb, S.B. and S.R. Shaw, 1994. Harmonic estimates for transient event detection, IEEE UPEC. Ong, C-M. 2998. Dynamic Simulation of Electric Machinery, Prentice-Hall, NJ. Park, R.H., 1929. A two-reaction theory of synchronous machines—generalized method of analysis, part I; AIEE Trans. 48(716-727). Sabersky, R.H. and A.J. Acosta, 1971. Fluid Flow, MacMillan, NY. Sen Gupta, D.P. and J.W. Lynn, 1980. Electrical Machine Dynamics, MacMillan. Shaw, S.R., and S.B.Leeb, 1999. Identification of Induction Motor Parameters from Transient Stator Current Measurements, IEEE Trans. Industrial Electronics, 46(1) pp.139- 149. Shaw, S.R., et al, 2000. A power quality prediction system, IEEE Trans. Industrial Electronics, 47(3) pp.511-517. Tavner, P.J. and J. Penman, 1987. Condition Monitoring of Electric Machines, Wiley. Velez-Reyes, M., et al, 1989. Recursive speed and parameter estimation for induction machines, Proc. IEEE-IAS Annual Meeting, pp.607-611. 93 Armstrong 20/29 Appendix A: Spectral Envelope Filter (after Laughman) The spectral envelope filter is an ideal demodulator for amplitude modulated signals. The ideal demodulator assumes a known carrier frequency. The carrier of interest for analysis of electric loads on the North American grid is, of course, 60Hz. For nonlinear loads, useful information may be lost with the 60Hz filter. In this case, signals obtained by spectral envelope filters applied at harmonics of 60Hz will contain additional information. If i(t) is the current of a load driven by a voltage, e(t) = Vsin(t), the direct and quadrature k th harmonic spectral envelopes are given by e e e 2 ) cos( ) ( 2 ) ( ) sin( ) ( 2 ) ( = = = ) ) ÷ ÷ T where d k x T t b d k x T t a t T t k t T t k Additional smoothing can be obtained (albeit with additional loss of information) by using integer multiples of T. {Matlab function to implement spectral envelope filter goes here.} Appendix B. Motor and Pump Data Table B-1. Bell and Gosset 1510 Series, Model 6E, 10.5” impeller. Data scaled from B&G published performance chart at 1770 rpm. Gpm ft.WC hp Eff 420 102 25 44 520 101.9 26.8 50 690 101.6 29.9 60 930 101 34.2 70 1070 99.6 36.5 75 1250 97.7 38.9 80 1465 94.2 41.6 85 1650 90.3 43.5 86 1895 82.5 45.9 87 2040 77 46.4 86 2125 73.5 46.7 85 2295 64 46.6 80 2415 57.5 46.3 75 Table B-2. Induction motor parameters; Cathay (1982, Table 4.10 in Krause), Ong (1998) Machine Rating T B I B R s X ls X M X´ lr R´ r J Hp V Rpm N-m A kg-m 2 V* X M hp 3 220 1710 11.9 5.8 .435 .754 26.13 .754 .816 .089 20 220 1748 79.2 49.7 .1062 .2145 5.834 .2145 .0764 2.8 50 460 1705 198 46.8 .087 .302 13.08 .302 .228 1.662 500 2300 1773 1980 93.6 .262 1.206 54.02 1.206 .187 11.06 2250 2300 1786 8900 421.2 .029 .226 13.04 .226 .022 63.87 94 Armstrong 21/29 Appendix C: Steady-State Motor Model: Equivalent circuit, torque equation, and torque-speed curves. In the qd domain a very simple equivalent circuit model, Figure B-1 (based on Figure 7) can be used to evaluate input and output power under steady state conditions. Although not useful for simulating certain details of transient response, its similarity to the standard transformer equivalent circuit does help one understand most of the important motor behaviors, such as locked rotor currents and how rotor resistance shapes the steady-state torque-speed curve. function [w,ssTe]=ssTcurve(p); % [w,ssTe]=ssTcurve(p) % given induction motor parameters (p=pIND*hp.m) return torque-speed % curve, [w ssTe]; note that friction and windage are neglected %p.{PP rs rr Xm Xls Xlr we J Bl vds vqs vqr vdr Tl Xss Xrr Y} D=1/p.YY; G=(p.rs^2 + p.Xss^2)/(p.rs^2*p.Xrr^2 - D); sstep=.01; w=[0:sstep:1]; w=p.we*w; for i=1:1/sstep s=1-(i-1)*sstep; x=3*p.PP*p.Xm^2*p.rr*s*p.vds^2/2/p.we; ssTe(i)=x/((p.rs*p.rr+s*D)^2 + (p.rr*p.Xss+s*p.rs*p.Xrr)^2); end ssTe(1+1/sstep)=0.0; function [smax,Temax]=ssTmax(p) % [smax,Temax]=ssTmax(p) % given induction motor parameters (p=pIND*hp.m) return slip, smax % at which maximum electromagnetic torque, Temax, is developed. %p.{PP rs rr Xm Xls Xlr we J Bl vds vqs vqr vdr Tl Xss Xrr Y} D=1/p.YY G=(p.rs^2 + p.Xss^2)/(p.rs^2*p.Xrr^2 - D) smax=p.rr*G Temax=3*p.PP*p.Xm^2*G*p.vds^2/2/p.we/((p.rs-G*D)^2 + (p.Xss+G*p.Xrr)^2) % pssIND1.m runs [p pINDname] = pind50hp;% set motor parameters [smax Tmax]=ssTmax(p); [w Tss]=ssTcurve(p); %subplot(2,1,1); p.rs=.087; p.rr=1.824;[w Tss]=ssTcurve(p);plot(w,Tss);hold on p.rr=.912;[w Tss]=ssTcurve(p);plot(w,Tss); p.rr=.456;[w Tss]=ssTcurve(p);plot(w,Tss); p.rr=.228;[w Tss]=ssTcurve(p);plot(w,Tss); p.rr=.114;[w Tss]=ssTcurve(p);plot(w,Tss); p.rr=.058;[w Tss]=ssTcurve(p);plot(w,Tss); p.rr=.029;[w Tss]=ssTcurve(p);plot(w,Tss); p.rr=.0145;[w Tss]=ssTcurve(p);plot(w,Tss); %ylabel('Torque (N-m)') %subplot(2,1,2); p.rs=.0435; p.rr=1.824;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.912;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.456;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.228;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.114;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.058;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.029;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); p.rr=.0145;[w Tss]=ssTcurve(p);plot(w,Tss,'g'); xlabel('Speed (rad/s)');ylabel('Torque (N-m)');hold off 95 Armstrong 22/29 0 50 100 150 200 250 300 350 400 0 100 200 300 400 500 600 700 800 900 1000 Speed (rad/s) T o r q u e ( N - m ) Figure C-1. Steady-state torque-speed curves for 50hp induction motor with no mechanical losses and rotor resistances (left to right) of r´ r = 1.824, 0.912, 0.456, 0.228, 0.114, 0.058, 0.029, 0.0145; stator resistances are r s = 0.087 (blue) and r s = 0.087 (green). (describe LRA and variation of Rr with slip to represent gyration to mechanical power. give torque eqn and explain gyration). 96 Armstrong 23/29 Appendix D. Simulation script and derivatives function for Model 1 %simpump02.m 2002.11.11pra pndPofndQ=[.174,0,-.000328,-.0000281];pndPofndQ=pndPofndQ(length(pndPofndQ):-1:1); pEffofndQ=[0,.2,-.0238,.00158,-.000053];pEffofndQ=pEffofndQ(length(pEffofndQ):-1:1); pTofW=[60,1.2,0,0,-1453e-9,182e-10,-635e-13];pTofW=pTofW(length(pTofW):-1:1); %pPofQ=[.0005 0 30];%arg in cfs, result in psi pPofQ=[.0005 0 0];%arg in cfs, result in psi I.motor=0.02;I.pump=0.03;I.fluid=8310;% ft-lb/2; lb-s2/ft5 C.coupl=0.05;%1/10 is compliant, 1/100 is stiff [ft-lb/radian] T0=.01*polyval(pndPofndQ,.01)/polyval(pEffofndQ,.01);%else Q/eff=undefined at speed=0 [t,y]=ode15s('motorpumpload02',[0 1.6],[0 0 0 0],[],1.94,1,I,C,T0,pTofW,pPofQ,pndPofndQ, pEffofndQ); subplot(3,1,1);plot(t,y(:,1));title('pump speed [1/s]'); subplot(3,1,2);plot(t,y(:,3));title('motor speed [1/s]'); subplot(3,1,3);plot(t,y(:,2));title('flow rate [cfs]');xlabel('time (s)'); function f=motorpumpload02(t,x,flag,rho,L,I,C,T0,pTofW,pPofQ,pndPofndQ,pEffofndQ) %x(1),f(1) = pump shaft speed [1/s] and its time derivative [1/s2] %x(2),f(2) = pump flow rate [cfs] and its time derivative [ft3/s2] %x(3),f(3) = motor shaft speed [1/s] and its time derivative [1/s2] %x(4),f(4) = shaft coupling deflection [rad] and its time derivative [rad/s] f=[0 0 0 0]'; if x(1)>0; flowCoeff=x(2)/x(1)/L^3;else;flowCoeff=0;end loadPressure=polyval(pPofQ,x(2))*144;%convert from psi to psf ndPressure=polyval(pndPofndQ,flowCoeff*10000);%g' pumpEfficiency=polyval(pEffofndQ,flowCoeff*10000);%eta if flowCoeff>.01; ndTorque=flowCoeff*ndPressure/pumpEfficiency;%10000f' else;ndTorque=T0;end moTorque=polyval(pTofW,x(3))/12;%convert from in-lb to ft-lb f(1)=(x(4)/C.coupl - rho*L^5*x(1)^2*ndTorque/10000)/I.pump; f(2)=(rho*(L*x(1))^2*ndPressure - loadPressure)/I.fluid;if f(2)<0; f(2)=0;end f(3)=(moTorque - x(4)/C.coupl)/I.motor; f(4)=x(3)-x(1); 97 Armstrong 24/29 Appendix E. Simulation script and derivatives function for Model 2 %simpump03.m 2002.11.12pra pndPofndQ=[.174,0,-.000328,-.0000281];pndPofndQ=pndPofndQ(length(pndPofndQ):-1:1); pEffofndQ=[0,.2,-.0238,.00158,-.000053];pEffofndQ=pEffofndQ(length(pEffofndQ):-1:1); pTofW=[60,1.2,0,0,-1453e-9,182e-10,-635e-13];pTofW=pTofW(length(pTofW):-1:1); %pPofQ=[.0005 0 30];%arg in cfs, result in psi pPofQ=[.0005 0 0];%arg in cfs, result in psi n=length(pTofW)-1; d1=pTofW(1:n);d1=d1.*[n:-1:1] p=d1;p=roots(p);k=0; for i=1:n-1;%eliminate infeasible roots if imag(p(i))==0 & real(p(i))>0; k=k+1; p(k)=p(i);end end wmax=min(p(1:k));maxT=polyval(pTofW,wmax); I.motor=0.02;I.pump=0.03;I.fluid=8310;% ft-lb/2; lb-s2/ft5 C.coupl=0.05;C.emag=0.05;%1/10 is compliant, 1/100 is stiff [ft-lb/radian] T0=.01*polyval(pndPofndQ,.01)/polyval(pEffofndQ,.01);%else Q/eff=undefined at speed=0 [t,y]=ode15s('motorpumpload03',[0 1.6],[0 0 0],[],1.94,1,I,C,T0,pTofW,maxT,pPofQ,pndPofndQ,pEffofndQ) % function f=motorpumpload01(t,x,flag,rho,L,I,T0,pTofW,pPofQ,pndPofndQ,pEffofndQ) subplot(3,1,1);plot(t,y(:,1));title('pump speed [1/s]'); subplot(3,1,2);plot(t,y(:,3));title('motor speed [1/s]'); subplot(3,1,3);plot(t,y(:,2));title('flow rate [cfs]');xlabel('time (s)'); function f=motorpumpload03(t,x,flag,rho,L,I,C,T0,pTofW,maxT,pPofQ,pndPofndQ,pEffofndQ) %x(1),f(1) = pump shaft speed [rad/s] and its time derivative [rad/s2] %x(2),f(2) = pump flow rate [cfs] and its time derivative [ft3/s2] %x(3),f(3) = field-rotor deflection [rad] and its time derivative [rad/s] %global wm;%need initial guess on first call (previous value is fine for subsequent calls) %another ploy that might work: use pump speed as the initial guess f=[0 0 0]'; if x(1)>0; flowCoeff=x(2)/x(1)/L^3;else;flowCoeff=0;end loadPressure=polyval(pPofQ,x(2))*144;%convert from psi to psf ndPressure=polyval(pndPofndQ,flowCoeff*10000);%g' pumpEfficiency=polyval(pEffofndQ,flowCoeff*10000);%eta if flowCoeff>.01; ndTorque=flowCoeff*ndPressure/pumpEfficiency;%10000f' else;ndTorque=T0;end moTorque=12*x(3)/C.emag; %convert from ft-lb to in-lb %=polyval(pTofW,x(3))/12; %use Matlab fcn ROOTS and fact that wm>wmprev during start transient p=pTofW; p(7)=p(7)-moTorque; p=roots(p); k=0; for i=1:6; %eliminate infeasible (neg & complex) roots if imag(p(i))==0 & real(p(i))>0; k=k+1; p(k)=p(i);end end [wm i]=min(p(1:k)); if moTorque>maxT; wm=min([p(1:i-1) p(i+1:k)]);end if length(wm)~=1; disp([moTorque maxT i]);disp(p);disp(wm);error('moTorque,maxT,i,p,wm (not scalar)');end %if x(1)==0; wm=0; end moTorque=moTorque/12 %SECANT METHOD will have trouble getting over torque peak %d1=pTofW(1:6);d1=d1.*[6:-1:1];%slope of polynomial %y=polyval(pTofW,w); yp=polyval(d1,w); dw=(y-moTorque)/yp; %while abs(dw)>0.0001; % w=w-dw; % y=polyval(pTofW,w); yp=polyval(d1,w); dw=(y-moTorque)/yp; %end %w=w-dw %f(1)=(x(3)/C.emag - rho*L^5*x(1)^2*ndTorque/10000)/(I.pump+I.motor); f(1)=(moTorque - rho*L^5*x(1)^2*ndTorque/10000)/(I.pump+I.motor); f(2)=(rho*(L*x(1))^2*ndPressure - loadPressure)/I.fluid;if f(2)<0; f(2)=0;end f(3)=wm-x(1); %wm is fcn of x(3) 98 Armstrong 25/29 Appendix F. Simulation script and functions for induction motor, Model 3 Model that assumes a linear magnetic system (linear flux-current relation) % punIND1.m runs ODE45 simulation of induction motor [p pINDname] = pind3Qhp;% set motor parameters [smax Tmax]=ssTmax(p); [w Tss]=ssTcurve(p); %plot(w,Tss);xlabel('Speed (rad/s)');ylabel('Torque (N-m)') %pause options = odeset('reltol',.0001,'abstol',1e-6*[1 1 1 1 1 1]); [tout,y] = ode45('pind1',[0,.5],[0 0 0 0 0 0],options,p); plot(tout,y(:,5)); xlabel('Time (s)'); ylabel('Rotor Speed (rad/s)'); title([date ' punIND1.m,pIND1.m, ' pINDname]); T = (3/2)*p.PP*(p.Y*p.Xm/p.we)*(y(:,1).*y(:,4) - y(:,3).*y(:,2)); figure(2); plot(y(:,5),T);hold; plot(w,Tss,'r'); xy=axis;xy(2)=p.we;axis(xy); xlabel('Rotor Speed (rad/s)'); ylabel('Torque (N-m)'); [m,m2]=pconvIND(tout,y, function [ydot] = pind1(t,y,flag,p) % [ydot] = pind1(t,y) % evaluate derivatives for standard model (Krause 2002, eqns 4.5-39,40 % and 4.6-6; see also Shaw & Leeb 1999) of a balanced, three phase % induction motor with linear magnetic system. The 5 state variables are % D and Q stator and rotor flux linkage rates, psi*, and rotor speed wr. % Displacement (th = y(6)) is not really part of the state vector but is % included for numerical consistency (same integration scheme). TEST NEED FOR THIS! % w is angular velocity of the arbitrary reference frame. w=we=synchronous % speed, e.g. 60*2pi, is most convenient for simulating the induction motor. %p.PP rs rr Xm Xls Xlr we J Bl vds vqs vqr vdr Tl Xss Xrr Y %Xss = p.Xls + p.Xm; Xrr = p.Xlr + p.Xm; Y = 1/(Xss*Xrr+Xm^2) psiqs = y(1); psids = y(2); psiqr = y(3); psidr = y(4); wr = y(5); %th = y(6); ydot = [0 0 0 0 0 0]'; ydot(1) = p.we*(p.vqs - p.rs*p.Y*(p.Xrr*psiqs - p.Xm*psiqr)) - p.w*psids; ydot(2) = p.we*(p.vds - p.rs*p.Y*(p.Xrr*psids - p.Xm*psidr)) + p.w*psiqs; ydot(3) = p.we*(p.vqr - p.rr*p.Y*(p.Xss*psiqr - p.Xm*psiqs)) - (p.w-wr)*psidr; ydot(4) = p.we*(p.vdr - p.rr*p.Y*(p.Xss*psidr - p.Xm*psids)) + (p.w-wr)*psiqr; % PP=P/2 is number of pole pairs T = (3/2)*p.PP*(p.Y*p.Xm/p.we)*(psiqs*psidr - psiqr*psids); ydot(5) = p.PP*(T - p.Tl)/p.J; ydot(6) = p.w; ; 99 Armstrong 26/29 Appendix E. Simulation script and derivatives function (X m solver in progress) for Model 3 with thermal and saturation effects % punIND1.m runs ODE45 simulation of induction motor; thermal states added 20021213 [p pINDname] = pind50hp;% set motor parameters [smax Tmax]=ssTmax(p); [w Tss]=ssTcurve(p); options = odeset('reltol',.0001,'abstol',1e-6*[1 1 1 1 1 1 1 1]); [tout,y] = ode45('pind1sat',[0,1.4],[0 0 0 0 0 20 20 0],options,p); figure(1); plot(tout,y(:,5)); xlabel('Time (s)'); ylabel('Rotor Speed (rad/s)'); title([date ' punIND1.m,pIND1.m, ' pINDname]); T = (3/2)*p.PP*(p.YY*p.Xm/p.we)*(y(:,1).*y(:,4) - y(:,3).*y(:,2)); figure(2); plot(y(:,5),T);hold; plot(w,Tss,'r'); xy=axis;xy(2)=1.04*p.we;axis(xy); xlabel('Rotor Speed (rad/s)'); ylabel('Torque (N-m)'); [m,m2]=pconvIND(tout,y,p); function [p,pINDname]=pIND50hp % pIND50hp.m loads motor parameters into struct p % Call pINDparm, then pIND1 via ODE45 to simulate motor start transient % These are the parameters for a typical 50 Hp, 480V (L-L) 3-phase motor % Krause (2002) p.165 after JJ Cathay et al IEEE PAS 92(1399-1406) 1973. pINDname='typ 50hp ind motor' p.PP = 2; % Number of pole pairs =P/2 p.vnmp = 460; % nameplate voltage (line-to-line rms) p.hpnmp = 50; % nameplate horsepower p.rpmn = 1705; % nameplate rpm %p.Vb= %p.Pb= %p.Tb= %p.Ib= %ORIGINAL VALUES p.rs = 0.087; % Stator resistance p.rr = 0.228; % Rotor resistance referred to stator (ractual*(Ns/Nr)^2) p.Xm = 13.08; % Magnetizing impedance, in Ohms on a 60 Hz base p.Xls = 0.302; % Stator leakage impedance, in Ohms on a 60 Hz base p.Xlr = 0.302; % Rotor leakage impedance p.we = 120*pi; % Base electrical frequency, rads per second (376.99 @60 Hz) p.J = 1.662; % Rotor inertia (kg-m-s2=N-m3?) p.Bl = 0; % Mechanical load damping coefficient %TEST VALUES p.rs = 0.087; p.rr = 0.057; p.J = 1.662; p.Bl = 0; p.vds = 391.9; % D axis stator voltage p.vqs = 0.0; % Q axis stator voltage p.vdr = 0.0; % D axis rotor voltage (used for starting large machines) p.vqr = 0.0; % Q axis rotor voltage p.Tl = 0.0; % Mechanical load torque p.w = p.we; % Angular velocity of the arbitrary reference frame p.Xss = p.Xls + p.Xm; p.Xrr = p.Xlr + p.Xm; p.YY = 1.0/(p.Xss*p.Xrr - p.Xm*p.Xm); %THERMAL MODEL PARMS (C=thermal capacitance, J/K or Btu/R) p.Cscslot= 0; % stator conductor shielded from air stream, J/K p.Cscair = 130; % stator conductor exposed to air stream p.Csi = 0; % stator iron (distributed RC and air coupling) p.Crcslot= 0; % rotor conductor shielded from air stream p.Crcair = 130; % rotor conductor exposed to air stream p.Cri = 0; % rotor iron (distributed RC and air coupling) p.Tamb = 20; % assume rs and rr given at Tamb p.salpha =.00393;% stator R coeff per degree temperature rise (copper:0.00393/K) p.ralpha =.00393;% rotor R coeff per degree temperature rise p.uscair = 30;%123; %W/K p.urcair = 30;%123; 100 Armstrong 27/29 function [ydot] = pind1sat(t,y,flag,p) % [ydot] = pind1sat(t,y) % evaluate derivatives for standard model (Krause 2002, eqns 4.14-1:17 % and 4.6-5; see also Shaw & Leeb 1999) of a balanced, three phase % induction motor with linear magnetic system. The 7 state variables are % D and Q stator and rotor flux linkage rates, psi*, rotor speed wr, and % stator and rotor winding temperatures. % Displacement (th = y(8)) is not really part of the state vector but is % included for numerical consistency (same integration scheme). TEST NEED FOR THIS! % w is angular velocity of the arbitrary reference frame. w=we=synchronous % speed, e.g. 60*2pi, is most convenient for simulating the induction motor. %p.PP rs rr Xm Xls Xlr we J Bl vds vqs vqr vdr Tl Xss Xrr YY %Xss = p.Xls + p.Xm; Xrr = p.Xlr + p.Xm; YY = 1/(Xss*Xrr+Xm^2) psiqs = y(1); psids = y(2); psiqr = y(3); psidr = y(4); wr = y(5); Ts = y(6); Tr = y(7); %th= y(8); rs = p.rs*(1 + p.salpha*(Ts-p.Tamb)); rr = p.rr*(1 + p.ralpha*(Tr-p.Tamb)); %HERE need to solve for variable Xm when core becomes saturated Xpl = 1 / (1/p.Xm + 1/p.Xls + 1/p.Xlr); psimq = Xpl *(psiqs/p.Xls + psiqr/p.Xlr); psimd = Xpl *(psids/p.Xls + psidr/p.Xlr); ydot = [0 0 0 0 0 0 0]'; ydot(1) = p.we*(p.vqs + rs*(psimq - psiqs)/p.Xls) - p.w*psids; ydot(2) = p.we*(p.vds + rs*(psimd - psids)/p.Xls) + p.w*psiqs; ydot(3) = p.we*(p.vqr + rr*(psimq - psiqr)/p.Xlr) - (p.w-wr)*psidr; ydot(4) = p.we*(p.vdr + rr*(psimd - psidr)/p.Xlr) + (p.w-wr)*psiqr; iqs = (psiqs - psimq)/p.Xls; ids = (psids - psimd)/p.Xls; iqr = (psiqr - psimq)/p.Xlr; idr = (psidr - psimd)/p.Xlr; % PP=P/2 is number of pole pairs %T = (3/2)*p.PP*(p.YY*p.Xm/p.we)*(psiqs*psidr - psiqr*psids); T = (3/2)*(p.PP/p.we)*(psiqr*idr - psidr*iqr); ydot(5) = p.PP*(T - p.Tl - p.Bl*wr)/p.J; ydot(6) = ((iqs^2+ids^2)*rs-(Ts-p.Tamb)*p.uscair)/p.Cscair; ydot(7) = ((iqr^2+idr^2)*rr-(Tr-p.Tamb)*p.urcair)/p.Crcair; ydot(8) = p.w; 101 Armstrong 28/29 Appendix F. Simulation script and derivatives function for combined models 1 & 3 % punINDpump2.m runs ODE45 simulation of induction motor,pump,load 20021207 %MOTOR PARMS [p pINDname] = pind3Qhp;% set motor parameters [smax Tmax]=ssTmax(p); [w Tss]=ssTcurve(p); %plot(w,Tss);xlabel('Speed (rad/s)');ylabel('Torque (N-m)') %pause %LOAD PARMS (I,C,...,curveParms) pndPofndQ=[.174,0,-.000328,-.0000281];pndPofndQ=pndPofndQ(length(pndPofndQ):-1:1); pEffofndQ=[0,.2,-.0238,.00158,-.000053];pEffofndQ=pEffofndQ(length(pEffofndQ):-1:1); %pPofQ=[.0005 0 30];%arg in cfs, result in psi pPofQ=[.0005 0 0];%arg in cfs, result in psi pump.rho = 1.94;%density of pump fluid (water) [lbm-s^2/ft] pump.L = 1.0; %impeller diameter [ft] pump.Imotor=0.042;pump.Ipump=0.008;pump.Ifluid=8310;% ft-lb/s2; lb-s2/ft5 pump.Ccoupl=0.001;%1/10 is compliant, 1/100 is stiff [ft-lb/radian] pump.Tp0=.01*polyval(pndPofndQ,.01)/polyval(pEffofndQ,.01)%else Q/eff=undefined at speed=0 y1=ones(1,9); y0=0*y1; opts = odeset('reltol',.0001,'abstol',1e-6*y1); [tout,y] = ode45('pindpump2',[0,1.6],y0,opts,p,pump,pPofQ,pndPofndQ,pEffofndQ); plot(tout,y(:,5)); xlabel('Time (s)'); ylabel('Rotor Speed (rad/s)'); title([date ' punINDpump2.m,pINDpump2.m, ' pINDname]); T = (3/2)*p.PP*(p.YY*p.Xm/p.we)*(y(:,1).*y(:,4) - y(:,3).*y(:,2)); figure(2); plot(y(:,5),T);hold; plot(w,Tss,'r'); xy=axis;xy(2)=p.we;axis(xy); xlabel('Rotor Speed (rad/s)'); ylabel('Torque (N-m)'); plpump2(tout,y,pump,pPofQ,pndPofndQ,pEffofndQ) y(:,6)=y(:,9); [m,m2]=pconvIND(tout,y,p); function [ydot] = pindpmp2(t,y,flag,p,pump,pPofQ,pndPofndQ,pEffofndQ) % [ydot] = pindpmp2(t,yflag,p,pump,p3,p4,p5,p6) % evaluate derivatives for standard model (Krause 2002, eqns 4.5-39,40 % and 4.6-6; see also Shaw & Leeb 1999) of a balanced, three phase % induction motor with linear magnetic system. The 5 motor state variables % are D and Q stator and rotor flux linkage rates, psi*, and rotor speed wr. % The 3 load state variables are coupling deflection, pump speed and flow. % Displacement (th = y(9)) is not really part of the state vector but is % included for numerical consistency (same integration scheme). TEST NEED FOR THIS! % w is angular velocity of the arbitrary reference frame. w=we=synchronous % speed, e.g. 60*2pi, is most convenient for simulating the induction motor. %p.PP rs rr Xm Xls Xlr we J Bl vds vqs vqr vdr Tl Xss Xrr YY %Xss = p.Xls + p.Xm; Xrr = p.Xlr + p.Xm; Y = 1/(Xss*Xrr+Xm^2) %pump.rho L Ccoupl Ipump Ifluid %pTofW, pPofQ, pndPofndQ, pEffofndQ are polynomial curve fit coeffs psiqs = y(1); psids = y(2); psiqr = y(3); psidr = y(4); wr = y(5); delta = y(6); wp = y(7); flow = y(8); %th = y(9); ydot = [0 0 0 0 0 0 0 0 0]'; 102 Armstrong 29/29 %INDUCTION MOTOR (SI units) ydot(1) = p.we*(p.vqs - p.rs*p.YY*(p.Xrr*psiqs - p.Xm*psiqr)) - p.w*psids; ydot(2) = p.we*(p.vds - p.rs*p.YY*(p.Xrr*psids - p.Xm*psidr)) + p.w*psiqs; ydot(3) = p.we*(p.vqr - p.rr*p.YY*(p.Xss*psiqr - p.Xm*psiqs)) - (p.w-wr)*psidr; ydot(4) = p.we*(p.vdr - p.rr*p.YY*(p.Xss*psidr - p.Xm*psids)) + (p.w-wr)*psiqr; % PP=P/2 is number of pole pairs moTorque= (3/2)*p.PP*(p.YY*p.Xm/p.we)*(psiqs*psidr - psiqr*psids); %N-m %ydot(5) = p.PP*(T - p.Tl)/p.J; ydot(9) = p.w; %PUMP & HYDRAULIC LOAD (change to Engrg units @motor-coupling interface) if wp>0; flowCoeff = flow/wp/pump.L^3; else; flowCoeff=0; end loadPressure = polyval(pPofQ,flow)*144; %convert from psi to psf ndPressure = polyval(pndPofndQ,flowCoeff*10000);%g' pumpEfficiency= polyval(pEffofndQ,flowCoeff*10000);%eta if flowCoeff~=0;%>.01; ndTorque = flowCoeff*ndPressure/pumpEfficiency;%10000f' else;ndTorque=pump.Tp0; end ydot(5) = p.PP*(moTorque - 1.3564*delta/pump.Ccoupl)/p.J; %convert ft-lbf to N-m ydot(6) = wr - wp; if wp<0; ydot(6)=0;end ydot(7) = (delta/pump.Ccoupl - pump.rho*pump.L^5*wp^2*ndTorque/10000)/pump.Ipump; %f(1)=(moTorque - rho*L^5*x(1)^2*ndTorque/10000)/pump.Ipump; ydot(8) = (pump.rho*(pump.L*wp)^2*ndPressure - loadPressure)/pump.Ifluid; if ydot(8)<0; ydot(8)=0;end 103 Parameterization, Analysis & Simulation of a Heat Gun Submitted by Thomas A. Bowers December 10, 2002 2.141: Modeling and Simulation of Dynamic Systems Fall 2002 Massachusetts Institute of Technology 104 1 Introduction This paper discusses the dynamic analysis and simulation of a heat gun. The system consists of an electric heating coil and a universal AC electric motor that drives a centrifugal fan in order to produce airflow. A diagram of the system is shown in Figure 1. Although the system looks relatively simple there are complex interactions between electrical, mechanical, thermal, and fluid domains. + - Heating Coil Fan AC Motor Centrifugal Figure 1: Schematic of Heat Gun 2 System Model Because the system operates in four domains there are several couplers required to convert energy from one domain into energy in another domain. 2.1 Electro-mechanical Coupling The coupling between electrical and mechanical domains is the universal AC motor. A diagram of the universal motor is shown in Figure 2. Because the windings on the rotor are connected in series with windings on the two poles of the stator, this motor is able to Schematic and graph removed due to copyright considerations. See reference [1]. Figure 2: Universal Motor Wiring Diagram and t-N Curve operate with an AC or DC power supply [1]. This allows the motor to be treated similarly to a simple DC motor, which is modeled as a linear gyrator. The motor constant, K m , can be determined experimentally by measuring the input voltage and current when driving the motor at a known speed. It is evident from part b of Figure 2 that AC operation is 1 105 even more linear than DC operation for this type of motor adding validity to the use of a linear gyrator. 2.2 Electro-thermal Coupling The interaction between the electrical domain and the fluid domain is a thermal coupling. To transfer energy to the fluid, the material of the electrical resistor must first heat up. The coupling is modeled as a non-conservative two-port resistor. This is due to the fact that electrical energy is converted to thermal energy, but thermal energy does not create electrical energy. The power dissipated in the resistor is equal to e 2 /R. This power is converted to thermal energy through the generation of entropy. Thermal energy is stored in the resistor, which acts as a thermal capacitor, and transferred to the air by convection. 2.3 Thermo-Fluid Coupling The thermal energy that is transferred through convection can be modeled using the HRS macro element that is presented in Brown [2]. This bond is used to model heat exchangers and allows the heater temperature to be much larger than the temperature of the fluid at the inlet or outlet port. 2.4 Mechanical-Fluid Coupling The centrifugal fan provides the coupling between mechanical and fluid domains. The geometry of the fan used in this heat gun is a forward-curved blade centrifugal blower, which is also known as a sirocco fan. As with the coupling between electrical and fluid domains, the mechanical to fluid transmission requires a non-conservative coupler modeled as a two-port resistor. Losses in the fan are due to vorticity, friction, and turbulence. An efficiency coefficient, q, can be used to account for these losses in the fan. 2.5 System Bond Graph After determining the necessary transmission elements of the system, it is possible to create the bond graph of the system, which is shown below in Figure 3. p e line e back t motor GY 0 1 1 I : J S e e i motor i coil t fan RS : R coil R : R motor R f T Q AP S R P amb 0 C 1 S e S C P loss S loss P,h 2 P,h 1 P amb HRS R 0 u S e Figure 3: Bond Graph Representation of Heat Gun 2 106 2.6 System Equations Because the heat gun has a parallel circuit, it is possible to analyze much of the system as two separate networks. The output of the motor-fan system, which is the airflow through the heat gun, can be determined independent of the thermal characteristics of the system. The thermal response of the system is dependent on the transient in the flow rate; however, the flow rate transient is much faster than the heat coil transient. The equations for the fan are the most difficult to derive, but they can be determined based on conservation of momentum [3]. The rate of change of fluid angular momentum is equal to the torque applied on it: ) ( r ) t fan = dH 0 = d (V × dm r = V m u 2 r 2 ÷V u 1 1 dt dt For the system blade geometry the velocity diagrams shown in Figure 4 are used to determine V u2 and V u1 . | 2 V 2 W 2 U 2 e W 1 V 1 = V r1 U 1 V r2 V u2 Figure 4: Velocity Diagrams for Inlet and Outlet Flow of Centrifugal Fan From Figure 4 it is evident that the V u2 is equal to: Qcot | 2 V u 2 =U 2 +V r 2 cot | = r 2 e + 2 b r 2 2t 2 where 2t b r 2 is the area of the fan outlet and Q is the volumetric flowrate. 2 The flow into the fan is assumed to be purely radial, which gives V u1 = 0. Substituting these velocities back into the torque equation results in the following expression: 2 2 t fan = r m 2 | r e + Qcot | | | | = µQr 2 | r e + Qcot | | | | 2 b r 2 . \ 2 2t 2 \ 2t 2 b r 2 . From this is clear that the fan torque is dependent both on volumetric flowrate and angular velocity. This is consistent with non-conservative two-port resistors, which supports the use of this type of coupling in the system bond graph. However, the torque equation does not account for losses in the fan. The forward-curved blade geometry is not 3 107 as efficient as a backward-curved blade or an airfoil [4]. Therefore, the fan torque needs to be divided by the fan efficiency in order to model its non-conservative nature. t fan µQr | r e + Qcot | | 2 2 = b r 2 . q \ 2 2t 2 | | The pressure rise in the fan is determined from the conservation of energy (with the efficiency loss taken into account). 2 t fan e = APQ ¬ AP = µer 2 | r e + Qcot | | | | b r 2 . q \ 2 2t 2 The pressure rise in the fan is equal to the pressure lost as the air travels through the heat gun. An adjustable orifice varies the inlet area allowing for control of the flowrate. Additional restrictions in the flow path include the heating coil and wall friction along the length of the flow path [5]. 2 | 2 2 2 n µ | | Q | | | ÷ | Q | | | | + _ f i µ L i | Q AP = µ V 2 ÷V 1 + h l = 2 \ \ A 2 . \ A 1 ( ) . | 2 . | i =2 2 u 2 D i \ A ( ) . | | u i 2 n 2 | A 2 ÷ A 1 2 L i 1 | = µ Q AP = µ Q 2 \ A 2 2 A 1 2 + _ f i 2 i =2 D i A i 2 . | | 2 A e 2 where, f i , L i , and D i are the friction factor, length, and diameter of the i th restriction section, respectively, and A e is the effective area of the entire system. The ratio L/D, also known as the effective length, is provided for various pipe geometries, valves, and other types of restrictions. The preceding expression for AP is in terms of Q only; however, AP found from the energy balance equation also included the angular velocity of the fan, e. Therefore, Q can be solved in terms of the angular velocity, e: 2 µer 2 | Qcot | | | | = µ Q ¬ | q | | | Q 2 ÷ cot | 2 Q÷er 2 = 0 2 2 2 2t 2 A r e 2 . 2t 2 q \ r e + b r 2 . 2 A e \ 2e 2 b r 2 cot | 2 | cot | 2 | 2 2q b r 2 + b r 2 . | | + A 2 2t 2 \ 2t 2 Q = e q A r 2 e 2 e The angular velocity is related to the state variable, p, by the following equation: 4 108 e = p J The remaining equations needed to solve for the system dynamics are provided by the motor equations and the state equation for angular momentum, p: e back = K e m e line ÷e back = R i motor motor m t motor = i K motor p =t ÷t fan motor The thermal component of the heat gun behavior is determined from the equations for the HRS macro element and the conversion of electrical energy into thermal energy. The thermal equations are as follows: S C ÷S 0 mc T coil = e T 0 2 2 P = e line e line T = coil S R ¬S R = R coil T coil R coil S loss 1÷T amb / T coil = 1÷T amb / T coil = 1/ H +1/ c p m 1/ H +1/ c p µQ S C = S R ÷ S loss 3 System Parameterization In order to simulate the heat gun, the system parameters were determined through measurement and experimentation. The following subsections discuss the methods used to determine the system parameters. 3.1 Fan Parameters Most of the parameters associated with the fan involved its geometry, as indicated by the velocity diagrams in Figure 4. The dimensions of the fan and its blades were measured, and the blade angle at the outlet estimated by assuming that the blade formed the arc of a circle. According to Logan [6], the maximum efficiency of centrifugal fans varies from 70 to 90 percent. Logan also provides a table relating centrifugal fan efficiency to volumetric flow rate. The logarithmic regression from his table provides the following relationship between flow rate and efficiency: q = ln( 0722 . 0 Q) + 5638 . 0 5 109 For this application the flow rate is probably on the order of 0.5 L/s, which leads to an efficiency of about 50%. Therefore, the fan parameters are as follows: %Fan Parameters r1=.03066; %Inlet Blade Raidus, m B1=59.27*pi/180; %Inlet Blade Angle, rad b1=.01; %Inlet Height, m r2=.0375; %Outlet Blade Radius, m B2=10*pi/180; %Outlet Blade Angle, rad b2=.01; %Outlet Height, m eta=.50; %Fan Efficiency 3.2 Flow Parameters The flow parameters were the most difficult to determine by measurement. While the geometries of the inlet and exit were easy to measure, the internal restrictions in the heat gun posed a problem. The inlet and outlet were treated as lossless elements in the flow path, only contributing to the calculation of velocity into and out of the system for Bernoulli’s equation. To determine the pressure losses due to the flow within the heat gun, two major sources of loss were considered. The first was the loss associated with the flow directional change due to the fan volute. Fox and McDonald [5] provide a graphical estimation for losses associated with 90º pipe bends that is based on the ratio of the radius of curvature to the diameter of the pipe, r/D. For this system the ratio is ~3, which results in an effective length, L e /D, of about 13. This value is multiplied by 3 to account for 270º of volute bend. The volute pressure loss is also proportional to the loss coefficient, which is determined by the Reynolds number and the Blasius correlation for turbulent flow in smooth pipes: 316 . 0 f = Re 25 . 0 The second, and more significant loss of pressure within the heat gun, is the restriction due to the heating coil. The coil significantly reduces the diameter of the pipe for a length of 10cm just before the outlet. In addition to this significant reduction in diameter, the coil also creates fully turbulent flow in this section of the heat gun. Both of these effects result in huge losses, which are difficult to determine analytically. Therefore, this parameter, which was identified as the effective length of the coil, was left as a variable to be optimized in the simulation. This parameter also accounts for other losses in the fan including the losses at the inlet and outlet that were neglected. The complete set of flow parameters used in the simulation are as follows: %Flow Parameters Douter=.0670; %Intake Outer Diameter, m Dinner=.0320; %Intake Inner Diameter, m Ain=pi*(Douter^2-Dinner^2)/4%Maximum Intake Area, m^2 A1=cos(6*theta)*Ain/2; %Restricted Intake Area, m^2 D2=.029; %Outlet Diameter, m A2=pi*D2^2/4; %Outlet Area, m^2 rho=1.19; %Assume Constant Density, kg/m^3 mu=2e-5; %Dynamic Viscosity of Air, Ns/m^2 Dduct=.0145; %Volute Diameter, m Aduct=pi*Dduct^2/4; %Duct Area, m^2 Re=rho*.01/Aduct*Dduct/mu; %Reynolds Number in Duct (Assume 1 L/s flow) 6 110 fduct=.316/Re^.25; %Duct Friction Factor (Blasius) fcoil=.015; %Coil Friction Factor Lduct=13*3; %Effective Length of Duct, L/D Lcoil=550; %Effective Length of Coil, L/D Acoil=A2/2; %Effective Coil Area, m^2 A=(A2^2-A1^2)/(A1^2*A2^2)+fduct*Lduct/Aduct^2+fcoil*Lcoil/Acoil^2; 3.3 Motor Parameters The motor parameters were fairly easy to determine, though measurements were made under several different operating conditions to fully characterize the system. The easiest parameter to measure was the resistance of the motor windings, though this varied from 70 to 75 depending on the angle of the motor shaft. Because the system incorporated a universal motor, its parameterization was simplified by utilizing a DC power supply. By applying a known voltage and recording steady-state current draw and shaft speed, it was possible to determine the motor constant, K m , and therewith the motor torque. The motor speed was measured with a timing gun. K m is plotted versus applied voltage in Figure 5 below for two operating conditions: no restriction (cover removed), and minimum restriction (A 1 =A in ). M o t o r C o n s t a n t , K m 0.6 0.5 0.4 0.3 0.2 0.1 0 No Restriction (Cover Removed) Minimum Restriction (A1=Ain) 0 10 20 30 40 50 60 70 80 90 DC Voltage (V) Figure 5: Motor Constant, K m , versus DC Voltage For increasing voltage the value of K m decays to about 0.9 Nm/A. However, it was shown in Figure 2b that AC operation yields slightly different values for K m . Therefore, the steady-state current and speed were also measured for 120VAC. This resulted in K m equal to 0.114 Nm/A, which is about 25% larger than the motor constant corresponding to DC operation. In reality there is a thermal transient in the motor as the rotor and stator heat up due to electrical losses. This results in a slight decay in the current draw that was observed to stabilize completely after about a minute. The decrease in current during this transient was not included in the simulation because it was very small (~10mA). Additionally, the 7 111 motor resistance was measured immediately after running in order to capture the true steady-state characteristics of the system. The DC measurements were also used in characterizing the low-speed characteristics of the motor. The steady-state operating torque was found for each of the points on Figure 5 and plotted versus the motor speed. This resulted in the torque-speed curve of the load, which is shown below in Figure 6. 0 ) i i i in) 0.02 0.04 0.06 0.08 0.1 0.12 T o r q u e ( N m No Restr ction (Cover Removed) Minimum Restr ct on (A1=A 0 1000 2000 3000 4000 5000 6000 Motor Speed (rpm) Figure 6: Load Torque versus Motor Speed This figure is very informative because of the information it contains about motor friction. At low speed operation, one would expect the contribution of torque due to flow rate to be negligible. Therefore, at low speeds the load torque is dominated by motor friction. Judging from the figure, the load torque at 500 to 1500 rpm is entirely due to kinetic friction; beyond this range, the air flow begins to add to the motor load at a quadratic rate. This provides the parameter for motor friction, which is assumed to provide a constant resistive torque (independent of motor speed). This value is not shown on the bond graph, but it would be represented by a constant effort source applied at the e 1-junction for t motor >t friction . The motor parameters are as follows: %Motor Parameters eline=120; %Line Voltage, Vrms Rmotor=72.5; %Motor Coil Resistance, Ohms Km=.114; %Motor Constant, Nm/Amp J=.0001; %Rotatioinal Inertia, kg/m^2 tau0=.03; %Motor Friction, Nm 3.4 Heat Coil Parameters The heat coil parameters were very easy to measure. An ohmmeter was used to find the resistance of the coil, which was then weighed to determine its mass. The coil was assumed to be Nichrome (80%-Ni, 20%-Cr) and the specific heat was found in Incropera 8 112 and DeWitt [7]. The value for the initial entropy of the coil does not affect the result. The only value dependent on the coil entropy is the coil temperature, and since temperature is dependent on the difference between the instantaneous entropy and the initial entropy the initial value is arbitrary. %Heat Coil Parameters Rcoil=6; m=.05; c=385; S0=0; %Heat Coil Resistance, Ohms %Heat Coil Mass, kg %NiChrome Specific Heat, J/kg_K %Initial Entropy Condition 3.5 Fluid Convection Parameters The inlet air was assumed to be at standard temperature and pressure. Therefore, the only variable that was needed for the simulation was the convection coefficient. In order to determine this constant, the thermal behavior of the heat gun was measured using a thermocouple. The following two plots, Figure 7 and Figure 8, show the temperature of the exhaust air for two operating conditions: maximum inlet area and minimum inlet area. 0 50 o C ) Trial 1 Trial 3 100 150 200 250 300 350 400 450 T e m p e r a t u r e ( Trial 2 Average 0 5 10 15 20 25 30 35 40 Time (s) Figure 7: Exhaust Air Temperature with Maximum Inlet Area 9 113 T e m p e r a t u r e ( o C ) 600 500 400 300 200 100 0 0 5 10 15 20 25 30 35 40 Trial 1 Trial 2 Trial 3 Average Time (s) Figure 8: Exhaust Air Temperature with Minimum Inlet Area From the preceding figures it is clear that reducing the inlet area increases the output temperature. This is due to the fact that convection is dependent on the mass flow of air and the flow rate depends on the inlet area. These two figures can be used to adjust the value of H in the simulation. %Fluid Convection Parameters Tamb=297; %Ambient Temperature, K Pamb=101325; %Ambient Pressure, Pa cp=1004; %Air Specific Heat @ Constant Pressure, J/kg_K H=4.5; %Bulk Convection Constant, J/K 4 System Simulation Using the parameters defined in Section 3 and the constitutive equations from Section 2.6, the following MATLAB input was used to simulate the dynamic behavior of the heat gun: %2.141 Term Project %Simualtion of a Heat Gun clear all global r2 eline Rmotor Rcoil Km Tamb S0 m c H cp rho J Vt2 eta A omegaout tauout tau0 %j=1; for j=1:2 theta=14*(j-1)*pi/180; %theta=input('What is the Restriction Plate Angle (0-14)?')*pi/180; Q1(j)=fzero(@solveQ,[1e-20 .02]); 10 114 omega1(j)=omegaout; tau1(j)=tauout; tau=[Km*eline/Rmotor 0]; omega=[0 eline/Km*30/pi]; %ODE Solver t=0:.1:40; [T,X]=ode45('heatgun_dot',t,[1e-10 S0]); %Shaft Angular Speed Omega(:,j)=X(:,1)/J; for i=1:length(X) %Volumetric Flow Rate Q(j,i)=(r2*Vt2+sqrt((r2*Vt2)^2+2*eta*A*r2^2))/(eta*A*J/X(i,1)); %Fan Load tau_fan(j,i)=rho*Q(j,i)*(r2*(r2*Omega(i,j)+Q(j,i)*Vt2))/eta; %Exhaust Air Temperature Tcoil(j,i)=Tamb*exp((X(i,2)-S0)/(m*c)); Sloss_dot(i)=(1-Tamb/Tcoil(j,i))/(1/H+1/(cp*rho*Q(j,i))); Qdot(i)=Tcoil(j,i)*Sloss_dot(i); Tair2(j,i)=Tamb+Qdot(i)/(cp*rho*Q(j,i))-273.15; %Motor Current eback(i)=Km*Omega(i,j); imotor(j,i)=(eline-eback(i))/Rmotor; end close all if j==2 figure plot(omega1(1)*30/pi,tau1(1),'bo',omega1(2)*30/pi,tau1(2),'ro') hold on plot(omega,tau,'k--') xlabel('Shaft Speed, (rpm)') ylabel('Torque, (Nm)') title('Torque-Speed Curve for Motor Indicating Steady-State Operation Points') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') figure plot(Omega(:,1)*30/pi,tau_fan(1,:),'b',Omega(:,2)*30/pi,tau_fan(2,:),'r',omega,tau- tau0,'k--') xlabel('Fan Speed, (rpm)') ylabel('Load Torque, Nm') title('Load Torque versus Speed for Minimum and Maximum Inlet Area') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') figure plot(T,Omega(:,1)*30/pi,'b',T,Omega(:,2)*30/pi,'r') xlabel('Time, (s)') ylabel('Shaft Speed, (rpm)') title('Motor Speed versus Time') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') figure plot(T,Q(1,:)*100,'b',T,Q(2,:)*100,'r') xlabel('Time, (s)') ylabel('Volumetric Flowrate, (L/s)') title('Volumetric Flowrate, Q, versus Time') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') figure plot(T,Tcoil(1,:)-273.15,'b',T,Tcoil(2,:)-273.15,'r') xlabel('Time, (s)') ylabel('Heat Coil Temperature, (decC)') 11 115 title('Heat Coil Temperature for Minimum and Maximum Inlet Area') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') figure plot(T,Tair2(1,:),'b',T,Tair2(2,:),'r') xlabel('Time, (s)') ylabel('Exhaust Air Temperature, (degC)') title('Exhaust Air Temperature for Minimum and Maximum Inlet Area') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') figure plot(T,imotor(1,:),'b',T,imotor(2,:),'r') xlabel('Time, (s)') ylabel('Motor Current, (A)') title('Motor Current for Minimum and Maximum Inlet Area') legend('\theta = 0 deg (max area)','\theta = 14 deg (min area)') end end The first function called on by the preceding file was an early solution to the problem that ignored the inertia of the motor and fan. It was, therefore, a zero order system providing the steady-state behavior of the motor and fan. The function ‘solveQ’ is shown below: function solveQ=f(Q) global r2 eline Rmotor Km rho Vt2 eta A omegaout tauout tau0 tauout=(r2^2*eline/Km+Q*r2*Vt2)/(eta/(rho*Q)+r2^2*Rmotor/Km^2)+tau0; omegaout=eline/Km-Rmotor*tauout/Km^2; P=rho*omegaout*(r2*(r2*omegaout+Q*Vt2))/eta; solveQ=sqrt((1/A)*2*P/rho)-Q; The second function that is solved by the main function is the set of differential equations for the system: function heatgun_dot=f(t,x) global r2 eline Rmotor Rcoil Km Tamb S0 m c H cp rho J Vt2 eta A tau0 heatgun_dot=[0 0]'; p=x(1); Scoil=x(2); w=p/J; Q=(r2*Vt2+sqrt((r2*Vt2)^2+2*eta*A*r2^2))/(eta*A/w); tau_fan=rho*Q*(r2*(r2*w+Q*Vt2))/eta; tau_motor=Km*(eline-Km*w)/Rmotor; p_dot=tau_motor-tau_fan-tau0; mdot=Q*rho; Tcoil=Tamb*exp((Scoil-S0)/(m*c)); Sloss_dot=(1-Tamb/Tcoil)/(1/H+1/(cp*mdot)); Sr_dot=eline^2/(Rcoil*Tcoil); Scoil_dot=Sr_dot-Sloss_dot; heatgun_dot(1)=p_dot; heatgun_dot(2)=Scoil_dot; 4.1 Simulation Results The simulation was run foru equal to 0° and 14°, which correspond to the maximum and minimum inlet areas achievable by adjustment of the restriction plate, respectively. The 12 116 following results show the outstanding correlation between the mathematical simulation and the recorded data. The first simulation output, Figure 9 below, shows the motor torque versus speed. This line can be drawn without simulation using the parameter K m , which was found through experimentation. Torque-Speed Curve for Motor Indicating Steady-State Operation Points 0.2 T o r q u e , ( N m ) 0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 0 2000 4000 6000 8000 10000 12000 u u i ) = 0 deg (max area) = 14 deg (mn area Shaft Speed, (rpm) Figure 9: Motor Torque-Speed Curve (Linear K m ) Including Steady-State Speeds for Minimum and Maximum Inlet Area The blue and red circles on the plot are also determined from the zero dynamics of the system. Because the system is first order, with only a very small inertance due to the motor rotor and fan, the majority of the system operation is at steady-state. The blue circle on the plot, which indicates the steady-state motor speed and torque for the maximum inlet area configuration, is at 6669 rpm. This is less than 0.3% larger than the measured steady-state speed of 6650 rpm. The red circle corresponds to a smaller inlet area and therefore a lower flow rate. Because less momentum is transferred from motor to fluid due to the reduced flow, the resulting torque is lower and the speed is higher. An experimental value of this motor speed could not be measured for this geometry because the restriction plate completely covered the fan preventing the use of the timing gun. However, the current drawn by the motor was measured for both of these conditions (it was used to determine the motor constant for the first configuration) and can be used to calculate the steady-state torque and speed. For the high speed operation with the restriction fully closed, the current draw was 530mA. This results in 0.06043 Nm of torque at a speed of 6832 rpm. The MATLAB output for this configuration was 6970 rpm, which is only 2% larger than the value calculated by the current measurement. 13 117 The following two plots, Figure 10 and Figure 11, show the transients in the motor speed and current. Because of the low motor and fan inertia the system is able to reach steady- state speed in just a few seconds. 0 5 10 15 20 25 30 35 40 0 f ) i u u i ) 1000 2000 3000 4000 5000 6000 7000 8000 9000 S h a t S p e e d , ( r p m Motor Speed versus Tme = 0 deg (max area) = 14 deg (mn area Time, (s) Figure 10: Motor Speed for Dynamic Simulation of Heat Gun Motor Current for Minimum and Maximum Inlet Area 1 2 ) u u i ) 0.4 0.6 0.8 1.2 1.4 1.6 1.8 M o t o r C u r r e n t , ( A = 0 deg (max area) = 14 deg (mn area 0 5 10 15 20 25 30 35 40 Time, (s) Figure 11: Motor Current for Dynamic Simulation of Heat Gun Figure 12 shows the volumetric flow rate of the heat gun in Liters per seconds for both operating regimes. Substituting the steady-state flow values back into the equation for 14 118 centrifugal fan efficiency from Section 3.1 gives efficiency values of 52% for the fully restricted fan and 53% for the fully opened fan. These values are only a fraction higher than the value of 50%, which was used in the simulation. Volumetric Flowrate, Q, versus Time 0 V o l i l ( ) u u (mi 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 u m e t r c F o w r a t e , L / s = 0 deg (max area) = 14 deg n area) 0 5 10 15 20 25 30 35 40 Time, (s) Figure 12: Volumetric Flow Rate for Dynamic Simulation of Heat Gun While an accurate measurement of flow rate was not made, the experimental load versus speed curve shown previously in Figure 6 illustrated how flow rate contributes quadratically to the load. A similar result is seen in Figure 13, which shows both the shaft load and the motor torque versus speed, which is directly proportional to flow rate. Load Torque versus Speed for Minimum and Maximum Inlet Area 0 u u (mi 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 L o a d T o r q u e , N m = 0 deg (max area) = 14 deg n area) 0 2000 4000 6000 8000 10000 12000 Fan Speed, (rpm) Figure 13: Load Curve and Motor Torque-Speed Curve 15 119 Finally, the thermal aspects of the heat gun were addressed. Figure 14 shows the simulated exhaust air temperature for each of the system configurations considered. The time constant of the thermal behavior was modified by adjusting the convection coefficient. The temperature could be scaled by changing the effective length of the flow restrictions in order to decrease volumetric flow rate. Of course this affects the steady- state current and speed, which were known values. The following plot represents the optimized model of the system, which was constrained to meet the known speed and current parameters while also attempting to simulate the thermal data shown in Figure 15. 0 50 i i i i l u u i ) 100 150 200 250 300 350 400 450 500 E x h a u s t A r T e m p e r a t u r e , ( d e g C ) Exhaust A r Temperature for Mnimum and Max mum In et Area = 0 deg (max area) = 14 deg (mn area 0 5 10 15 20 25 30 35 40 Time, (s) Figure 14: Exhaust Air Temperature for Minimum and Maximum Inlet Area 500 450 400 350 o T e m p e r a t u r e ( C ) 300 250 200 150 100 50 0 0 5 10 15 20 25 30 35 40 Time (s) Figure 15: Experimental Thermal Data for Minimum and Maximum Inlet Area 16 120 It is apparent from Figure 14 that the magnitude of the simulated temperature was only about 25-30°C less than the recorded temperature—an error of 5-7%. The simulation demonstrates a remarkable correlation with the data. There are several possible sources for the error in the simulation with the most likely source being temperature rise in the fluid due to flow friction. This explanation seems very feasible especially when examining the thermocouple data. In the data presented in Figure 7, which provided the thermal data for the fully opened configuration, it was seen that the initial temperature of the air was about 50°C. Each of these tests was run with the motor already at steady state, which suggests that the temperature rise of the fluid as it flowed through the heat gun was about 25°C before it reached the heating element. This would provide the 25°C bias that is seen in the simulation versus the actual data. This hypothesis is also supported by the use of dissipative elements in the model of the fluid circuit without also including their entropy contribution to the system. With a 5-7% effect on the result of the simulation, it appears that these losses are important to the thermal dynamics of the system. The final variable considered in order to validate the model was the temperature of the coil. Although no data was taken for the coil temperature, the color of the wire can be used to qualitatively approximate the steady-state coil temperature. It was observed during thermal testing that the coil glowed in the medium to light orange color range. The following table provided by Process Associates of America indicates the possible range of temperatures corresponding to this color range [8]. Table 1: Metal Color versus Temperature Table removed due to copyright considerations. See reference [8]. From this table it is evident that the temperature of the coil was probably in the range of 890-940°C. The temperature of the coil as predicted by the mathematical model is shown in Figure 16. With a bulk heat transfer coefficient of 4.5 W/K the simulation predicts coil temperatures of 850 and 910°C for the fully opened and fully restricted flows respectively. According to Table 1, these temperatures would result in a coil color in the range of salmon to medium orange, which is indeed the case. 17 121 Heat Coil Temperature for Minimum and Maximum Inlet Area 1100 1000 900 800 H e a t C o i l T e m p e r a t u r e , ( d e c C ) 700 600 500 400 300 200 100 0 0 5 10 15 20 25 30 35 40 u u i ) = 0 deg (max area) = 14 deg (mn area Time, (s) Figure 16: Temperature of Heating Coil for Minimum and Maximum Inlet Area 5 Conclusion Although the heat gun initially seemed very complicated due to the coupling between multiple domains, it was found that the lack of energy storage elements in the system resulted in a deficiency of system dynamics. From the constitutive equations for the heat gun it is apparent that the fluid flow exhibits first order behavior and that the thermal behavior depends on both the time constant of the flow rate and on the thermal time constant of the heat coil. However, the time constant of the flow rate is so small compared to the thermal time constant that its contribution is not even apparent in the thermal results of the system model. Because of this, the mechanical, electrical, and fluid flow elements of the system can essentially be treated as a zero order system with complicated static coupling between domains. While the dynamics of the system were relatively uninteresting, the coupling between domains led to a rigorous analysis of domain interactions and provided tremendous insight into non-conservative couplers. In addition, the inability of the model to fully predict the exhaust gas temperatures led to further insight about system dynamics that were not included in the model. A more complete model would attempt to alleviate this problem by including the entropy generated by flow restrictions in the overall temperature increase in the system. Overall, the dynamic model was very successful at demonstrating the behavior of this device in all four domains. Unfortunately, the system could not be characterized entirely through measurement and analysis and some values, such as the effective length of the 18 122 flow restriction and the convection coefficient, were determined simply through trial and error. However, experimental results agreed completely with the final simulated results, with the only major discrepancy being the exhaust air temperature as discussed previously. With this major discrepancy between the results accounted for, the model appears to provide a complete picture of the system. References [1] “Universal Motors”. University of Michigan, Mechanical Engineering, ME 350 Webpage. Accessed 12/04/2002. http://www.engin.umich.edu/labs/csdl/ME350/motors/ac/universal/index.html. [2] Brown, Forbes T. Engineering System Dynamics: A Unified Graph-Centered Approach. New York: Marcel Dekker Inc. 2001. [3] Wright, Terry. Fluid Machinery: Performance, Analysis, and Design. New York: CRC Press. 1999. [4] “Centrifugal Wheel Designs”. Lau Industries Inc., Barry Blower Central Webpage. Accessed 12/08/2002. http://www.barryblower.com/centrifugal.htm. [5] Fox, R.W. and A.T. McDonald. Introduction to Fluid Mechanics: Fifth Edition. New York: John Wiley and Sons, Inc. 1997. [6] Logan, Earl Jr. Turbomachinery: Basic Theory and Applications, Second Edition. New York: Marcel Dekker, Inc. 1993. [7] Incropera, F.P and D.P. DeWitt. Fundamentals of Heat and Mass Transfer: Fourth Edition. New York: John Wiley & Sons. 1996. [8] “Metal Temperature by Color,” Process Associates of America. 1995-2002. Website accessed 12/09/02. http://www.processassociates.com/process/heat/metcolor.htm 19 123 Stirling Engine Marten Byl 12/12/02 1 124 x Te Th Tc =0 R Figure 1: Schematic of Stirling Engine with key variables noted. Introduction In the undergraduate class 2.670 at M.I.T., the students explore basic manufacturing tech- niques by building a stirling engine. The class is concluded by all of the students running their engines at the same time. As the students discover, the stirling engine is very sensi- tive to manufacturing tolerance, specifically the fit of the components determines both the friction in the engine and air leakage out of the engine. The purpose of this project was to develop a model of the stirling engine that accurately predicts the effects of leakage and friction on engine performance. 1 Stirling Engines Figure 1 shows a simple schematic of a stirling engine with key parameters noted. The concept of a stirling engine is fairly simple. The engine consist of heat source, in our case an alcohol flame, and a heat sink, ambient air, an enclosed cylinder, a ”heat” piston, a ”power” piston, and a flywheel connected to the two pistons by a set of linkages. The concept is that the heat flowing through the air in the enclosed cylinder is modulate by the position of the ”heat” piston. When the ”heat” piston is located directly over the flame the heat flow into the engine is minimized while the heat flow out of the cylinder to the heat sink is maximized. Similarly, when heat flow in is maximize, heat flow out is minimized. While the ”heat” piston is moving, the ”power” piston is also moving thus converting the thermal energy being captured by the air into mechanical motion. The flywheel then stores this mechanical energy, thus allowing the mechanical power to flow both in and out of the engine. The geometry of the linkages determines the relationship between the motion of the ”power” piston and the ”heat” piston. Figure 2 shows an animation of the stirling engine in operation. Frame A shows the engine in the starting angular position. In the starting position, we see that the ”heat” cylinder 125 S S S S S S A B C D E F Figure 2: Animation of stirling engine in operation. is positioned to maximize the heat in-flow while at the same time the ”power” piston is positioned to maximize output power. In frame B, we see the engine has rotated such that output power is minimized while the heat input area is reducing. In frame C, we see that heat outflow is nearing maximum while mechanical power may actually be flowing back into the engine. Frames D and E, show the transition back to heat in flow and mechanical power outflow. Frame F shows the engine moving back into the maximum thermal power in and mechanical power out position. I would like to thank Katherine Lilienkamp ([email protected]) for allowing me to use her matlab code to generate these animations. From the 2.670 class notes by Prof. David Hart [1], the stirling engine built in the class operates with a hot temperature, T h , of 600 K and a cold temperature, T c , of 300 K. The typical engine will produce 1 W at 400 Rpm. The typical engine will operate at between 400-600 Rpm, with exceptional engines running at speeds up to 1200 Rpm. 126 C 1 TF 1 MTF 1 R I 0 0 R Se:Pa R Se:Th R Se:Ta R*cos x Ap V Sa Na Te Sh Sc Ah( ) Ac( ) Figure 3: Bond graph model of stirling engine. 2 Stirling Engine Model Figure 3 shows the bond graph model developed for the stirling engine. The heat source is modelled a constant temperature effort source, T h , which transfers entropy to the air in the cylinder, modelled as a multi-port capacitor, through a variable resistor. Similarly, the heat sink is modelled as a constant temperature effort source, T c , which also transfers entropy to the air in the cylinder through a different variable resistor. As mentioned earlier, the air in the cylinder is modelled as a multi-port capacitor. Since leakage from the cylinder is impor- tant, one port on the multi-port capacitor tracks the mass loss through a resitor to ambient conditions, modelled as a constant pressure effort source. A second port on the capacitor tracks the entropy flows too and from the heat sources and the entropy loss due to mass flow. The final port on the capacitor is associated with the volume change. The pressure in the cylinder acts upon the power piston which is modelled as a constant transformer. The piston then acts upon the linkage to the flywheel, modelled as a modulated transformer. Finally, the flywheel is modelled as an inertia, while all of the friction losses in the system are modelled as a resistor with damping b. The major modelling assumption used in this bond graph are: • No power transfer through the ”heat” piston. • Mass-less pistons. • Uniform temperature for air in engine. • Lumped friction element to govern engine speed. • Uniform constant temperature sources. • All leakage from engine through power cylinder. • Motion of ”heat” piston is sinusoid 90 degrees ahead of power piston. There are four state variable in this system. 127 θ the angular position of the flywheel ˙ θ the angular velocity of the flywheel S e the total entropy of the air in the cylinder N e the number of mols of air contained in the cylinder The model results in the following formulation equations: x = R e (1 + sin θ) A h = A sc (1 + cos θ) A c = A sc (1 −cos θ) + P ps x ˙ S h = A h µ(T h −T e ) T e ˙ S c = A c µ(T e −T c ) T e ˙ N e = −A l 2ρ e (P e −P a ) or A l 2ρ a (P a −P e ) ˙ S a = S e N e ˙ N a ˙ S e = ˙ S h − ˙ S c + ˙ S a V e = V c + A p x ¯ v e = V e mN e T e = T o ¯ v e ¯ v o −R Cv exp ¯ s e − ¯ s o C v P e = P o ¯ v e ¯ v o − ( −R Cv +1 ) exp ¯ s e − ¯ s o C v F e = (P e −P a )A p τ e = F e R e cos θ τ I = τ e −b ˙ θ ¨ θ = τ e −b ˙ θ I Where: 128 R e = 1.25 cm = Radius of linkage pivot on flywheel A sc = 40 cm 2 = Heat transfer surface area of cylinder P pc = 4.9 cm = Perimeter of power piston A h = variable = Hot heat transfer area A c = variable = Cold heat transfer area T e = variable = temp of air in engine µ = 100000 W/m 2 = heat transfer constant of cylinder, this was calculated as the thermal conductance of steel with the cylinder wall thickness A l = nominally 0.06 mm 2 = area of leak A p = 1.9 cm 2 = area of power piston V c = 40 cm 3 = volume of air cylinder m = 29 kg/kmol = molar mass of air R = 287 J/kg = mass gas constant air ¯ s o = 2800 J/K*kg = specific entropy of air at T=300 K T o = 300 K = starting temp of air in cylinder P o = P a = 1e5 Pa = ambient pressure C v = 717 J/kg*K = constant volume specific heat b = 0.7e-3 N/rad/s = damping constant I = 4 kg ∗ cm 2 = inertia of flywheel Values for thermal conductance from [2]. Values for thermal dynamic properties from [3]. Initial Conditions The model stirling engine was initialize at ambient thermal and pressure conditions. To start the stirling engine in motion, a short duration (0.1 s) and small magnitude external torque (0.1 N*m = 0.6 lbf*in) was applied to the flywheel. This is sufficient to accelerate the flywheel to a nominal velocity of 500 Rpm. Further exploration showed that a nominal motor would start with an initial velocity of 60 Rpm but the rise time to steady state operation was on the order of 15 seconds. This long rise time resulted in impractically long computational runs, thus for convenience the large initial velocity is used. Both large and small initial velocities resulted in the same steady state velocity. 3 Results For this project, I evaluated the two variables that the 2.670 students have the most control over when building their motors. First is to evaluate the fit of the power piston into the power cylinder (Note: in my model I assumed all of the leakage occurred at this interface but leakage occurs at other spots). I ran the simulation with no leakage, a leakage area of 0.06 mm 2 (the equivalent to having to having a 2.54 µm gap between the piston and cylinder wall), and a leakage area of 0.1 mm 2 (5 µm gap). While these gaps are smaller than I would expect in actual motors, my model does not take into account the flow resistance caused by 129 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 100 200 300 400 500 600 700 800 900 Rotational Speed vs Time Time (s) S p e e d ( R p m ) No leakage Leakage Area = 0.06 mm Leakage Area = 0.1 mm 2 2 Figure 4: Rotational Speed vs Time for various leakage areas. long interface between the piston and the cylinder (length of interface ≈ 1000x gap). My modelling assumptions thus result in very small gaps allowing large flow. The results for these three different leakage areas are shown in figures 4, 5, and 6. In figure 4, we see rotational speed of the motor in Rpm vs time. As we expected, the steady state velocity of the motor drops as leakage increases. In fact, a very small leakage results in the motor not running. The no leak case has a terminal speed of 800 rpm with an output power of 3.5 W. While the nominal motor, leakage area of 0.06 mm 2 , has a steady state speed of 400 rpm with a corresponding power output of 1 W. Figure 5 shows the plot of temperature vs time for all three leakage cases. All of the motors operate between 550 K and 300 K. The size of this oscillation is most likely greater than that in an actual motor because the actual motor has quite a bit of additional thermal storage that is not included in my model. Figure 6 shows the pressure vs time relationship for the three leakage cases. As expected, the motor with no leakage operates at much higher pressures than the two motors with leakage. Also as expected, the motor with leakage operate around ambient pressure. The second variable that I evaluated was to change the friction in the engine by changing the viscous damping on the flywheel. Figure 7 shows the rotation speed vs time plots for the nominal leakage motor for nominal friction, 50% friction, and 200% friction. As the plot shows the terminal speed for the 50% friction motor is 950 rpm with a corresponding power output of 2.4 W. While the 200% friction motor does not spin. 130 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 200 400 600 Temperature vs Time for various leakage areas No Leakage T e m p e r a t u r e ( K ) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 200 400 600 Leakage Area = 0.06 mm 2 T e m p e r a t u r e ( K ) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 200 400 600 Leakage Area = 0.1 mm 2 Time (s) T e m p e r a t u r e ( K ) Figure 5: Temperature vs Time for various leakage areas. Conclusions In general, I am pleased with the results of this term project. The simulated system responds appropriately to changes in the leakage and system friction. The model would likely by improved with the incorporation of additional thermal storage elements and a better model of leakage (thus allowing more realistic air gaps). Specifically, I would try to determine how to add the thermal storage of the air in the ”heat” cylinder. Given the difficulties that I had in setting up the simulated system with both the correct constitutive equations and system parameters (I still do not know what was wrong with the .m file that failed with no heat input), I am extremely pleased with these results. References [1] Hart, D. P. Stirling Engine Analysis 2.670 Course notes, Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, MA, 1999 and 1997. [2] White, Frank. Heat and Mass Transfer Addison-Wesley Publications, 1988. [3] Van Wylen, G., Sonntag, R. Fundamentals of Classical Thermodynamics 3rd John Wiley & Sons, Inc, New York, 1986. 131 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.5 1 1.5 2 x 10 5 Pressure vs Time for various leakage areas No Leakage P r e s s u r e ( P a ) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.5 1 1.5 2 x 10 5 Leakage area = 0.06 mm 2 P r e s s u r e ( P a ) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.5 1 1.5 2 x 10 5 Leakage area = 0.1 mm 2 Time (s) P r e s s u r e ( P a ) Figure 6: Pressure vs Time for various leakage areas. A Matlab file for model parameters % %Stirling Engine Simulation %Th=Hot temp, Tc=ambient temp, Tg=gas temp, Ae=end area, As=cylinder area T %Asc=area power cylinder, Al=leakage area, rf=radius of pivot, mu=heat transfer clear all; close all; global Th Tc Ae As Al rf mu Apc Veo R Cv Pa so Pp If b M Ts qa qo vo rf=0.5*2.5/100; %radius of power linkage Ae=(1.25*2.5)^2*pi/(4*100^2); %area of heat cylinder end plate As=(1.25*2.5*pi*1.5*2.5/100^2); %area of oscillating portion Apc=(0.625*2.5)^2*pi/(4*100^2); %power cylinder area Pp=(0.625*2.5*pi/100); %power cylinder perimeter mu=100000; %heat transfer coefficient W/K*m^2 Veo=pi*((3.69*2.5*(1.225*2.5/2)^2-2.5*2*(1.060*2.5/2)^2))/100^3; %base volume for engine R=287; Cv=716; Pa=1e5; so=2800; Al=pi*((2.5*.625)^2-(2.5*.6249)^2)/(4*100^2); %leak area If=4/(100^2);%kg*m^2 inertia of flywheel Th=600; %hot temp 132 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 100 200 300 400 500 600 700 800 900 1000 Rotation Speed vs Time for three different frictional losses Leakage Area = 0.06 mm 2 Time(s) R o t a t i o n a l S p e e d ( R p m ) Nominal Friction Friction 1/2 nominal Friction twice nominal Figure 7: Rotational Speed vs Time for various motor frictions. Tc=300; %cold temp Ts=300; b=560e-6; %rotational damping N*m/rad/s M=29; ni=Pa*(Veo+Apc*12.5e-3)/(Ts*R*M); Veo=(Veo+Apc*12.5e-3); vo=Veo/(ni*M); si=ni*M*so; qa=Pa/(R*M*Tc); yi=[0 si ni 0]; options=odeset(’MaxStep’,0.001); [t,y]=ode15s(@stirling,[0 5],yi,options); B Matlab function for Stirling Engine simulation function dy=stirling(t,y); global Th Tc Ae As Asc Al rf mu Apc Veo R Cv Pa so Pp If b M Ts qa qo vo %y(1)=angle y(2)=engine entropy y(3)=moles gas y(4)=flywheel speed dy=zeros(4,1); x=12.5e-3*(1+sin(y(1))); ve=Veo+x*Apc; 133 Ah=As*(1+cos(y(1))); Ac=As*(1-cos(y(1)))+Pp*x+Ae; ve=ve/(y(3)*M); se=y(2)/(y(3)*M); Te=Ts*(ve/vo)^(-R/Cv)*exp((se-so)/Cv); Pe=Pa*(ve/vo)^(-(R/Cv+1))*exp((se-so)/Cv); dsh=mu*Ah*(Th-Te)/Te; dsc=mu*Ac*(Te-Tc)/Te; if Pe>Pa dn=-Al*sqrt(2*(Pe-Pa)/ve); else dn=Al*sqrt(2*(Pa-Pe)*qa); end if (t>0)&(t<0.1) tau=0.1; else tau=0; end dsa=(y(2)/(M*y(3)))*dn; dy(2)=dsh-dsc+dsa; dy(3)=dn; dy(4)=((Pe-Pa)*Apc*rf*cos(y(1))-b*y(4)+tau)/If; dy(1)=y(4); 134 Massachusetts Institute of Technology Department of Mechanical Engineering Some useful definitions - System: That which is to be described, analyzed & controlled—anything of interest which is to be described in detail. Often defined as a collection of objects enclosed by a boundary, but this is not essential and the boundary may be conceptual rather than tangible. - Environment: All that is external to the system 1 . Everything else of interest, but which will not be described in detail. Commonly conceived as external to the system, but again, this is not essential. - Open, closed: The behavior of an open system may depend upon its environment; i.e., the two interact. A closed system does not interact with its environment. - System Variable: A quantity, used to describe the system, which may change with time (or space). - System Input: A quantity that is prescribed or imposed on the system by the environment; i.e. an independent variable. - System Output: Any system variable of interest. - State Determined Systems (SDS): A class of systems fully determined by a finite set of state variables ( ) n x x x , , 2 1 . - State: A minimal, complete and independent set of state variables ( ) n x x x , , 2 1 that uniquely describe the system. - State Equations: To describe a state-determined system’s behavior uniquely for all t>t 0 it is sufficient to have: (i) Values of a finite set of variables ( ) n x x x , , 2 1 at t 0 , (ii) Values of a finite set system inputs ( ) r u u u , , 2 1 for all t>t 0 , and (iii) A set of state equations: ( ) ( ) ( ) t u u u x x x f dt dx t u u u x x x f dt dx t u u u x x x f dt dx r n n n r n r n , , , , , , , , , , , , , , , , , , 2 1 2 1 2 1 2 1 2 2 2 1 2 1 1 1 = = = - Output equations: Any output variables ( ) m y y y , , 2 1 of a state-determined system may be expressed as functions of its state and input variables: ( ) ( ) ( ) t u u u x x x g y t u u u x x x g y t u u u x x x g y r n m m r n r n , , , , , , , , , , , , , , , , , , 2 1 2 1 2 1 2 1 2 2 2 1 2 1 1 1 = = = 1 It should be clear that the distinction between "system" and "environment" is not a property of the real world, but a matter of descriptive convenience. Any given object may be described as part of a system in one situation, and part of the environment in another. 135 Massachusetts Institute of Technology Department of Mechanical Engineering Vector notation A more compact notation is as follows. - State Space: An abstract n-dimensional space defined by the state variables. - State Vector: A point in state space defined by a complete set of state variables ( ) t n x x x , , 2 1 = x . - Input Space: An abstract r-dimensional space defined by the input variables. - Input Vector: A point in input space defined by a complete set of input variables ( ) t r u u u , , 2 1 = u . - Output Space: An abstract m-dimensional space defined by the output variables. - Output Vector: A point in output space defined by a complete set of output variables ( ) t m y y y , , 2 1 = y . - State Equations: ( ) t dt d , , u x f x = - Output Equations: ( ) t , , u x g y = 136 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Bond Graph notation for physical system models One of our first concerns in developing a modelling formalism is notation. To allow a concise representation of our models, we will use pictorial diagrams, similar to the electrical network diagram used above. Several well-established sets of pictorial symbols already exist for depicting electrical systems, mechanical systems, and so on, but none of these notations are adequate to be applied to all energetic systems. To emphasize the generality of energetic considerations we will use a notation introduced by Paynter in 1959 and developed by him and his students in the sixties — bond graphs. The net power flow between two interacting systems results in an interdependence between the energetic states of the two systems: it bonds the two systems together into one. Consequently, the basic symbol of the bond graph notation is a line called a bond (somewhat reminiscent of the way chemical bonds are represented). It depicts the exchange of power between the two systems or subsystems or elements at each end of the bond. Figure 3.6: A bond denoting energetic interaction between two systems. Other elements of the bond graph notation are depicted by letters and/or numbers placed at the ends of the bond. We will introduce these symbols as we encounter the corresponding modelling elements. The power flow between the two systems will usually be represented as a product of two real variables, an effort and a flow. As needed, the corresponding symbols are written adjacent to the bond as shown below. For clarity and efficiency, we omit the box or outline representing the boundary of each system (which is purely conceptual, anyway) and represent the interaction as follows. e f Figure 3.7: A basic bond graph. Bond Graphs and Block Diagrams The most important feature of the bond graph notation is that a bond explicitly represents power flow or energy transport and distinguishes it from signal flow, the transfer of information. Generally, the behavior of an element or system will be described mathematically as an operation on an input variable to produce a corresponding output variable. The operator may be dynamic, acting on one time function to produce another, or it may be static (algebraic), simply mapping one number onto another. Those mathematical operations may be represented as the flow of signals (e.g. the input and output) and the transformation of one into the other. 137 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 To represent signal flow, a familiar and versatile graphical notation is available: block diagrams. Indeed, we have used block diagrams freely up to this point and will continue to use them. However, before proceeding further we must recognize that block diagrams do not provide a suitable notation for depicting physical system models because not all block diagrams represent physical processes. One of the most important consequences of energetic coupling between elements or systems is that energy exchange implies interaction; a bilateral, two-way influence of each system on the other. In contrast, block diagrams fundamentally depict a unilateral influence of one system on another. If we wish to describe energetic interaction of two systems or elements in terms of signal flow, then the output of one must be the input to the other and vice versa. Usually the input to an element will be one of the power dual variables, effort or flow (though, to reiterate, that is not essential). Consequently, because the bond represents power flow which is determined by both of the power dual variables, the output must be the dual or conjugate of the input; if effort is input, flow must be output and vice versa. Then, when two systems interact energetically, we must have the situation represented by the block diagram shown in figure 3.8 (or its converse, obtained by switching A and B). A B effort flow Figure 3.8: Block diagram of energetic interaction. In contrast, the block diagrams shown in figure 3.9 might represent possible operations on signals or information, but neither represents any possible energetic interaction between two physical systems. A B effort flow A B effort flow flow Figure 3.9: Two block diagrams with no physical counterparts. Causality Because energetic interaction is a function of two variables, when we come to describe a system in terms of mathematical operations on numbers (i.e. signals), there are two possible choices for 138 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 the input and output of each element (or subsystem). In making these choices we are assigning one variable to the role of cause (or input) and the other to the role of effect (or output), so this choice is referred to as causality assignment. To represent this choice on a bond graph we add a causal stroke at one end of the bond 1 as shown in figure 3.10. Figure 3.10: Representation of causal assignment. This graphical symbol means that the system nearest the causal stroke has effort impressed on it as input and produces flow as output. Of necessity, the system at the other end of the bond has flow imposed on it as input and produces effort as output. In terms of signal flow, the bond graph of figure 3.10 is equivalent to the block diagram of figure 3.8. We refer to the two ways of describing an element's behavior (e.g. effort in, flow out vs. flow in, effort out) as different causal forms. Note that the two alternative causal forms may, in general, require quite different mathematical operations. As we will see later, the causal form we use, i.e. which variable we select as input and which we select as output, can make a lot of difference. For example, the required mathematical operations may be well-defined in one causal form, but not defined at all in the other. Sign Convention One of the important details of formulating any model is establishing a sign convention. While this may appear to be a trivial detail, and there are no associated conceptual problems, in practice sign conventions require considerable attention. We know that when we depress the accelerator pedal, the speed of a car should change, and it usually does. But that information is insufficient for almost all purposes, and especially if we wish to design a control system. It is important to know the sign of the change: whether the car speeds up or slows down. In complex systems, determining the sign of the effect of a given cause can require careful attention. The toil can be minimized by working with a consistent sign convention for all energy domains, a sign convention based on power flow. In a bond graph, the direction of positive power flow is indicated by a half-arrow at the appropriate end of the bond. 1 If you need a mnemonic to help you to remember the convention, try this: with a little imagination the bond with the causal stroke looks like a nail. Obviously, you would push — exert effort — with the flat end of the nail. 139 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 Figure 3.11: Representation of power sign convention This symbol means that the direction of positive power flow is out of system A and into system B. Active vs. Passive Systems The key to our modelling formalism is to keep track of the flow of energy. We will find it useful to distinguish between active and passive systems or elements. The precise mathematical definition of passivity is quite subtle, but the basic idea is that a passive system (or subsystem) is one which cannot supply an infinite amount of energy to its environment. In contrast, an active system may supply energy to its environment indefinitely. Of course, you should realize from this definition that an active system is an idealization, a fiction; but, as we will see, it is a very useful one. The sign convention we will use is that power is positive into a passive element. Much needless confusion can be avoided by adhering to this convention. Referring to figure 3.11, by this convention, A cannot represent a passive system, whereas B may. The sign convention we will use for active elements is more flexible. Usually, power will be positive out of an active element, as that makes the most physical sense; the power into the passive elements has to come from somewhere, usually out of an active element. However, on occasion we may wish to model an active element which is a sink, not a source, an element which removes energy from the system independent of the system's internal state. In those cases, we may represent power as positive into an active element. Augmented Bond Graphs Because the basic bond graph (figure 3.7) depicts energetic interaction independent of choice of sign convention or assignment of causality, a graph with half-arrows (for power sign convention) and/or strokes (for causal assignment) is sometimes referred to as an augmented bond graph. Note that the choices — power sign convention and causal assignment — are quite independent of each other. 140 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Ideal Dissipative Elements Energy cannot be destroyed, but it can be lost or dissipated, irreversibly removed from a system. Loss or dissipation of energy may be characterized by an ideal dissipative element, ideal in the sense that it does not store, supply or transmit energy, but simply removes it from the system. The bond graph symbol is: Figure 3.12: Bond graph symbol for an ideal dissipative element. Any phenomenon characterized by an algebraic relation (possibly nonlinear) between effort and flow will exhibit this behavior, provided the relation exists only in the first and third quadrants, as indicated in the following diagram. flow, f Figure 3.13 Sketch of a possible ideally dissipative characteristic. The restriction to the first and third quadrants ensures that the product of effort and flow is never negative 1 , and that, by our sign convention, ensures that power always flows into the element and never out of it; this is a passive element. We may augment the bond graph with a half-arrow denoting the power sign convention. Figure 3.14: Bond graph symbol for an ideal dissipator with sign convention. 1 Here we are assuming the usual convention that positive values are upwards on the vertical axis and to the right on the horizontal axis. 141 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Dissipative elements come in either of two causal forms. If the dissipator accepts flow as input, it has resistance causality. An ideal resistor is defined as an element for which effort is a single- valued algebraic function of flow. e = R(f) (3.13) Adding the causal stroke to the bond graph: Figure 3.15: Bond graph for an ideal dissipator with sign convention and resistance causality. The algebraic function R(·) is the constitutive equation for this element. The term refers to the fact that this equation describes and is defined by the physical constitution of the device or phenomenon being modeled. Using the constitutive equation, the power into this element may be written as a function of the input flow alone. P(f) = R(f) f (3.14) Thus the power dissipated by this element is a function of its input, which is determined by the rest of the system. An ideal resistor may have a nonlinear constitutive relation. An ideal linear resistor is defined to have a linear constitutive relation. e = R f (3.15) The power flow into an ideal linear resistor is a quadratic function of the input flow. P(f) = R f 2 (3.16) The parameter R is the resistance of this ideal linear dissipative element. In bond graph notation, parameter values may be written next to the corresponding symbol as shown in the next figure 2 , but this is optional. :R Figure 3.16: Bond graph convention for denoting parameter values. 2 This apparently redundant practice will make more sense when we need to distinguish between similar types of element, e.g. resistors, with different parameter values, e.g. R 1 , R 2 , etc. 142 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 An example of an ideal linear resistor is the familiar (idealized) electrical resistor, characterized by Ohm's law, which states that the voltage drop, e, across the resistor (an effort) is proportional to the current, i, through the resistor (a flow). e = R i (3.17) Because the ideal linear electrical resistor is the commonest example of a dissipative element, an ideal (nonlinear) dissipator is sometimes called a generalized resistor. If the dissipator has the dual causal form and accepts effort as input it has conductance causality. This is an element which has a constitutive equation defining flow as a single-valued algebraic function of effort. f = G(e) (3.18) Adding the causal stroke to the bond graph: Figure 3.17: Bond graph for an ideal dissipator with sign convention and conductance causality. The power flow into this element may be written as a function of the input effort alone. P(e) = e G(e) (3.19) An ideal linear conductance has a linear constitutive relation. f = G e = e/R (3.20) The power flow into this element is a quadratic function of the input effort. P(e) = G e 2 = e 2 /R (3.21) The parameter G = 1/R is the conductance of this ideal linear dissipative element. Again, it may optionally be written next to the symbol for the element as shown in figure 3.18. :G Figure 3.18: Dissipator in conductance causality with associated parameter. Causal Constraints A word of caution is appropriate at this point. If we are dealing with a linear element, we are assured that its constitutive equation can be inverted, (with the possible exception of the degenerate cases R = 0 or G = 0) and so the element can assume either causal form. A linear 143 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 dissipator can be modelled equally well as a resistance or as a conductance; the choice of which may be determined by considering other elements in the system. For a nonlinear element the matter may not be so simple. Frequently, a nonlinear constitutive equation will have a well-defined inverse (as in the sketch in figure 3.13), but this will not always be the case. Consider, for example, Coulomb's friction law, commonly used to represent dry friction. F friction = B sgn(v) (3.22) where B is a constant and sgn( . ) is the signum function. sgn(v) A — — ¹ ¦ ´ ¦ ¦ 1 if v > 0 0 if v = 0 -1 if v < 0 (3.23) Figure 3.19: Dissipative characteristic of a Coulomb friction element. Equating force with effort and speed with flow as in Table 3.1, we can see that this object is an ideal dissipator which has resistance causality. An appropriate bond graph would be figure 3.15 and its constitutive equation is depicted in the figure 3.19. However, this constitutive equation cannot be inverted. Whereas there is an unique force corresponding to every speed, there is only one value of force, F = 0, which determines an unique speed; two values of force, F = +B and F = -B correspond to infinite sets of speeds ( v > 0 and v < 0 respectively); and all other values of force do not correspond to any defined speed at all. This model cannot be represented in conductance causality. If the nonlinear constitutive equation of a dissipator is not invertible, the element is causally constrained. It must be represented in whichever causal form will result in a well-defined function which operates on the input to produce an unique output. 144 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Ideal Power Sources The ideal dissipative element defined above should be clearly distinguished from another type of system element which can also model the removal of energy from a system. The power dissipated in an ideal generalized resistor is a function of the state of the system; it is a passive element which responds to changes within the system. It is also useful to define active elements which may remove energy from a system but which are independent of its state. Being independent of the system state, these elements may also supply energy to the system, and thus we are led to define ideal power sources. As we might expect from our discussion of conjugate power variables, there are two types. An ideal effort source is an element which produces an effort independent of the flow from the element. The bond graph symbol is: S e Figure 3.20: Bond graph symbol for an ideal effort source. An ideal flow source is an element which produces a flow independent of the effort on the element. The bond graph symbol is: S f Figure 3.21: Bond graph symbol for an ideal flow source. Causal Constraints By definition, an ideal effort source always imposes effort on the rest of the system. Consequently, the causal assignment for an effort source must be as shown in figure 3.22. The dual causal assignment would be logically incompatible with the definition of the element. S e Figure 3.22: Causal assignment for an ideal effort source. Similarly, an ideal flow source always imposes flow on the rest of the system and its causal assignment must be as shown in figure 3.23. The dual causal assignment would be logically incompatible with the definition of the element. S f Figure 3.23: Causal assignment for an ideal flow source. 145 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Note that in both cases the half-arrow depicting the direction of power flow is not determined by the definition of the element. This is because these elements may be used to model sinks (positive power in) as well as sources (positive power out). Modulated Power Sources To control something it is necessary to act on it. For a physical system this requires the addition or removal of energy, but that is not sufficient; it is also important to be able to change one's actions to respond to changing events. For a physical system this requires that the energy supplied or removed be modulated, usually as a function of feedback information. Real devices which perform this function will be examined in depth in a subsequent chapter. For the present we will confine ourselves to defining idealized elements which would serve this purpose: modulated power sources. An ideal modulated power source is defined as an ideal power source with an output variable which may be modulated as a function of an input or set of inputs. Internal Modulation is Prohibited This definition may seem straightforward enough, but an important restriction is that, when used in a physical system model, the input variable(s) which modulate the power source output must not depend directly on the internal state of the system. To understand the reason for this restriction it is important to recognize that an ideal power source, by virtue of its definition, does not come under the jurisdiction the first law of thermodynamics. If we were to include a power source within the boundary of a system, then the total energy within that boundary need not be constant. Ideal power sources are boundary elements; they describe the flow of power into and out of a system, due to external influences, and belong on the system boundary, not within it. If an ideal power source were to be modulated by a variable internal to the system, then it would be possible to describe all phenomena as modulated power sources. For example, a generalized resistor could be described as a modulated effort source, with output, e = R(f), modulated by an input flow, f. But that description would not retain the special properties of a dissipator. In particular, a modulated power source is not necessarily passive, whereas a dissipator must be. Of course, it would be possible to add those special properties as restrictions on the modulating function for that particular source, but that would defeat the purpose of defining idealized elements in the first place. We would, in effect, deny ourselves access to a considerable body of knowledge which has been amassed about the behavior of physical systems. For example, without knowing the details of their constitutive equations, we may conclude that the total energy in a system composed only of passive elements may never increase 1 . If those passive 1 The proof of this statement requires a deeper look at the definition of passivity. This is postponed until chapter X. 146 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 elements were described instead as modulated sources, we could not draw that conclusion a priori; we would have to examine the modulating functions of each of the elements. This prohibition of internal modulation of power sources does not preclude the modulation of a power source via feedback based on a measurement of the internal state of a system. Indeed, we will frequently encounter that situation in later chapters. However, the feedback modulating functions will not, in general, be subject to the kinds of restrictions which apply to the internal elements of a system. For example, a feedback-modulated power source will usually be an active system, theoretically capable of injecting infinite energy into a system. That is one of the reasons why feedback control systems are intrinsically prone to instability. But we are getting ahead of ourselves; that is the topic of a later chapter. The main point to be taken from this discussion is that modulated sources should be used only to describe external influences on a system. Power Bonds vs. Signal Flow Paths As emphasized above, an ideal power source is a boundary element. An ideal modulated power source also serves as a "boundary" between the description of power flow or energetic interaction (described by bond graph notation) and signal flow or informational interaction (which may be described by block diagram notation). To emphasize the distinction between the two, we reserve the thick line for power flow (and use a half-arrow to denote power sign convention), and use a thin line with a full arrow for signal flow (the usual block diagram convention). Note that the full arrow on a block diagram connection does not specify any sign convention, but rather the input to a block (and the output from the connected block); it corresponds to the causal stroke on a bond graph. A modulated power source is thus represented graphically as shown in figure 3.24. The labels near the power bond and the signal flow path are optional; they need only be included as necessary for clarification. u(t) S e e(t) u(t) S f f(t) (a) (b) Figure 3.24: Bond graph symbols for modulated power sources with signal flow and power flow explicitly represented. (a): Modulated effort source. (b): Modulated flow source. Explicit representation of the signal flow path is not essential, especially if the source of the input signal is not included. In that case, the modulating input, u(t), may be written next to the symbol for the ideal power source as shown in figure 3.25. e(t) u(t): S e u(t): S f f(t) (a) (b) 147 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 Figure 3.25: Bond graph symbols for modulated power sources without signal flow explicitly represented. (a): Modulated effort source. (b): Modulated flow source. 148 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Ideal Symmetric Junction Elements Having defined elements to model where energy can come from (sources) and where it can go to (dissipators) we next need to describe the ways these pieces can be assembled. To this end we will define ideal junction elements. Junction elements will describe the distribution of energy between other elements, so, in keeping with the philosophy of lumped-parameter modelling, we will assume that they don't do anything else. They don't supply, store or dissipate energy; they are isenergic 1 . The elements we have defined so far have a single interaction port, and one conjugate pair of variables was sufficient to describe their energetic transactions with their environments. The junction elements will be used to describe the interchange of energy between sets of elements, and therefore will have many ports (obviously, a one-port junction element would be pretty useless; it couldn't join anything to anything else). These multi-port elements obey the principle of conservation of energy which we will apply in differential form as a power balance condition, equation 3.4. For a junction element, stored energy and dissipated power are both zero, so summing across all ports results in the following condition. _ i=1 n e i f i = 0 (3.24) where n is the number of ports. Now, if we add a symmetry 2 condition and require that each of the ports of a junction element be the same as each of the other ports, then it can be shown (see appendix) that there are only two possible symmetric junction elements, and both are linear. They are as follows. Common Effort Junction: Type Zero A common-effort or type zero junction has an unique effort associated with all connected bonds. A multi-port element with n ports is characterized by n constitutive equations. The n constitutive equations for an n-port zero junction are: e i = e for i = 1 to n (3.25) where e is the unique effort associated with the junction and the subscripts refer to labels on the n bonds or ports. The bond graph symbol for a four-port zero junction is shown in figure 3.26. 1 Sometimes junction elements are called non-energic, but that is an odd and misleading term for one of the most important primitives of an energy-based modelling formalism. Junction elements are defined by the requirement that energy be constant; hence the term iso-energic or isenergic. 2 The symmetry referred to is an invariance of properties under the operation of exchanging or permuting the interaction ports. See Appendix. 149 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Identifying the unique (common) effort next to the zero symbol as in figure 3.26 is optional, but highly recommended. 0 Figure 3.26: Bond graph symbol for a common effort or type zero junction. Applying the power balance condition, equation 3.24, the net power flow into the junction is identically zero, e _ i=1 n f i + 0 (3.26) where the common effort, e, has been brought outside the summation. But this identity must hold for all values of the common effort, therefore this junction gives rise to a flow continuity equation. _ i=1 n f i = 0 (3.27) This flow continuity equation is a generalization of Kirchhoff's current law, encountered in electrical circuit theory. In the electrical domain, a zero junction corresponds to a parallel connection. Sign Convention Equation 3.27 is implicitly based on an assumption that the half arrows on all bonds point inwards as shown below. 0 Figure 3.27: Sign convention assumed in flow continuity equation 3.27. That is strictly in accordance with our sign convention for passive elements, but it is not always convenient. Because the junction element will be used to model the transmission and distribution of power between elements, it is more useful to allow the half arrow on any bond to point in either direction as circumstances dictate. Whatever the orientation of the half arrows, 150 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 the constitutive equations for the junction require the efforts on all bonds to be identical, and that includes sign. Therefore, for a zero-junction, the half arrow denoting power sign convention also determines the sign of the flow. For any bond with an outward-pointing half arrow the corresponding flow in the continuity equation must have a negative sign, as in the following example. f 1 - f 2 + f 3 - f 4 = 0 (3.28) 0 1 2 3 4 Figure 3.28: Sign convention assumed in flow continuity equation 3.28. Causal Constraints As the junction element has multiple interaction ports, it has multiple inputs and multiple outputs. However, the choice of what can be input and what can be output is subject to a strict constraint. By definition, the effort of the zero junction is common to all bonds, therefore one and only one bond may impress an effort on the junction element as shown in the following example. 0 Figure 3.29: Example of a permissible causal assignment for a zero junction. Any bond may impress the input effort, but only one may do so. Note that as soon as a single bond has been chosen to determine the effort on the junction, no further choice is available; all other bonds must provide an input flow. Conversely, that one bond must output a flow; and all others must output the common effort. This is known as a strong causal assignment. On the other hand, provided no bond has been chosen to impress an effort, as many as n-1 bonds may be chosen to impose a flow on the junction. But if n-1 flows are imposed on the junction, the n th flow is determined through the flow continuity equation. Therefore the n th bond must determine the effort on the junction. This is known as a weak casual assignment. Common Flow Junction: Type One If you understand the common effort junction, then you also understand the common flow junction, which is simply its dual. A common-flow or type one junction element has an unique 151 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 flow associated with all connected bonds. The n constitutive equations for an n-port one junction are: f i = f for i = 1 to n (3.29) where f is the unique flow associated with the junction. The bond graph symbol for a four-port one junction is shown in figure 3.30. 1 Figure 3.30: Bond graph symbol for a common flow or type zero junction. The power balance condition for this junction element is as follows. f _ i=1 n e i + 0 (3.30) where the common flow, f, has been brought outside the summation. Arguing as before, this identity must hold for all values of the common flow, therefore this junction gives rise to an effort compatibility equation. _ i=1 n e i = 0 (3.31) This effort compatibility equation is a generalization of Kirchhoff's voltage law, encountered in electrical circuit theory. In the electrical domain, a one junction corresponds to a series connection. Sign Convention Equation 3.31 is implicitly based on an assumption that the half arrows on all bonds point inwards. As with the zero junction, it is more useful to let the half arrow on any bond point in either direction as circumstances dictate. Arguing as before, whatever the orientation of the half arrows, the constitutive equations for the junction require the flows on all bonds to be identical including sign. Therefore, for a one-junction, the half arrow denoting power sign convention also determines the sign of the effort. For any bond with an outward-pointing half arrow the corresponding effort in the compatibility equation must have a negative sign, as in the following example. - e 1 - e 2 + e 3 + e 4 = 0 (3.32) 152 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 1 1 2 3 4 Figure 3.31: Sign convention assumed in effort compatibility equation 3.32. Causal Constraints As with the zero junction, the choice of what can be input to and output from a one junction is subject to a strict constraint. By definition, the flow is common to all bonds, therefore one and only one bond may impose a flow on the junction element as shown in the following example. 1 Figure 3.32: Example of a permissible causal assignment for a one junction. The choice of any single bond to determine the flow on the junction is a strong causal assignment, and no further choice is available; all other bonds must provide an input effort. Conversely, that one bond must output a effort; and all others must output the common flow. If no bond is selected to impose a flow, as many as n-1 bonds may be chosen to impress an effort on the junction. A weak causal assignment may be made by choosing n-1 efforts to be impressed on the junction. Because the n th effort is determined through the effort compatibility equation, the n th bond must determine the flow on the junction. 153 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Generalized Energy Variables Energetic interactions are mediated by the flow of power. Power flow through an interaction port may be expressed as the product of two real-valued variables, an effort and a flow, and all instantaneous interactions between systems or elements may be described in terms of these conjugate power variables. However, to define the energy stored in a system (i.e. its instantaneous energetic state) it is necessary to define energy variables. Just as we may define two power variables, we may define two dual or conjugate energy variables, obtained by integrating the power variables with respect to time. The first of these is generalized momentum 1 , p. p A_ _ ] ( t o t e(t)dt + p(t o ) (4.1) It will be associated with kinetic energy storage. This relation may be differentiated. dp = e dt (4.2) The conjugate variable is generalized displacement, q. q A_ _ ] ( t o t f(t)dt + q(t o ) (4.3) It will be associated with potential energy storage. This relation, too, may be differentiated. dq = f dt (4.4) The following tables provide a partial list of energy variables and notation we will use for several energetic media. 1 Sometimes known as impulse. 154 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Table 4.1 Generalized Momenta and Notation ENERGY MEDIUM EFFORT SYMBOL MOMENTUM SYMBOL General e p Mechanical translation force F momentum or impulse p Fixed-axis mechanical rotation torque or moment t (or µ) angular momentum q Electrical voltage or potential difference e (or v) flux linkage 2 ì Magnetic magnetomotive force F not defined Incompressible fluid flow pressure difference P pressure momentum I Compressible fluid flow enthalpy h not defined Thermal temperature u (or T) not defined Note that generalized momentum is not defined in some media. The reason is because there is no known kinetic energy storage phenomenon in those media. We will return to this point in a subsequently. 2 Strictly speaking, this variable should be associated with displacement in the magnetic medium. 155 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 Table 4.2 Generalized Displacements and Notation ENERGY MEDIUM FLOW SYMBOL DISPLACEMENT SYMBOL General f q Mechanical translation speed or velocity v position or deflection x Fixed-axis mechanical rotation angular speed or velocity e (or O) angle u Electrical current i charge q Magnetic flux rate ¢˙ flux ¢ Incompressible fluid flow volumetric flow rate Q volume V Compressible fluid flow mass flow rate m˙ mass m Thermal entropy flow rate s˙ entropy s, Ideal Energy-Storage Elements We are now in a position to define ideal energy-storage elements. (Ideal in the sense of not being contaminated by dissipation or any other non-storage phenomenon). The energy in a system may be determined from the power flux across its boundaries 3 . E = ] ( t o t Pdt + E(t o ) (4.5) Using equations 4.2 and 4.4 and the definition of effort and flow, this may be rewritten in the following ways. 3 Once again we assume the convention that power is positive inwards. 156 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 E - E(t o ) = ] ( t o t e f dt = ] ( t o t e dq = ] ( t o t f dp (4.6) Now, if we encounter a phenomenon characterized by a relation which permits the integral in either of the latter two forms to be evaluated so that it is not an explicit function of time, that phenomenon may be regarded as energy storage. As you might expect, there are two possibilities. Generalized Capacitor A ideal generalized capacitor is defined as any phenomenon characterized by an algebraic relation (possibly nonlinear) for which effort is an integrable (single-valued) function of displacement. e = u(q) (4.7) The algebraic function u(·) is the constitutive equation for this element. Note that although we will use energy storage elements to describe dynamic behavior, this constitutive equation is a static or memory-less function. The constitutive equation permits us to evaluate the generalized potential energy, E p E p A_ _ ] ( e dq = ] ( u(q) dq = E p (q) (4.8) For this element, potential energy is a function of displacement alone. It is a generalized potential energy storage element. The displacement, q, plays the same role as the specific entropy and specific volume do for a pure thermodynamic substance: it is sufficient to define the energy in the system. By convention we will define E p = 0 at q = 0 as shown in figure 4.1. It will turn out to be important to distinguish potential energy from a related quantity, (generalized) potential co-energy, E * p , which is a function of effort. E * p A_ _ ] ( q de = E * p (e) (4.9) Energy and co-energy are related by a Legendre transformation: E * p (e) = e q - E p (q) (4.10) 157 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 displacement, q * E p E p Figure 4.1: Sketch of a possible potential energy storage constitutive equation. The relation between the two quantities is illustrated in figure 4.1. We will postpone further discussion of co-energy until later. Taken together, the definitions of generalized displacement and the constitutive equation for a generalized capacitor specify a set of relations between flow, displacement and effort. These are represented by the symbol shown in figure 4.2. Note that as this is a passive element, power flow has been depicted as positive into it. Figure 4.2: Bond graph symbol for an ideal capacitor. An ideal capacitor may have a nonlinear constitutive relation. We may also define an ideal linear capacitor, one with a linear constitutive relation. e = q/C (4.11) The parameter C is termed the capacitance of this ideal linear element. The potential energy stored in an ideal linear capacitor is a quadratic function of displacement. E p = q 2 /2C (4.12) Aside: The reason for writing equation 4.11 with the proportionality constant, C, dividing the argument is largely historical. The generalized capacitor is based on an electrical capacitor, usually described by a linear relation between charge, q, (displacement) and voltage, e, (effort). q = C e (4.13) However, in the nonlinear case it may not always be possible to express charge as a function of voltage; the fundamental definition is the one which permits the energy integral to be evaluated: voltage as a function of charge as in equation 4.7. To be consistent the fundamental definition yet acknowledge historical precedent, the parameter, C, has been written as in equation 4.11. 158 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 6 Another example of an ideal linear capacitor is the common model of a mechanical spring. If its deflections are small and occur at modest rates of change, it may be well described by Hooke's law: F = k Ax (4.14) where F is force, k is stiffness and Ax is deflection. Equating deflection, Ax, with displacement, q, and force, F, with effort, e, this model provides a mechanical example of an ideal linear potential energy storage element with capacitance 1/k. A bond graph symbol with the parameter included is shown in figure 4.3. Figure 4.3: Bond graph symbol for an ideal linear potential energy storage element with capacitance 1/k. For large length changes, the force-deflection relation for typical mechanical spring departs from linear and the device provides a mechanical example of an ideal capacitor. In either case, the important point is that the stored potential energy is a function only of the displacement, q. Generalized Inertia In an exactly dual manner, an ideal generalized inertia is defined as any phenomenon characterized by an algebraic relation (possibly nonlinear) for which flow is an integrable (single-valued) function of momentum. f = +(p) (4.15) The algebraic function +(·) is the constitutive equation for this element. Again, it is a static or memory-less function. This constitutive equation permits us to evaluate the generalized kinetic energy, E k . E k A_ _ ] ( f dp = ] ( +(p) dp = E k (p) (4.16) For this element the potential energy is a function of the momentum alone. It is a generalized kinetic energy storage element. By convention we will define E k = 0 at p = 0 as shown in figure 4.5. 159 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 7 momentum, p * E k E k Figure 4.4: Sketch of a possible kinetic energy storage constitutive equation. It will turn out to be important to distinguish kinetic energy from a related quantity, (generalized) kinetic co-energy, E * k , which is a function of flow. E * k A_ _ ] ( p df = E * k (f) (4.17) Once again, energy and co-energy are related by a Legendre transformation: E * k (f) = f p - E k (p) (4.18) The relation between the two quantities is illustrated in figure 4.4. Taken together, the definitions of generalized momentum and the constitutive equation for a generalized inertia specify a set of relations between effort, momentum and flow. These may be represented by the symbol shown in figure 4.5. Again, as this is a passive element, power flow has been depicted as positive into it. Figure 4.5: Bond graph symbol for an ideal inertia. An ideal inertia may have a nonlinear constitutive relation. We may also define an ideal linear inertia, one with a linear constitutive relation. f = p/I (4.19) The parameter I is termed the inertance of this ideal linear element. The kinetic energy stored in an ideal linear inertia is a quadratic function of momentum. E k = p 2 /2I (4.20) 160 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 8 A common mechanical example of an ideal linear kinetic energy storage element is a body in motion. If the deflections of the body are small enough that it may be regarded as rigid, it may be characterized by a linear relation between velocity and momentum. v = p/m (4.21) where v is velocity (flow), p is momentum and m is mass (inertance). A bond graph symbol with the parameter included is shown in figure 4.6. Figure 4.6: Bond graph symbol for an ideal linear kinetic energy storage element with inertance m. Once again, the reason for writing equation 4.19 with the proportionality constant, I, dividing the argument is to be consistent with historical precedent. Equation 4.21 is frequently written with momentum as function of velocity as follows. p = m v (4.22) However, in a general, nonlinear case it may not be possible to express generalized momentum as a function of effort and the fundamental definition is the one which permits the energy integral to be evaluated: flow as a function of momentum as in equation 4.15. Another practice sustained by historical precedent is to equate kinetic energy with the integral of equation 4.22 with respect to velocity. This is regrettable as this quantity is properly termed kinetic co-energy. E * k = m v 2 /2 (4.23) At this point we do no more than note the point of confusion and proceed. The importance of the distinction between energy and co-energy will be discussed in depth later. Modulated Energy Storage is Prohibited Previously we encountered the use of modulated power sources to describe how a control system might influence the energy supplied to or removed from a system. When we consider energy- storage elements, an important restriction must be emphasized: modulation of energy storage elements is prohibited. The reason for this restriction is that a modulated energy-storage element would mean that the total energy in a system would be a function of the modulating input or set of inputs. Consequently, the total energy in the system would not be equal to the net power flow in across the system boundaries.. The system equations would not be guaranteed to satisfy the first law of thermodynamics; but that was to be the objective of our energy-based formalism. 161 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 9 The basic idea behind lumped-parameter modelling is to identify different phenomena with different elements. To be consistent with the energy-based formalism, the function of adding or removing energy from a system must be represented only by elements defined for that purpose: power sources and dissipators. The point to remember is: power sources and dissipators may be modulated; energy-storage elements may not. 162 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Basic Bond Graph Notation — power bond energetic interaction e.g. A — B between (sub)systems S e — effort source boundary condition inpedendent variable S f — flow source boundary condition inpedendent variable R — (generalized) dissipator irreversible energy removal C — (generalized) capacitor (generalized) potential energy storage I — (generalized) inertia (generalized) kinetic energy storage —0 — zero junction (generalized) continuity | equation —1 — one junction (generalized) compatibility | equation Fundamental quantities and relations P power e effort f flow P = e·f E energy E = (e·dq = (f·dp ] ] p (generalized) momentum q (generalized) displacement 163 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Examples: First-Order Systems Energy storage elements provide the basis of the state equations we will derive to describe the dynamic processes occurring in a system. Of course, an energy storage element does not by itself define a dynamic process — it needs an input. That input will arise from the interaction with other system components as we will see in the following examples. A simple fluid system The sketch in figure 4.7 depicts an open-topped cylindrical container of water with a section of circular pipe connected at the bottom through which may leave. We wish to predict the time- course of the volume of fluid as the container empties. circular pipe diameter d, length L depth, h water cylindrical container area A c Figure 4.7: A simple fluid system. At any cross section of interest, the instantaneous power transmitted is the product of the flow velocity and the force exerted on the cross section 1 . The force, F, exerted on the cross section is the product of pressure 2 , P g , and area, A F = P g A (4.24) The volumetric flow rate, Q, is the product of flow velocity, v, and area. Q = v A (4.25) Thus the power transmitted is the product of pressure and volumetric flow rate. 1 Strictly speaking, we may only refer to an unique flow velocity, force, pressure, etc. if the cross-section has infinitesimal area. If the area is finite, each of the quantities represents an average over the cross section. 2 For this example, it is assumed that we are dealing with gage pressures, i.e. that atmospheric pressure is taken to be zero, hence the subscript g. 164 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 P = F v = P g Q (4.26) Within the cylindrical container, the pressure at any depth, h, (e.g. the depth of the entry to the pipe) is due to the weight, W, of the water above. P c = W/A c = µghA c /A c = µgh (4.27) where subscript c denotes the container, µ is the density of water, g is gravitational acceleration and A c is the area of the container. We may rewrite this relation in terms of the volume of water, V c , above a given depth. P c = µg A c V c (4.28) Consequently, we may describe the container as an ideal linear capacitor, characterized by a linear relation between pressure (effort) and volume (displacement). The capacitance, C, is determined by the geometric and material properties of the container, the fluid, etc. C = A c µg (4.29) The energy stored in the capacitor is determined by the displacement variable, V c , and the system parameters. E c = µg 2A c V 2 c (4.30) The flow rate through the pipe is determined by the pressure difference between its ends. If we assume the pipe is horizontal with a constant circular cross-section and the flow is laminar, we may describe the pressure/flow-rate relation using the Hagen-Poiseuille law. P p = 128Lµ ad 4 Q p (4.31) where P is the gage pressure at the end of the pipe next to the container, subscript p denotes the pipe, L is the length of the pipe, d is its diameter and µ is the absolute viscosity of water. This is the equation of an ideal linear resistor with resistance, R, determined by the geometric and material properties of the pipe, the fluid, etc. R = 128Lµ ad 4 (4.32) The remaining idealized element in this system is the junction between the pipe and the container. If we assume the pressure at the entry to the pipe is the same as the pressure in the container at that depth, that common pressure defines a zero junction and a bond graph is as shown in figure 4.8. 165 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 0 R C µg A c 128Lµ šd 4 P = P c p : : Figure 4.8: Bond graph of the simple fluid system. Given this graph it is straightforward to determine state equations. We are interested in the rate of the change of the volume in the container which (from the definition of a generalized displacement) is due to the flow rate into it. dV c /dt = Q c (4.33) That flow rate is determined from the continuity equation of the zero junction. Reading signs from the half arrows on the graph: Q c = - Q p (4.34) The flow rate in the pipe is determined by inverting equation 4.30. Q p = P p /R (4.35) From the zero junction definition, the pipe pressure is the same as the container pressure. P p = P c (4.36) The container pressure is determined from equation 4.28. Successively substituting (4.28 into 4.35 into 4.34 into 4.33 into 4.32) yields a first-order linear state equation. dV c /dt = -V c /RC (4.37) Note that this simple system has one energy-storage element and is characterized by a first-order state equation. The state variable, V c , is directly related to the stored energy. This simple state equation may readily be integrated. ] ( t o t dV c /V c = ] ( t o t -dt/RC (4.38) ln{V c (t)/V c (t o )} = (t o - t)/RC (4.39) V c (t) = V c (t o )e (t o - t)/RC (4.40) Note that to predict the behavior of this first-order system we require one initial condition which is related to the energy stored at time t o . 166 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 A simple electrical system Figure 4.9 shows a diagram of a simple electrical circuit consisting of a capacitor connected to a resistor. C R e 0 R C e Bond graph Electrical circuit diagram Figure 4.9: A simple electrical system and a corresponding bond graph. Assuming a common voltage drop across the resistor and capacitor, a corresponding bond graph is also shown, which is the same as the bond graph of the previous example. This is no accident: the two systems exhibit analogous energetic behavior. Assuming an ideal linear capacitor, its charge, q, is proportional to the voltage difference, e, across it. e = q/C (4.41) where the capacitance, C, is a physical property of the device. The rate of change of charge is the current, i C , flowing into the capacitor. dq/dt = i C (4.42) Assuming an ideal linear electrical resistor, the current through it, i R , is proportional to the voltage difference across it, e, as determined by Ohm's law i R = e/R (4.43) where the resistance, R, is a physical property of the device. A zero junction describes the interaction between these elements. Its associated flow continuity equation requires that the current into the capacitor is the negative of the current into the resistor. i C = -i R (4.44) One reasonable choice of state variable is the charge on the capacitor. A first-order state equation is obtained by substitution (4.41 into 4.43 into 4.44 into 4.42). dq/dt = -e/RC (4.45) 167 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 Assuming the obvious analogy between like quantities (voltage drop and pressure difference are both effort variables, charge and volume are both displacement variables, etc.) equations 4.45 and 4.37 are clearly similar; the time response for discharging the capacitor is of the same form as the time response for emptying the fluid container and will be described by an equation similar to equation 4.40. The analogous dynamic behavior of these two systems is represented by their similar bond graphs. In this regard, bond graphs provide a unified notation for depicting different physical systems. A nonlinear fluid system In this example consider a system similar to that of figure 4.7 but instead of a container with a constant cross-sectional area, assume the radius, r, of the circular cross-section varies with height as follows. r = ah n n > 0 (4.46) where n is a positive constant and a is a scaling factor. The volume varies with depth as follows. V c = ] ( o h aa 2 h 2n dh = aa 2 h 2n+1 2n+1 (4.47) Pressure varies with depth as before (equation 4.27) and the relation between pressure and volume may be obtained by substitution. P c = µg ¸ ¸ ( ( 2n+1 aa 2 V c 1 2n+1 (4.48) Whereas in the previous example the container was described as an ideal linear capacitor, in this example it may be described as an ideal capacitor. The same bond graph (figure 4.8 or 4.9) may be used to represent this system too, the only difference being that the capacitor is characterized by a nonlinear constitutive equation. Indeed, the special case n = 0 corresponds to a container with a constant cross-sectional area; substituting n = 0, equation 4.48 reduces to equation 4.28. A state equation may be obtained following the same procedure as above but using equation 4.48 in place of equation 4.28. dV c dt = - µg R ¸ ¸ ( ( 2n+1 aa 2 V c 1 2n+1 (4.49) As before, this system contains a single (nonlinear) energy-storage element and is characterized by a first-order (nonlinear) state equation. Once again, it is again straightforward to integrate. The case n = 0 has been considered above. For n > 0 the result is as follows. 168 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 6 ] ( ( t o t V c - 1 2n+1 dV c = ] ( ( ( t o t - µg R ¸ ¸ ( ( 2n+1 aa 2 1 2n+1 dt (4.50) V c (t) 2n 2n+1 = V c (t o ) 2n 2n+1 + 2n 2n+1 µg R ¸ ¸ ( ( 2n+1 aa 2 1 2n+1 (t o - t) (4.51) V c (t) = ¹ ¦ ´ ¦ ¦ ) ¦ ` ¦ ¹ V c (t o ) 2n 2n+1 + 2n 2n+1 µg R ¸ ¸ ( ( 2n+1 aa 2 1 2n+1 (t o - t) 2n+1 2n (4.52) Though rather clumsy-looking, this equation, valid for V c _ 0, yields the simple behavior shown in figure 4.10. Comparing these examples, some points should be noted: A bond graph such as that of figure 4.8 or 4.9 is an abstract representation of a family of systems. Until the constitutive equations of the elements have been specified, state equations cannot be determined. However, all members of the family of systems represented by the bond graph share certain qualitative features. In the above examples, first-order state equations are sufficient to describe energetic transactions within the system; that is a consequence of the single energy- storage element. The behavior of the systems are qualitatively similar; all exhibit a non- oscillating decay to the equilibrium state V c = 0 or q = 0. 169 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 7 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Volume time n=1 n=1/2 n=1/8 n=1/4 n=1/16 n=0 Volume vs. time as the container empties for several values of the exponent n. Figure 4.10: Time course of emptying for different container shapes. Fixed parameters have been assigned arbitrary values. It is a common misconception to regard linear and nonlinear systems as radically different. A more enlightened perspective is to consider linear systems as special cases with certain interesting properties. In the examples above all of the nonlinear systems make the reasonable prediction that a finite time is required to empty the container. From equation 4.51 the time to empty is given by t empty = t o + V c (t o ) 2n 2n+1 2n 2n+1 µg R ¸ ¸ ( ( 2n+1 aa 2 1 2n+1 n = 0 (4.53) In contrast, it can be seen from equation 4.39 that the time for the linear system to empty is infinite. But as the exponent n becomes small the time to empty becomes large and the same result could be obtained by taking n to zero in the limit in equation 4.53. 170 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 8 lim t empty = · n C 0 (4.54) Another interesting property of the linear case is that the responses starting from different initial conditions are all of the same form; that is, they are identical when multiplied by a scaling factor inversely proportional to the initial condition. This is clearly not the case for the nonlinear systems: The value of a response at each point in time can be regarded as the initial condition for that response for all future time; yet for sufficiently large times the nonlinear responses are identically zero and therefore cannot be scaled to match any non-zero response. 171 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Real Power Sources: Static Characteristics The primitive elements which have been defined so far are intended to describe aspects of energetic behavior and they can be used in the construction of detailed models of specific physical systems. We should also be able to use energetic considerations to draw some general conclusions about behavior common to all physical systems; after all, that is one of the advertised benefits of an energy-based formalism. In this section we will briefly consider some physical limitations of real power sources (or sinks). Real power sources will exhibit dynamic behavior, but we will confine ourselves for the present to their static characteristics. As defined above, an ideal effort source could sustain that effort even if its output flow grew to infinity and therefore, in principle, it could deliver infinite power; the same is true for an ideal flow source. This is unlikely, to say the least; the amount of power which can be delivered by any real device built to date is limited. The static behavior of any power supply may be characterized as a relation between effort and flow. The maximum power output defines a hyperbola on the effort-flow plane e f = P max (3.33) where P max is the maximum power output. If its output power is limited, the effort-flow relation of the power source must remain within this bound, therefore the effort developed by any real power supply must decline as the output flow becomes sufficiently large. Conversely, the flow produced must decline as the effort required grows sufficiently large. e f = P flow, f Maximum power limit Maximum effort max limit Effort-flow Maximum flow relation must limit remain in this region. Figure 3.33: Common limitations of real power sources. The limitation on power output does not preclude infinite effort at zero flow, nor infinite flow at zero effort, but no real device can generate infinite flow or withstand infinite effort. The maximum effort limit is a horizontal line on the effort-flow plane; the maximum flow limit is a 172 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 vertical line. All of these limitations are illustrated in figure 3.33 for the first quadrant 1 . The effort-flow relation of the power source must stay below and to the left of these bounds. Similar limitations apply in the other quadrants. Reasoning this way, we may conclude that all real power sources must exhibit some dependence of effort on flow (or vice versa). A relation between effort and flow suggests a dissipator and if the effort-flow relation is limited as described above, it is possible to describe this aspect of any real power source by a combination of an ideal source and a dissipator as we did for the battery. There are two possible forms. One is a combination of an ideal effort source and a resistor coupled to a one junction as shown in figure 3.34. This is a generalized Thevenin equivalent network and the figure is a bond graph version of figure 3.3. S e R Figure 3.34: Bond graph symbol for a Thevenin equivalent network. In this bond graph, the half arrows indicate that positive power is directed out of the ideal source element and also out of the network; a reasonable choice for a model of a power source. Following convention, positive power is directed into the dissipator. The other form uses an ideal flow source and an ideal resistor coupled to a zero junction as shown in figure 3.35. This is a generalized Norton equivalent network. 0 S f R Figure 3.35: Bond graph symbol for a Norton equivalent network. These two networks provide versatile, general-purpose models for describing the static characteristics of real power sources. Note that if the constitutive equation of the dissipator in these models can be inverted, the two networks are duals and may be interchanged freely. 1 Note that in this figure we have assumed that positive power flow is directed out of the device. 173 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Hamiltonian Systems: Ideal Oscillators Consider a system composed of an ideal capacitor and an ideal inertia interacting through a common-flow junction as shown in figure 4.11. A physical example would be a rigid body connected to a spring (so that a point on the spring and a point on the rigid body move at the same velocity) as shown schematically in figure 4.11. rigid body elastic spring rigid support C Bond graph velocity (v or f) displacement (x or q) 1 f Mechanical schematic Figure 4.11: Mechanical schematic and corresponding bond graph for an ideal oscillator. In general notation, the constitutive equations for the capacitor and the inertia (subscripts c and i, respectively) are e c = u(q c ) (4.55) f i = +(p i ) (4.56) The differential form of their defining equations suggest that displacement and momentum may be used as state variables. dq c /dt = f c (4.57) dp i /dt = e i (4.58) By definition, the flows of the elements connected to the one-junction are identical. f c = f i (4.59) From the associated compatibility equation the efforts sum to zero. e i = -e c (4.60) 174 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 By direct substitution (4.56 into 4.59 into 4.57 and 4.55 into 4.60 into 4.58) these equations may be assembled into a pair of first-order differential equations. dq c /dt = +(p i ) (4.61) dp i /dt = -u(q c ) (4.62) These are state equations for this system. The reason that q c and p i are a good choice for state variables (other choices are possible — see below) is that they define the energetic state of the system. In fact, any set of numbers from which we can define the energetic state of a system will suffice as state variables. And the reason for that is because the dynamic behavior of a physical system is due to the movement of energy within the system. It is informative to rewrite these equations in terms of the total energy, H(p i ,q c ), of the system. H(p i ,q c ) A_ _ E k (p i ) + E p (q c ) (4.63) From the definitions of potential and kinetic energy we may write the constitutive equations of the capacitor and inertia as partial derivatives of the total energy. +(p i ) = cH/cp i (4.64) u(q c ) = cH/cq c (4.65) The state equations may then be written as follows. dq c /dt = cH/cp i (4.66) dp i /dt = -cH/cq c (4.67) This is the Hamiltonian form of the state equations (named after Sir William Rowan Hamilton, the illustrious Dublinman who first formulated it) and the generalized displacement and momentum are known as Hamiltonian or energy state variables. Hamiltonian systems play a fundamental role in modern physics, and have application at all scales, from celestial mechanics to quantum mechanics. A more compact way to write these equations, which leads to further physical insight, is to write the displacement, q c , and the momentum, p i , as a state vector, r. r A_ _ ¸ ¸ ( ( ( q c p i (4.68) The state equations then become: 175 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 d dt ¸ ¸ ( ( ( q c p i = ¸ ¸ ( ( ( 0 1 -1 0 ¸ ¸ ( ( cH/cq c cH/cp i (4.69) Alternatively: ' r = -J cH/cr (4.70) where J A_ _ ¸ ¸ ( ( ( 0 -1 1 0 (4.71) This form of the equations is known as the symplectic form. The symplectic matrix, J, has an interesting property: J -1 = J t = -J (4.72) It is therefore one of the square roots of minus unity. J J = -1 (4.73) Experience with simple differential equations will have taught you to associate the square root of minus one with oscillatory behavior, and this case is no exception. The oscillatory character of Hamiltonian systems is easiest to see if we consider a linear case by replacing the pure capacitor and inertia with an ideal capacitor and inertia, for example a Hookeian spring of stiffness k and a Newtonian rigid body of mass m (e.g. as shown in figure 4.11). The state equations are then as follows: d dt ¸ ¸ ( ( ( q c p i = ¸ ¸ ( ( ( 0 1/m -k 0 ¸ ¸ ( ( ( q c p i (4.74) Double-differentiating q c and substituting for ˙ p i yields the familiar second-order differential equation of a simple harmonic oscillator. '' q c + e 2 q c = 0 (4.75) where e A_ _ k/m (4.76) 176 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 The same result can be derived directly from the first-order symplectic form, and that yields some further insight and understanding of how the oscillatory behavior arises. To simplify matters 1 , let us suppose that k = m = 1. The state equations become: d dt ¸ ¸ ( ( ( q c p i = ¸ ¸ ( ( ( 0 1 -1 0 ¸ ¸ ( ( ( q c p i (4.77) or ' r = -J r (4.78) Equation 4.77 shows that the rate of change of the state vector is always orthogonal to the state vector — their inner product is zero. r t ' r = -r t J r = 0 (4.79) Now remember that the behavior of a state-determined system can be represented geometrically as the motion of a point in an abstract state-space. Geometrically, the tangent to the path of the point representing the state is always at right angles to the line joining that point to the origin, as shown in figure 4.12. If the rate of change is non-zero, that is only possible if the state trajectory is a circle. A state trajectory which closes on itself represents periodic behavior. A circular state trajectory means that the magnitude of the state vector is constant. From equation 4.77 the magnitude of the rate vector is proportional to the magnitude of the state vector, hence the circle is traversed at constant speed. As a result, q c executes a sinusoidal oscillation. 1 More fundamentally, we could reason that because the fundamental form of the system behavior should not depend on an arbitrary choice of the units for time or space, we may choose them so that k and m are numerically equal to unity. 177 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 r p i q c ' r state vector rate vector state trajectory Figure 4.12: Geometric representation of the behavior of equation 4.77. The orthogonality of the rate and state vectors means that the state trajectory must be circular. This result can be derived formally by integrating the matrix differential equations but the value of these manipulations is to illustrate that the oscillatory behavior comes from the symplectic form of the equations, rather than from any particular choice of parameter values. Consequently, we expect oscillatory behavior in a nonlinear system with this symplectic form, and this is in fact the case. Indeed, one of the reasons why Hamiltonian systems are so ubiquitous is because they are fundamentally oscillators, and oscillatory phenomena are found at all scales, from stellar objects to subatomic particles. If we return to the derivation of the equations, we can see that in this case the symplectic form is due to the interaction between two energy storage elements of dual type through a one-junction. As we will see shortly, interaction between two energy storage elements of the same type through a junction structure composed of zero- and one-junctions does not lead to the symplectic structure and therefore does not lead to oscillatory behavior. 178 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Hamiltonian Systems: Ideal Oscillators Consider a system composed of an ideal capacitor and an ideal inertia interacting through a common-flow junction as shown in figure 4.11. A physical example would be a rigid body connected to a spring (so that a point on the spring and a point on the rigid body move at the same velocity) as shown schematically in figure 4.11. rigid body elastic spring rigid support C Bond graph velocity (v or f) displacement (x or q) 1 f Mechanical schematic Figure 4.11: Mechanical schematic and corresponding bond graph for an ideal oscillator. In general notation, the constitutive equations for the capacitor and the inertia (subscripts c and i, respectively) are e c = u(q c ) (4.55) f i = +(p i ) (4.56) The differential form of their defining equations suggest that displacement and momentum may be used as state variables. dq c /dt = f c (4.57) dp i /dt = e i (4.58) By definition, the flows of the elements connected to the one-junction are identical. f c = f i (4.59) From the associated compatibility equation the efforts sum to zero. e i = -e c (4.60) 179 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 By direct substitution (4.56 into 4.59 into 4.57 and 4.55 into 4.60 into 4.58) these equations may be assembled into a pair of first-order differential equations. dq c /dt = +(p i ) (4.61) dp i /dt = -u(q c ) (4.62) These are state equations for this system. The reason that q c and p i are a good choice for state variables (other choices are possible — see below) is that they define the energetic state of the system. In fact, any set of numbers from which we can define the energetic state of a system will suffice as state variables. And the reason for that is because the dynamic behavior of a physical system is due to the movement of energy within the system. It is informative to rewrite these equations in terms of the total energy, H(p i ,q c ), of the system. H(p i ,q c ) A_ _ E k (p i ) + E p (q c ) (4.63) From the definitions of potential and kinetic energy we may write the constitutive equations of the capacitor and inertia as partial derivatives of the total energy. +(p i ) = cH/cp i (4.64) u(q c ) = cH/cq c (4.65) The state equations may then be written as follows. dq c /dt = cH/cp i (4.66) dp i /dt = -cH/cq c (4.67) This is the Hamiltonian form of the state equations (named after Sir William Rowan Hamilton, the illustrious Dublinman who first formulated it) and the generalized displacement and momentum are known as Hamiltonian or energy state variables. Hamiltonian systems play a fundamental role in modern physics, and have application at all scales, from celestial mechanics to quantum mechanics. A more compact way to write these equations, which leads to further physical insight, is to write the displacement, q c , and the momentum, p i , as a state vector, r. r A_ _ ¸ ¸ ( ( ( q c p i (4.68) The state equations then become: 180 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 d dt ¸ ¸ ( ( ( q c p i = ¸ ¸ ( ( ( 0 1 -1 0 ¸ ¸ ( ( cH/cq c cH/cp i (4.69) Alternatively: ' r = -J cH/cr (4.70) where J A_ _ ¸ ¸ ( ( ( 0 -1 1 0 (4.71) This form of the equations is known as the symplectic form. The symplectic matrix, J, has an interesting property: J -1 = J t = -J (4.72) It is therefore one of the square roots of minus unity. J J = -1 (4.73) Experience with simple differential equations will have taught you to associate the square root of minus one with oscillatory behavior, and this case is no exception. The oscillatory character of Hamiltonian systems is easiest to see if we consider a linear case by replacing the pure capacitor and inertia with an ideal capacitor and inertia, for example a Hookeian spring of stiffness k and a Newtonian rigid body of mass m (e.g. as shown in figure 4.11). The state equations are then as follows: d dt ¸ ¸ ( ( ( q c p i = ¸ ¸ ( ( ( 0 1/m -k 0 ¸ ¸ ( ( ( q c p i (4.74) Double-differentiating q c and substituting for ˙ p i yields the familiar second-order differential equation of a simple harmonic oscillator. '' q c + e 2 q c = 0 (4.75) where e A_ _ k/m (4.76) 181 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 The same result can be derived directly from the first-order symplectic form, and that yields some further insight and understanding of how the oscillatory behavior arises. To simplify matters 1 , let us suppose that k = m = 1. The state equations become: d dt ¸ ¸ ( ( ( q c p i = ¸ ¸ ( ( ( 0 1 -1 0 ¸ ¸ ( ( ( q c p i (4.77) or ' r = -J r (4.78) Equation 4.77 shows that the rate of change of the state vector is always orthogonal to the state vector — their inner product is zero. r t ' r = -r t J r = 0 (4.79) Now remember that the behavior of a state-determined system can be represented geometrically as the motion of a point in an abstract state-space. Geometrically, the tangent to the path of the point representing the state is always at right angles to the line joining that point to the origin, as shown in figure 4.12. If the rate of change is non-zero, that is only possible if the state trajectory is a circle. A state trajectory which closes on itself represents periodic behavior. A circular state trajectory means that the magnitude of the state vector is constant. From equation 4.77 the magnitude of the rate vector is proportional to the magnitude of the state vector, hence the circle is traversed at constant speed. As a result, q c executes a sinusoidal oscillation. 1 More fundamentally, we could reason that because the fundamental form of the system behavior should not depend on an arbitrary choice of the units for time or space, we may choose them so that k and m are numerically equal to unity. 182 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 r p i q c ' r state vector rate vector state trajectory Figure 4.12: Geometric representation of the behavior of equation 4.77. The orthogonality of the rate and state vectors means that the state trajectory must be circular. This result can be derived formally by integrating the matrix differential equations but the value of these manipulations is to illustrate that the oscillatory behavior comes from the symplectic form of the equations, rather than from any particular choice of parameter values. Consequently, we expect oscillatory behavior in a nonlinear system with this symplectic form, and this is in fact the case. Indeed, one of the reasons why Hamiltonian systems are so ubiquitous is because they are fundamentally oscillators, and oscillatory phenomena are found at all scales, from stellar objects to subatomic particles. If we return to the derivation of the equations, we can see that in this case the symplectic form is due to the interaction between two energy storage elements of dual type through a one-junction. As we will see shortly, interaction between two energy storage elements of the same type through a junction structure composed of zero- and one-junctions does not lead to the symplectic structure and therefore does not lead to oscillatory behavior. 183 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Choice of State Variables It must be kept in mind that there is no unique set of state variables (though as we will see, in many cases, some choices are clearly superior to others). Referring again to the system depicted in figure 4.11 and assuming an ideal (linear) capacitor and inertia, their constitutive equations may be differentiated as follows. F c = k x c (4.80) dF c /dt = k v c (4.81) v i = p i /m (4.82) dv i /dt = F i /m (4.83) This suggests that F c and v i would be a reasonable choice for state variables. Equations 4.81 and 4.83 may be combined using the constitutive equations of the one-junction. v c = v i (4.84) F i = -F c (4.85) The following state equations result. F c ( ( ( ¸ F c 0 k ( ( ( ( d (4.86) = dt ¸ ¸ ( ¸ ¸ ( ¸ -1/m 0 v i v i At first glance (comparing equation 4.86 with equation 4.74) it might appear that because the system matrix in these equations has changed, we have a different model. Not so; we have merely expressed it in different coordinates. Double differentiating the velocity and substituting ' for F c again yields the equation of a simple harmonic oscillator. '' v i + e 2 v i = 0 (4.87) where e = k/m as before. In fact we can readily recover the original state equations. Denote the new state variables as r*. ( ( ( ¸ F c ¸ _ r* A (4.88) _ v i Write the state equations in matrix notation. r ' * = A r* (4.89) 184 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 where A is the system matrix of equation 4.86. The new state variables, r*, are related to the original state variables, r, by a constant, linear transformation matrix M. F c k 0 ( ( q c( ( ( ¸ ( ( (4.90) = ¸ ¸ ( ¸ ¸ ( ¸ 0 1/m v i p i or r* = M r (4.91) The transformation matrix is non-singular, so the inverse transformation exists. F c( ( ( ¸ 1/k 0 ( ( q c ( ( (4.92) = ¸ ¸ ( ¸ ¸ ( ¸ 0 m p i v i or r = M -1 r* (4.93) Differentiating 4.93 and substituting equations 4.89 and 4.91 we recover the equation 4.74, our original state equations. ' r = M -1 r ' * = M -1 A r* = M -1 A M r (4.94) 1/k 0 0 k k 0 q c ¸ ( ( ( ¸ ( ( ( ¸ q c ( ( ( ( ( ( d (4.95) = ¸ ¸ ( ¸ ¸ ( ¸ ¸ ( ¸ dt 0 m -1/m 0 0 1/m p i p i 0 1/m ¸ ( ( ( ¸ q c ( ( q c ( ( d (4.74) = ¸ ¸ ( ¸ ( ¸ dt -k 0 p i p i If we regard state variables as coordinates of a state space, equation 4.91 describes a (non- singular) transformation of coordinates. But as there are an infinity of non-singular constant transformation matrices, we can see why state variables, which uniquely characterize a system's state, are not themselves unique — the infinity of equivalent sets of state variables corresponds to the infinity of possible choices of coordinates. Clearly the underlying physical behavior should remain the same under a change of coordinates, though the details of the description will be different in the different coordinate frames, hence the difference between equations 4.74 and 4.86. On the other hand, experience with choice of coordinates in mathematical analysis should have taught you that for certain purposes, some coordinate choices are more convenient than others. Because effort (force) and flow (velocity) together define the power into an element, this second choice is known as power state variables. It is a particularly convenient choice if the elements 185 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 are linear, and is widely used in electric circuit theory (where the power variables are voltage and current); hence these variables are sometimes called circuit state variables. Whichever state variables we choose, the minimum number necessary to describe energetic transactions in this system is two. Physically, there are two independent energy storing elements in the system, and the dynamic process arises from exchange of energy between them. Therefore, to define the energetic state of the system at any moment we require two quantities. Another way to think of this: each of the two energy storage elements embodies a differential equation which is integrated to determine the solution, the state trajectory of the system. Two initial conditions will be needed to determine the constants of integration, and this is another way of saying that we require two numbers to determine the energetic state of the system. 186 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Example: A Non-Oscillating Second-Order System The sketch below depicts two plates which can slide relative to one another in the horizontal direction. Figure 4.21: Mechanical system with two movable plates Assuming the plates are rigid and that we only care about horizontal motion, each plate may be represented by an ideal inertia. We will assume two sources of energy dissipation. One is the friction due to the motion of the top plate relative to the bottom; the other is the friction due to the motion of the bottom plate relative to ground. For simplicity, we will assume linear friction in both cases (although Coulomb friction would probably be a better model). Identifying the Junction Structure Having identified the energic elements in the system, it remains to establish the junction structure. This step usually requires the most care. To describe the system we will need to know the motion of each plate, so it is reasonable to associate a one-junction with each inertia; the flow on each one-junction represents the velocity of the corresponding plate. Figure 4.22: Bond graph elements representing the two plates. We can proceed from here in two ways. Perhaps the most fundamental way is to look for efforts or flows that are common to two or more elements, because they will define zero- (common effort) and one- (common flow) junctions respectively. The friction between the lower plate and ground is due to the velocity of the lower plate, so that dissipative element is attached to the one- junction on the corresponding inertia. 187 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Figure 4.23: Bond graph of the two plates with friction due to motion relative to ground. Any force exerted on the upper plate is generated by friction between the two plates. Thus the other dissipative element is attached to a zero junction between the two one-junctions and the complete system bond graph is as shown below. Figure 4.24: Bond graph of the two plates with both sources of friction. Two of the bonds in this graph are without half arrows. How can they be chosen? Well, the friction between the two plates acts to transmit energy between the plates. Therefore, it makes reasonable sense to depict power as flowing from one plate to the other, for example, as follows. Figure 4.25: Complete system bond graph with power signs included. But we are being a little glib; the choice of power signs bears closer inspection. Suppose, for some reason, we elected to depict the half arrows as follows: 1 0 1 R R : b 1 m 1 b 2 m 2 : : : v 1 v 2 v 3 188 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 Figure 4.26: Complete system bond with alternative power signs. At first glance, this would seem to be equally reasonable. However, a little care is required. The force generated by the friction between the plates is a function of the relative velocity of the plates. But from the signs in the bond graph above, the continuity equation associated with the zero junction would be: v 1 + v 2 + v 3 = 0 (4.103) This would mean that the relative velocity would be expressed as the negative sum of the plate velocities. v 3 = -(v 1 + v 2 ) (4.104) Can that be right? Strictly speaking, it is not wrong. However, it would require that the direction of positive velocity for one plate be defined as opposite to that for the other plate. The direction of positive velocity is purely a matter of convention and there is nothing physically wrong with doing this, but it would be perverse. Worse, it would be confusing. A much saner approach is to use the same convention for both inertias and use a junction structure which represents the relative velocity correctly. Write down a one-junction for each of the three distinct velocities, v 1 , v 2 , and the relative velocity, v 3 , then add a zero-junction between them. The signs on the bonds should be chosen so that the continuity equation yields the required definition of relative velocity. Figure 4.27: A junction structure which identifies v 3 as a relative velocity. Taking the signs into account, the continuity equation for the zero junction is v 1 - v 2 - v 3 = 0 (4.105) Now the relative velocity is the difference of the plate velocities, which makes much better sense. v 3 = v 1 - v 2 (4.106) Following this line of reasoning, a clumsier but surer procedure for identifying the junction structure is as follows. Identify all relevant velocities (including the velocity of the inertial reference frame, v o ) and assign a one-junction to each. Next identify all relevant relative velocities and assign a separate one-junction to each of these. Now write an appropriately signed zero-junction which will correctly represent each relative velocity as follows. 189 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 Figure 4.28: A junction structure for the two-plate system identifying all relevant velocities. Now it easy to see that the dissipative elements are associated with the relative velocities. The ground is always at zero velocity, (by assumption) independent of what the system does, so attach a flow source to the corresponding one-junction. The inertias go on the remaining one- junctions as before. Figure 4.25: Complete system bond graph with reference velocity explicitly represented. Next simplify this graph by eliminating superfluous elements. If the effort or flow on any bond or element is identically zero, the associated power flow is always zero and the bond may be removed. This eliminates the flow source, the two immediately adjacent bonds and the one- junction between them. If any two-port one- or zero-junction has one half arrow pointing inwards and one pointing outwards, it may be eliminated because both the efforts and the flows on the two bonds are identical. The resulting graph is shown below. It is essentially the same as in figure 4.25. 190 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 Figure 4.26: Complete system bond graph after superfluous elements are removed. Aside: There is nothing incorrect about including superfluous elements, but it usually simplifies matters to eliminate them; however, sometimes it is helpful to include a superfluous junction to identify a particular effort or flow. The important point to note here is that choosing the power signs (orienting the half arrows) is an important part of identifying the junction structure. Assigning Causality Next we add causal strokes to the graph. Remember, we want as many energy storage elements as possible to have integral causality. We begin with the inertia of the lower plate. In integral causality, it accepts a force input an determines a velocity as its output. That velocity is the input to the one-junction and therefore determines the causal assignment on the other two bonds of that one-junction as shown in figure 4.27. Figure 4.27: Partial causal assignment for the two-plate system. This is an example of a strong causal assignment. The same velocity is input to the friction between the lower plate and ground (b 1 ) and the zero-junction. But the velocity input to the zero-junction does not uniquely determine the outputs of the zero-junction, and the consequences of this choice of integral causality on the lower inertia do not propagate any further. Thus we are free to choose integral causality for the second inertia. When we do so, the zero junction receives a second velocity input. In this situation, the consequences of this causal choice do propagate through the zero-junction. Two of the three bonds of the zero junction input velocities, therefore the remaining bond must input a force as shown in figure 4.28. This is because something — in this case the friction due to relative motion — must determine the unique effort on the junction. This is an example of a weak causal assignment. 191 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 6 Figure 4.28: Complete causal assignment for the two-plate system. Deriving State Equations To obtain state equations, we begin by choosing state variables. As the energy storage elements give rise to the system's dynamic behavior, we choose variables associated with the independent energy storage elements. In this case, one such choice would be the momenta of the two inertias; they will suffice to define the energy in the system. But remember, state variables are not unique. Another choice would be the velocities of the two inertias. As all elements in this system are linear, we will use the latter. Differentiating the constitutive equations for the inertias and using the differential definitions of momentum, we obtain: dv 1 /dt = (1/m 1 ) F 1 (4.107) dv 2 /dt = (1/m 2 ) F 2 (4.108) Expressions for the forces F 1 and F 2 are obtained by combining the junction structure equations and the constitutive equations of the dissipative elements. There are numerous ways to write and combine these equations, but we need expressions in causal form, as assignment statements (e.g. equation 4.106 rather than equation 4.105). The required form, complete with correct signs, can be found by reading the equations directly from the half arrows and causal strokes on the bond graph. Thus the causal strokes on the zero junction show that the force F 2 is determined solely by the dissipator b 2 . Subscripting variables in an obvious way: F 2 = F R2 = b 2 v R2 (4.109) The velocity v R2 is determined from the junction structure as follows. v R2 = (v 1 - v 2 ) (4.110) Substituting 4.110 into 4.109 into 4.108 yields a state equation. dv 2 /dt = (1/m 2 )b 2 (v 1 - v 2 ) (4.111) The causal strokes on the one junction indicate that the force F 1 is determined by both dissipators. Again, using obvious subscripts: 192 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 7 F 1 = -(F R1 + F R2 ) (4.112) The dissipator b 2 is in resistance causal form. F R1 = b 1 v R1 = b 1 v 1 (4.113) Substituting 4.110 into 4.109, 4.113 and 4.109 into 4.112, and 4.112 into 4.108 yields another state equation. dv 1 /dt = -(1/m 1 )(b 1 v 1 + b 2 (v 1 - v 2 )) (4.114) Equations 4.114 and 4.111 may be re-written in the standard integrable form ' x = Ax. d dt ¸ ¸ ( ( ( v 1 v 2 = ¸ ¸ ( ( ( -(b 1 + b 2 )/m 1 b 2 /m 1 b 2 /m 2 -b 2 /m 2 ¸ ¸ ( ( ( v 1 v 2 (4.115) As the system has two independent energy storage elements, it is second order. However, if we examine the system matrix, A, we can see that the off diagonal elements may be unequal but have the same sign. The antisymmetry of signs we found in the harmonic oscillator is absent. That sign antisymmetry was what gave rise to oscillatory behavior and we might expect that this system will not exhibit oscillatory behavior. In fact, this result can be derived formally by integrating the matrix differential equations. In general, energy storage elements of the same type connected by one- and zero-junctions do not give rise to oscillation. Sign Convention for Energy Storage Elements Note that once sign and causality have been determined on the bond graph, equations follow directly. The basic sign convention, that power flow is positive into all passive elements, should not be violated, as it is what allows us to write the constitutive equations of the elements in the usual way. By this sign convention, a positive effort applied to an inertia will cause its flow to increase. In the mechanical domain, a positive force will cause positive acceleration. In contrast, if the half arrow were to point away from the element, this would mean that power was positive out of the element. In the mechanical domain, positive force would cause deceleration, not acceleration. Note the negative sign in the constitutive equation shown in figure 4.29. Figure 4.29: Consequence of power-positive-out convention on an energy-storage element. If you were perverse and were to follow this sign convention consistently, everything would be fine; but being human and therefore prone to error, there's a good chance that you might forget to reverse the sign on the constitutive equations. Don't do it! Stay with the convention that power is positive into a passive element. 193 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Dependent Energy Storage Elements In the foregoing examples we found that one state variable was associated with the energy stored in each energy storage element. Will every energy storage element give rise to an unique state variable? Not necessarily, as we will see below when we consider two energy storage elements of the same type connected by a simple junction. Suppose we wish to model one dimension of the motion of two space vehicles in a vacuum under free-fall conditions (i.e. zero net gravitational effects). As we are only concerned with their overall motion it seems reasonable to model each vehicle as an ideal Newtonian rigid body. But then the two vehicles dock with one another such that they subsequently move with the same velocity. As the two now move with a common velocity (flow), they interact through a one junction and a bond graph for one-dimensional motion of the system would be as in figure 4.13. I 1 : m 1 2 : m v I Figure 4.13: Bond graph representation of common-velocity coupling between two inertias. Because these are ideal (linear) inertias, we may differentiate their constitutive equations as before: dv 1 /dt = F 1 /m 1 (4.96) dv 2 /dt = F 2 /m 2 (4.97) The equations associated with the one-junction may be written as follows: v 1 = v 2 (4.98) F 1 = -F 2 (4.99) In this case, simple substitution does not lead to state equations; a little manipulation is required. dv 1 /dt = -(m 2 dv 1 /dt)/m 1 (4.100) Note this is an implicit equation. Rearranging: (m 1 + m 2 )dv 1 /dt = 0 (4.101) Assuming m 1 and m 2 are non-zero constants, dv 1 /dt = 0. Integrating, the system state trajectory is: v 1 (t) = constant (4.102) This system only requires one constant of integration, and therefore only one state variable. Yet the model had two storage elements. Why doesn't it require two state variables as in the previous example? Because the two energy storage elements in this model are not independent. Because of the one-junction, the velocity or momentum of one determines the velocity or momentum of the other; given the masses of both bodies, knowing the energy of one is sufficient to determine the energy of the other. Therefore only one state variable is needed; only one initial condition is required to determine the single constant of integration. 194 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Causality and Dependent Energy Storage Elements In previous examples, state equations were obtained by a simple process of substitution, yet in the simple example above, further algebraic manipulation was required. This is a typical consequence of dependent energy storage elements and, as one might expect, in more complex systems the algebraic manipulations can become formidable, even prohibitively so. It would be useful to know about dependent energy-storage elements before attempting to derive equations. How may we do so? The inter-dependence of energy storage elements is easily discovered by considering causality. It refers to the choice of input and output which must be made when we come to describe a system in terms of mathematical operations 1 on numbers. As we will see below, it is also closely related to cause and effect in the usual temporal meaning of those words, though in this context the consequences of causality may be unfamiliar. It is generally accepted that physical systems must be causal. If a cause produces an effect, then that effect may not precede the cause in time. Said another way, if a system is causal, its present output cannot depend on its future input. Integral Causality If we describe a system in terms of an input and an output then the equations used to model the system can be usefully treated as a sequence of operators which act upon an input to produce an output. Thus, in the first-order examples above, an ideal capacitor comprises two operations: an input flow is integrated to yield an output displacement; that displacement in turn determines an effort. This integral causal form of the capacitor equations may be represented by the operational block diagram shown in figure 4.14. Conversely, for an inertia, an input effort may be integrated to yield an output momentum; that momentum in turn determines a flow. The integral causal form of the inertia equations may be represented by the operational block diagram shown in figure 4.15. C Figure 4.14: Block diagram representation of the mathematical operations associated with the integral causal form of a generalized capacitor. A corresponding bond graph is also shown. 1 For that reason it is sometimes referred to as operational causality. 195 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 I Figure 4.15: Block diagram representation of the mathematical operations associated with the integral causal form of a generalized inertia. A corresponding bond graph is also shown. When an inertia is coupled to a capacitor through a one-junction, (as in the Hamiltonian system above) the operations required to model the complete system may be represented by combining the operational block diagrams for each of the pieces as shown in figure 4.16. The defining equation for the one-junction (the identity of the flow variables) is simply a connection in the block diagram; The associated compatibility equation is (in this simple case) a sign change operation. This diagram shows that the complete system comprises a closed loop of operations, a chain of causes and effects. Both the inertia and the capacitor are in integral causal form and that means that all of the operations well-defined. Figure 4.16: Block diagram representation of the mathematical operations associated with a capacitor and an inertia connected through a one-junction. An equivalent bond graph is also shown. Derivative Causality If we attempt to represent two inertias and a one junction (as in the example of the space vehicles) using block diagrams, we encounter a difficulty: one of the inertias has to accept flow as input and 196 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 produce effort as output, the opposite of what is shown in figure 4.15. Simply reversing the orientation of the arrows on the block diagram is meaningless; one of the two inertias (it doesn't matter which) must be represented using a different set of operations, the inverse of those depicted in figure 4.15. The constitutive equation for the inertia must be replaced by its inverse; the time- integration operator must be replaced with a time-derivative operator, as shown in figure 4.17. As in this figure, an asterisk will often be used to call attention to the time-derivative operator. An energy- storage element which is represented by a time-derivative operation is said to be in derivative causal form. Figure 4.17 shows that, as before, the complete system comprises a closed loop of operations, a chain of causes and effects. However, unlike figure 4.16, the derivative causality of one of the inertias in figure 4.17 means that all of the operations in the loop may not be well-defined. First, the inverse of the constitutive equation for the inertia may not exist. In the example above it does, as the constitutive equation is linear, but in general a well-defined function may not have a well-defined inverse. A second and more profound problem stems from the time-derivative operation; it is highly undesirable because time differentiation is a physically impossible operation. Figure 4.17: Block diagram representation of the mathematical operations associated with two inertias connected through a one-junction. An equivalent bond graph is also shown. There are several ways to understand this. The most fundamental is based on the definition of a time derivative as the limit of the difference of two values of a function divided by their time separation as the latter goes toward zero. To obtain the time derivative of a function at any point in time one must take the difference of two values of the function, one in the immediate past, and one in the immediate future. This is illustrated in figure 4.18. The output of a time derivative operation depends on future values of its input, and therefore it is non-causal in the usual physical sense. An alternative argument: The input of any operational block may be any well-defined function of time, including a discontinuous function such as a step function. For the integration operator, discontinuities present no problem: the input force applied to an inertial object may change abruptly; its velocity, momentum and stored kinetic energy will not. 197 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 Figure 4.18: Diagram illustrating the fundamental definition of a derivative operation. In contrast, the derivative operator will generate infinite output in response to a discontinuous input, and this infinity is a departure from the physically realizable. A discontinuous change in the velocity of an inertial object would require an infinite force. Furthermore, the kinetic energy stored in the inertial object would also change discontinuously, and that would require an infinite power flow. Neither of these infinities is physically possible. For these reasons, a time differentiation operation cannot be physically realized; it can at best be approximated. How much does this matter? We know from thermodynamics that real macroscopic phenomena are not reversible. The time sequencing of events is a fundamental aspect of physical system behavior. Therefore causality is an important aspect of our description of macroscopic physical phenomena. The point is that to use a time derivative operation in a model of a physical system is a major misrepresentation of physical reality as we understand it and it should avoided. Derivative Causality Indicates Dependent Energy Storage However, to put this discussion in perspective, every mathematical model falls short of reality, usually very far short; for example, the idea of a rigid body (which we used in the example with the two space vehicles) is itself a fiction, albeit a very convenient one — a typical space vehicle is anything but rigid. It can then be argued that a time-derivative operation is intrinsically no worse than this and other modelling approximations. Indeed, when two inertias are coupled by a one-junction, the resulting derivative causality really means that the two inertias are one object, one rigid body. That is the true meaning of inter- dependence of energy storage elements: in the model they are not distinct energy storage elements, despite appearances to the contrary. These two modelling approximations — rigid-body models and time-derivative operations — are intimately related. If we consider an object undergoing translational motion along one dimension in space, the assumption that it is rigid is equivalent to assuming that the (infinitely) large collection of fictional "mass points" which comprise the object all move with a common speed. In bond-graph notation, if each mass point were represented by an inertia, all would be connected to a single one-junction representing the common speed. But in terms of mathematical operations, only one of those inertias can determine the common speed associated with the one-junction; only one inertia can have integral causality. All the others must accept that speed as an input; they must be in derivative causality. The entire collection of mass points is a single independent energy storage element; a 198 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 6 single number (the common momentum or common speed) is sufficient to determine the stored energy. Preferred Causality for Energy Storage Elements A point to be taken from this discussion is that, if possible, energy-storage elements should be independent and have integral causality. But why? Why does it matter if an energy-storage element is dependent? If it is regarded as merely another modelling approximation, why should the time-derivative operation be avoided? Heed the admonition of the the fourteenth-century scholar William of Ockham 2 : "Entia praeter necessitatem non sunt multiplicanda" In the modelling context: Approximations must not be made needlessly; wherever possible, the integration operation should be used rather than the derivative operation. Every energy-storage element which can be described using an integration operator should be. It will require one initial condition to determine its constant of integration, and therefore will give rise to one state variable; energy storage elements which have integral causality are independent. Conversely, any energy storage element which must be described using a derivative operation will not require an independent initial condition and therefore will not give rise to a state variable; energy storage elements which have derivative causality are dependent. A final comment: Elements such as power sources are subject to causal constraints; their definition will only permit one causal form. In contrast, in certain situations which we will discuss later, derivative causality is unavoidable, or the steps required to eliminate it would obscure the insight that the model is supposed to provide. The integral causal form is the preferred causal form for energy-storage elements; it is not essential. 2 If I have my Latin correct, this translates as "Entities are not to be multiplied beyond necessity"; it is popularly known as Occam's Razor. 199 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Causality and Model Formulation Causal considerations can be extremely helpful at the stage of formulating a model because they can identify the consequences of particular choices. Consider the following example. Two large, identical electrical capacitors, initially charged to different voltages, are to be connected together by two bars of copper as shown conceptually in the figure 4.30. This system would be a reasonable laboratory simulation of one of the problems of switching in an electrical power system. The copper bars are large, several square centimeters in cross sectional area. We wish to formulate the simplest possible model which is competent to describe the events that happen as the contact is closed. The simplest model for this system would appear to be two capacitors and to keep things simple we may assume they are ideal. The resistance of the connector would appear to be negligible. Ignoring surface contact phenomena, the resistance of a conductor is the product of resistivity and length divided by cross-sectional area R = µl/A (4.116) where R is resistance, µ is resistivity, l is length and A is area. Copper is an excellent conductor; the resistivity of drawn copper is listed as 1.724 µO-cm. Even if the copper bar is tens of centimeters long, its resistance will be measured in micro-ohms. Figure 4.30: Conceptual sketch of a capacitor and switch system. When the switch is closed the two capacitors share a common voltage, which implies that their connection should be modeled as a zero-junction. As a first step in thinking about this model it is useful to assign causality. As a result, without deriving equations (admittedly a simple enough 200 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 task in this case) it is clear that the two capacitors are not independent; assign either capacitor integral causality and, because of the zero-junction, the other must have derivative causality as shown in figure 4.31. Does this make physical sense? Yes. Given the capacitance of both capacitors, a knowledge of the energy stored in one is sufficient to determine its charge or voltage; the voltage is the same for both capacitors, hence the energy stored in the second capacitor can be computed. Therefore only one state variable is needed to characterize the energetic state of this system. 0 C C Figure 4.31: Model of two capacitors connected. Simple though this system is, there is plenty of room for confusion: before connection the charges on the two capacitors could be specified independently; afterwards, our model says they cannot. Do we need two initial conditions or one? The confusion arises because we have two structurally different models for the system before and after connection. That is, different mathematical operations will be used to describe the capacitors in the two different situations. This is easily seen by comparing the causality of the capacitors in the two situations. A model of the unconnected capacitors would be as shown in figure 4.32. By assumption, no current may flow from the capacitors, so this boundary condition is modeled by a flow source imposing zero flow on the capacitor. Before connection, each of the two capacitors is independent and we are free to assign integral causality to each. C C S f : 0 0 : S f Figure 4.32: Model of the two unconnected capacitors. There is nothing wrong with using two structurally different models to describe different operating conditions for a system, though care must be taken to ensure that the models are mutually consistent and do not contradict each other. Is there some contradiction or inconsistency here? No. In the absence of charge leakage, the total charge in the system must be the same before and after connection. Using obvious subscripts: (q 1 + q 2 ) before = (q 1 + q 2 ) after (4.117) (Ce 1 + Ce 2 ) before = (Ce 1 + Ce 2 ) after = 2Ce after (4.118) Thus the two models may be made mutually consistent by deriving the single initial condition for the model after connection from the state of the model just before connection. e after = | \ e 1 + e 2 2 | | . before (4.119) However, if we attempt to concatenate the two pieces of the model valid before connection (figure 4.32) to obtain a model for the same two capacitors after connection (figure 4.31) we 201 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 discover a "causal conflict". This "causal conflict" means that the pieces cannot just be assembled, the entire model for the coupled system has to be reformulated; new equations have to be written. In this case, the new equations require a different number of state variables than the old. But you don't have to work out the equations to see this; it is evident from the change of causal form for one of the capacitors from integral causality to derivative causality. In this case we were asked for a model to describe the events that happen as the contact is closed. But if we want to describe the events at switch closure, we have a problem: How do we reconcile the different descriptions in the two regimes? The key to understanding the dynamics of an energetic system is to keep track of energy. In this linear system, energy is numerically equal to co-energy. Before contact: * 1 E p before = 2 C(e 1 2 + e 2 2 ) before (4.120) After contact: * 1 E p after = 2 C(e 1 2 + e 2 2 ) after = C | \ e 1 + e 2 2 | | . 2 (4.121) before Therefore: * * | \ E p before - E p after = C e 1 e 2 - 2 2 | | . 2 _ 0 (4.122) Energy is lost during switch closure. To model this phenomenon we must include at least one dissipative element 1 , however small. The point to be taken here is that by considering causality we uncover an aspect of the model which we may want to modify. It is important to stress that "causal conflicts" are not errors. They simply spell out consequences of modelling assumptions which may otherwise be obscure. Think of them as waving a banner which reads "Are you sure you really meant your model to have this property?" Junction Structure Suppose we decide to include a dissipator. The simplest model we can assemble will include two capacitors and an ideal dissipator. How should they be connected? Any current that leaves one capacitor flows through the dissipator and enters the other capacitor; this indicates a one- junction and suggests the bond graph shown in figure 4.33. Assigning causality we see that both 1 In practice, the dissipative phenomenon encountered on switch closure is dominated by an arc which "jumps the gap" before metal-to-metal contact is made rather than by the resistance of the metallic components. 202 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 capacitors may be given integral causality and the dissipator has conductance causality; it determines the current as a function of the voltages on the two capacitors. C 1 C R Figure 4.33: Model of the two capacitors including dissipation. But this graph has a serious flaw. We expect the dissipator to determine current as a function of the potential difference across its terminals. Yet the compatibility equation (Kirchhoff's voltage law) associated with the junction in figure 4.33 would be written as follows (using obvious subscripts and reading the signs and the required causal form from the graph): e R = -(e 1 + e 2 ) (4.123) Thus the equation for the common current would depend on the negative sum of the capacitor voltages. e 1 + e 2 i R = - R (4.124) This is the same problem we encountered previously. In fact, it is possible to define a sign convention for the capacitor and dissipator voltages so that equation 4.124 is correct, but it is quite counter-intuitive. A better course of action is to revise the junction structure so that the expected potential difference is identified as shown in figure 4.34. 1 0 C 0 C e 1 e 2 0 : e 1 - e 2 R Figure 4.34: Revised junction structure which identifies the potential difference across the dissipator. The zero junction in the center may be eliminated, because, reading the half-arrows, its equations merely state that the efforts and flows on the associated bonds are identical. The same is true for the zero junction on the right, but not for that on the left. The compatibility equation for this junction states that the flows on the two associated bonds are equal and opposite; this junction cannot be eliminated. A simplified bond graph is shown in figure 4.35. 203 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 C 0 1 C 1 4 2 3 R Figure 4.35: Simplified bond graph for the two capacitors with dissipation. The numbers written beside the bonds in figure 4.35 are not strictly necessary. They are included solely to facilitate the following derivation. State Equations At this point, deriving state equations is straightforward. As all elements in this system are linear we might use circuit variables (capacitor voltages in this case) for state variables. But we may equally well use energy variables (capacitor charges in this case) and, in fact, will do so. The definition of generalized displacement yields: dq 1 /dt = i 1 (4.125) dq 2 /dt = i 2 (4.126) Reading the causal strokes on the graph, i 1 is determined by i 4 which is in turn determined by i 3 ; i 2 is identical to i 3 . Reading the signs from the half arrows: i 1 = -i 4 = -i 3 (4.127) The dissipator determines i 3 . 1 i 3 = R e 3 (4.128) e 3 is determined by e 2 and e 4 ; e 4 is identical to e 1 . Reading the signs from the half arrows: e 3 = e 4 - e 2 (4.129) e 4 = e 1 (4.130) The capacitors determine e 1 and e 2 . q 1 e 1 = C (4.131) q 2 e 2 = C (4.132) State equations are now obtained by substitution. 204 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 6 dq 1 /dt = - 1 R | \ | | . (4.133) q 1 q 2 - C C dq 2 /dt = 1 R | \ | | . q 1 q 2 (4.134) - C C These equations may be written in standard integrable matrix form x ' = Ax. -1/RC 1/RC ( ( ( ( q 1 q 1 ( ( d (4.135) = ¸ ¸ ( ¸ ¸ ( ¸ ( ¸ dt 1/RC -1/RC q 2 q 2 Notice that the off-diagonal terms in the system matrix lack the sign asymmetry required in an oscillator. In fact, all elements in the system matrix have the same magnitude and the pattern of signs comes from the fact that i 1 = -i 2 . It is instructive to change coordinates and use the sum and difference of the two charges as state variables. The transformation equation is as follows. 1 -1 ( ( ( ( q d q 1 ( ( (4.136) = ¸ ¸ ( ¸ ¸ ( ¸ ( ¸ 1 1 q s q 2 The inverse transformation also exists. 1/2 1/2 ( ( ( ( q 1 q d ( ( (4.137) = ¸ ¸ ( ¸ ¸ ( ¸ ( ¸ -1/2 1/2 q 2 q s From equation 4.94, the transformed state equations are: 1 -1 -1/RC 1/RC 1/2 1/2 ( ( ( ( q d q d ( ( ( ( ( ( d (4.138) = ¸ ¸ ( ¸ ¸ ( ¸ ¸ ( ¸ ¸ ( ¸ ( ¸ dt 1 1 1/RC -1/RC -1/2 1/2 q s q s -2/RC 0 ( ( ( ( q d q d ( ( d (4.139) = ¸ ¸ ( ¸ ¸ ( ¸ ( ¸ dt 0 0 q s q s With this change of coordinates the state equations are trivial to integrate. The solutions are: q d (t) = q d (t o )e (t o - t)2/RC (4.140) q s (t) = q s (t o ) (4.141) Hence the second state equation merely reiterates the fact that charge is conserved in this system. As we expect from the pattern of signs in the system matrix, this system cannot exhibit oscillatory behavior. 205 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 7 A Switch Behaves as a Modulated Dissipator As a final remark on this simple example, if we compare figures 4.32 and 4.35 we may note that the causality on capacitors is same in both cases, before and after connection. This indicates that it is possible to represent the models for both regimes as a single bond graph, shown in figure 4.36. The important addition is that the switch has been represented as a modulated dissipator. A suitable constitutive equation would be: u i R = R e R (4.142) where u is a modulating or control input which can assume only two real values, 1 or 0, corresponding to connected and disconnected, respectively. i R = ¦ ¦ ´ ¦ ¹ 0 if u = 0 e R R if u = 1 (4.143) C 0 1 C modulating or control input, u R Figure 4.36: Model with modulated dissipator representing connected and unconnected behavior. Note that this modulated dissipator is subject to a strict causal constraint: it must have conductance causality, determining an output flow (current) as a function of an input effort (voltage). The inverse of equation 4.142 is not defined for u = 0. Modulated dissipators will be encountered frequently in describing control systems where signal-level events influence power-level behavior. 206 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Examples Ideal asymmetric two-port junction elements, gyrators and transformers, describe the transmission or transduction of power, one of the most fundamental aspects of all machinery. An abundant variety of devices that may be modeled using transformers or gyrators. Rotation to Translation Almost all mechanisms involve some relation between rotation and translation, as the following examples illustrate. Pivoted Lever Arm One of the simplest examples is the pivoted lever arm shown in figure 5.7. Rotation of the shaft results in motion of the end of the arm. Assuming the arm is rigid and the pivot constrains it to motion about a fixed axis, a relation between displacements (the angle of the arm, u, and the horizontal position of its end, x) may be derived from simple geometry. x = l sin(u) (5.19) A relation between flows (the angular speed of the arm, e, and the horizontal linear velocity of the end, v) may be obtained by differentiating with respect to time. v = l cos(u) e (5.20) u arm of length l shaft pivots about fixed axis x Figure 5.7: Diagram of a simple rotation-to-translation transformer. This is one of the equations of a transformer relating the translational and rotational domains. The other equation relates horizontal force, F, at the end to moment, µ, about the axis. The moment arm is the perpendicular distance from the end to the axis. µ = l cos(u) F (5.21) 207 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Equations 5.20 and 5.21 define a modulated transformer (e.g figure 5.6) with a single parameter T = l cos (u) modulated, in this case, by the angle u. As required by the definition of a pure transformer, power in equals power out (Fv = µe). Note that this fundamentally stems from the fact that the same geometric quantity, the perpendicular distance from the end to the axis, is the parameter in both equations. Scotch Yoke When the shaft in figure 5.7 rotates, the vertical position of the end of the arm also changes, a fact ignored in equation 5.19. It may seem that an important part of the mechanical behavior has been ignored, but it should be recognized that there are numerous contrivances by which the vertical position may be made irrelevant. One example is the Scotch Yoke shown in figure 5.8. u wheel slider bearing roller x Figure 5.8: Another rotation-to-translation transformer: a Scotch Yoke. 208 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Wheel It should not be assumed that linearity may only be achieved for small displacements. Appropriate design can result in devices which are linear over large ranges of displacements. Perhaps the simplest and most ubiquitous example is the wheel. Figure 5.9 depicts a wheel which may rotate about a fixed axis. The rim of the wheel contacts a sliding bar. Due to friction between the rim and the slider, when the wheel rotates, the slider translates. Assuming the wheel and slider are rigid and there is no slip between rim and slider, simple geometry provides a relation between displacements. x = r (5.22) where r is the radius of the wheel. As before, a relation between flows (speed of the slider, v, and angular speed of the wheel, ) may be obtained by differentiating with respect to time. v = r (5.23) The relation between force, F, on the slider to the moment, µ, about the shaft is also linear. µ = r F (5.24) These are the equations of a linear transformer (e.g. figure 5.3) relating the translational and rotational domains. Again, the same geometric parameter, r, shows up in both equations. wheel x shaft slider Figure 5.9: A linear rotation-to-translation transformer composed of a wheel and slider. Nevertheless, it is well to keep in mind that in general, a transformer may be modulated, embodying a nonlinear relation between efforts and flows. As any practical machine designer knows, the radius of any real wheel will not be constant. The wheel may be non-circular or may be mounted eccentrically. In either case, its radius will be a function of its angle. If these effects are sufficiently important, they may modeled by describing the device as a modulated transformer as depicted in figure 5.6. 209 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Parasitic Dynamics As with other elements of a lumped-parameter model, an ideal transformer such as that of figure 5.3 or 5.6 describes a single phenomenon: power transmission. Real power transmission devices such as those sketched in figures 5.7 to 5.9 also store and dissipate energy. Being secondary (and somewhat athwart the primary function of power transmission) these phenomena are often referred to as parasitic. A competent model of a real power transmission device may need to include these "parasitic" effects. This may readily be accomplished using the energy storage and dissipation elements defined earlier. For example, the wheel and slider of figure 5.9 might exhibit substantial energic phenomena in both domains. Likely candidates in the rotational domain are the inertia of the wheel and shaft and the friction due to the bearings which support the shaft (not shown in figure 5.9). All of the rotational phenomena are associated with a single speed, so the inertia, friction and the rotational side of the transformer are connected by a one-junction. friction friction bearing R R R friction between and slider between rim slider and frame 1 r v rim TF 1 0 1 v slider wheel and shaft mass of inertia I I slider Rotational domain Translational domain Figure 5.10: Bond graph of a model of the wheel and slider of figure 5.9 including plausible energic effects in the rotational and translational domains. Inertia and friction are also plausible in the translational domain. There are two distinct frictional components: One is due to the motion of the slider relative to the frame. That component and the inertia of the slider are associated with the same velocity, v slider , so the two elements are connected to a one junction. The other component is due to the possibility of relative motion between the slider and the wheel rim. The translational velocity of the wheel rim at the point of contact, v rim , is the translation port of the transformer. The frictional element is connected to a zero junction because the force on it is the same as that transmitted to the slider. Including these phenomena, a model of the system might be as shown in figure 5.10. Of course, it should not be assumed that figure 5.10 is necessarily a better model. For some purposes, to include these phenomena may be to add superfluous detail and an ideal transformer may be the best model. 210 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Rotation to Fluid Flow Another common phenomenon in machinery is the transport of fluids. Pumps are among the more widespread machine components, as may be seen by examining a typical automobile engine. Direction of rotation Input shaft Fluid input port Fluid output port Fluid flow direction Entrapped volume Direction of rotation Figure 5.11: A sketch of a gear pump is shown on the left. A cross section is shown on the right. Gear Pump One of the more robust and versatile types of pump is the gear pump. As illustrated in figure 5.11, two meshing gears rotate inside a close-fitting housing. Fluid entrapped between the gears and the housing is drawn from the input port to the output port. A gear pump is generally used to pump incompressible fluids and, by design, is a positive-displacement pump: aside from a small amount of leakage between the gears and the housing, a fixed volume of fluid is pumped per unit rotation of the drive shaft. If we neglect any leakage in the device sketched in figure 5.11, for every sixty degrees of shaft rotation, each of the two gears pumps the volume, V e , entrapped between one gear- tooth and the housing. The relation between displacements is as follows. 6 V e V = = K (5.25) where V is fluid volume, is shaft angle in radians and K is a constant. A relation between flows is obtained by differentiating this relation between displacements with respect to time. Q = K (5.26) where Q is volumetric flow rate and is angular speed. This is one of the equations of a transformer relating the fluid flow and rotational domains. The other equation relates the pressure difference, P, between the input and output fluid ports to the moment, µ applied to the drive shaft. It could be obtained by calculating the pressure-induced forces on the faces of the gear teeth and taking moments about the shaft axis, but that requires careful thought and tedious calculation. A far 211 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 simpler method is to remember that power into an ideal transformer must equal power out of it (PQ = µ) thus the relation between efforts must be as follows. µ = K P (5.27) Hydraulic Motor An important feature of all power transduction devices is that, in general, power may be transferred in both directions. A transformer or gyrator is fundamentally a two-way device. Consequently, the positive-displacement gear pump depicted in figure 5.11, designed to transmit power from the rotational to the fluid domain, is also one of the basic designs for a hydraulic motor which transmits power from the fluid to the rotational domain 1 . Parasitic Dynamics In addition to the power transmission it is designed for, a real pump or motor is likely to exhibit energy storage and dissipation phenomena. In the rotational domain, friction and the inertia of the rotating components may be significant. A single speed is associated with the inertia, friction and the rotational side of the pump and these elements are connected by a one-junction. In the fluid domain, there are at least three distinct sources of energy dissipation, each associated with a different flow rate. The first is due to leakage, e.g. between the gears and the housing end- plates. This leakage flow is not proportional to the speed of the shaft, but is related to the pressure difference between input and output. If the leakage flow is not zero, then the total flow though the pump is different from the component due to shaft rotation, and each of these two flows may have a distinct dissipation phenomenon associated with it. For example, the orifice head losses at entry and exit are related to total flow while the head loss due to flow past the circular walls of the chamber is related to the flow component due to shaft rotation. The required junction structure is shown in figure 5.12. 1 However, a motor and a pump may differ in important aspects of their design details. 212 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 chamber leakage orifice losses losses losses 1 TF I R 1 inertia of gears and shaft bearing friction 1 0 R R 0 0 R P Q input pump total output P P Q K : Rotational domain Fluid domain Figure 5.12: Bond graph of a model of the gear pump of figure 5.11 including plausible energic phenomena in the rotational and fluid domains. As before, this is not necessarily a better model than the ideal transformer, merely more detailed. The extra detail comes with a price: to be useful, the parameters of the additional elements must be determined and that may require considerable effort. 213 © Neville Hogan page 1 Magnetic electro-mechanical machines Lorentz Force A magnetic field exerts force on a moving charge. The Lorentz equation: f = q(E + v × B) where f: force exerted on charge q E: electric field strength v: velocity of the moving charge B: magnetic flux density Consider a stationary straight conductor perpendicular to a vertically-oriented magnetic field. Magnetic flux B F i Current Force Stationary conductor Motion of charges An electric field is oriented parallel to the wire. As charges move along the wire, the magnetic field makes them try to move sideways, exerting a force on the wire. The lateral force due to all the charge in the wire is: f = µAl (v × B) where µ: density of charge in the wire (charge per unit volume) l: length of the wire in the magnetic field A: its cross-sectional area The moving charges constitute a current, i i = µAv The lateral force on the wire is proportional to the current flowing in it. f = l (i × B) For the orthogonal orientations shown in the figure, the vectors may be represented by their magnitudes. f = l B i 214 © Neville Hogan page 2 This is one of a pair of equations that describe how electromagnetic phenomena can transfer power between mechanical and electrical systems. The same physical phenomenon also relates velocity and voltage. Consider the same wire perpendicular to the same magnetic field, but moving as shown Magnetic flux B Moving conductor Electro-motive force Motion of charges Velocity v A component of charge motion is the same as the wire motion. The magnetic field makes charges try to move along the length of the wire from left to right. The resulting electromotive force (emf) opposes the current and is known as back-emf. The size of the back-emf may be deduced as follows. Voltage between two points is the work required to move a unit charge from one to the other. If a unit charge moves along the wire from right to left the work done against the electromagnetic force is e = v B l This is the other of the pair of equations that describe how electromagnetic phenomena can transfer power between mechanical and electrical systems. Two important points: 1. The interaction is bi-lateral (i.e., two-way). If an electrical current generates a mechanical force mechanical velocity generates a back-emf. 2. The interaction is power-continuous. Power is transferred from one domain to the other; no power is dissipated; no energy is stored; electrical power in equals mechanical power out (and v.v.). P electrical = e i = (v B l) i = v (B l i) = v f = P mechanical Power continuity is not a modeling approximation. It arises from the underlying physics. The same physical quantity (magnetic flux density times wire length) is the parameter of the force- current relation and the voltage-velocity relation 215 © Neville Hogan page 3 D'Arsonval Galvanometer Many electrical instruments (ammeters, voltmeters, etc.) are variants of the D'Arsonval galvanometer. A rectangular coil of wire pivots in a magnetic field as shown in figure 2. Restraining spring Magnetic field Square coil + - S N u Figure 2: Sketch of a D'Arsonval galvanometer. Current flowing parallel to the axis of rotation generates a torque to rotate the coil. Current flowing in the ends of the coil generates a force along the axis. Assuming the magnetic flux is vertical across the length and width of the coil, the total torque about the axis is: t = 2NBlh cos(u) i where t: clockwise torque about the axis N: number of turns of wire B: magnetic flux density l: length of the coil h: half its height. This torque is counteracted by a rotational spring As the coils rotate, a back-emf is generated. e = 2NBlh cos(u) e where e: angular speed of the coil. Note that the same parameter, 2NBlh cos(u), shows up in both equations. 216 © Neville Hogan page 4 Direct Current Permanent Magnet Electric Motor With a different geometry the dependence on angle can be substantially reduced or eliminated Radial flux lines Stationary permeable core Rotating cylindrical coil (much of the magnetic field is not shown) Stationary permanent magnets N S Figure 3: Schematic end-view of a D'Arsonval galvanometer modified to reduce the angle- dependence of the transduction equations. Features: - rotating cylindrical coil - stationary permeable core - shaped permanent magnets - constant radial gap between magnets and core - the magnetic field in the gap is oriented radially If all turns of the coil are in the radial field the torque due to a current in the coil is independent of angle. 0° 30° 60° -60° 90° -30° -90° Angle Figure 4: Sketch of half of the constant-current torque/angle relation resulting from the design of figure 3. From symmetry the torque/angle relation for the other half of the circle is the negative of that shown above. The reversal of torque can be eliminated by reversing the current when the angle passes through ±90°. A mechanical commutator is sketched in figure 5. 217 © Neville Hogan page 5 N S Split-ring commutator rotates with coil Brushes are stationary electrical contacts Electrical connection to the coil is through the commutator + - Figure 5: Schematic of the mechanical commutation system used in a direct-current permanent magnet motor. Electrical connection is through a set of stationary conductors called brushes 1 . They contact a split ring called a commutator that rotates with the coil. This commutator design is used in a direct-current permanent-magnet motor (DCPMM). The same effect may be achieved electronically. That approach is used in a brushless DCPMM. Assuming perfect commutation the relation between torque and current for a DCPMM is t = K t i K t : torque constant, a parameter determined by the mechanical, magnetic and electrical configuration of the device. There is also a corresponding relation between voltage and rotational speed. e = K e e K e : back-emf or voltage constant. Excerpt from a manufacturer's specification sheet for a direct-current permanent-magnet motor. MOTOR CONSTANTS: (at 25 deg C) SYMBOL UNITS torque constant KT oz in/amp 5.03 back emf constant KE volts/krpm 3.72 terminal resistance RT ohms 1.400 armature resistance RA ohms 1.120 average friction torque TF oz in 3.0 1 The reason for this terminology is historical — the earliest successful designs used wire brushes for this purpose. 218 © Neville Hogan page 6 viscous damping constant KD oz in/krpm 0.59 moment of inertia JM oz in sec-sec 0.0028 armature inductance L micro henry <100.0 temperature coefficient of KE C %/deg c rise -0.02 These specifications imply that the torque and back-emf constants are distinct parameters. different symbols: KT, KE different units: oz-in/amp, volts/krpm But if we express both constants in mks units K t = 0.0355 N-m/amp K e = 0.0355 volt-sec/rad As required by the physics of electromagnetic power transduction, the constants are in fact identical. “Parasitic” Dynamics The Lorentz force yields two equations describing power-continuous electro-mechanical transduction. A practical electric motor also includes energy storage and/or power dissipation in the electrical and mechanical domains. A competent model may include these effects. Electrical side - inductance of the coil - resistance of the coil - resistance of electrical connectors ("terminal resistance") Mechanical side - inertia of the rotating components (coil, shaft, etc.) - friction of brushes sliding on the commutator - viscous drag due to entrained air Connections: Electrical side: - only one distinct current—series connection of inductor, resistor & model element for back-emf due to velocity Mechanical side: - only one distinct speed—common-velocity connection of inertia, friction & model element for torque due to motor current 219 © Neville Hogan page 7 coil inductance coil & terminal resistance rotor inertia friction losses back emf motor torque electrical domain rotational domain Schematic diagram of DCPMM model with “parasitic” dynamics Motor vs. Generator Lorentz-force electro-mechanical transduction is bi-lateral. An electric motor uses it to convert electrical power into rotational power. An electrical generator uses it to convert rotational power into electrical power. Tachometer Lorentz-force electro-mechanical transduction is also used for sensing. A DC permanent magnet tachometer generates voltage proportional to angular velocity as described by the velocity- voltage equation above. 220 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 PHYSICAL BASIS OF ANALOGIES Equivalences: Transforming Through Transformers and Gyrators One useful way to enhance your understanding of a system involving multiple energy domains is to develop an equivalent dynamic system in a single domain of your choice. To do this, any asymmetric junction elements (transformers or gyrators) coupling the multiple domains must be eliminated, if possible. Indeed, even if the original system involves only a single energy medium, identifying an equivalent system without transformers and gyrators can be useful. In effect, it is helpful to know how an element (or even an entire subsystem) would appear to behave if it were observed through a transformer or gyrator. By loose analogy with viewing an object through a lens, we speak of transforming elements or subsystems through transformers or gyrators. Transforming Through a Transformer Any element or subsystem transformed through a transformer retains its dynamic character, though its parameter values change. That is, a zero junction still appears to behave as a zero junction, a capacitor still appears to behave as a capacitor, a flow source still appears to behave as a flow source, and so forth. This is true whether the elements are linear or nonlinear. The equivalent (transformed) active source elements are related to the original source elements in proportion to the transformer coefficient, as summarized in table 5.2. The relation for two-port asymmetric junctions (transformers and gyrators) is equally straightforward and is summarized in table 5.3. The other possible causal assignments are treated similarly. Multi-port symmetric junctions (one and zero junctions) may also be transformed as summarized in table 5.4. Representative causal assignments are shown. The other possible causal assignments are treated similarly. Table 5.2 Transforming active elements through a transformer. Initial bond graph: Equivalent bond graph: Constitutive equations: Constitutive equations: 1 TF S : e(t) 2 e T 1 S e : Te(t) e 1 = T e 2 ; e 2 = e(t) e 1 = Te(t) 1 TF 2 1/T S f : f(t) 1 S f : f(t)/T f 1 = (1/T) f 2 ; f 2 = f(t) f 1 = f(t)/T 221 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Initial bond graph: Table 5.3 Transforming asymmetric junction elements through a transformer. Equivalent bond graph: Constitutive equations: Constitutive equations: TF 1 2 3 T 1 TF T 2 TF T 1 T 2 1 3 e 1 = T 1 e 2 ; e 2 = T 2 e 3 e 1 = T 1 T 2 e 3 f 2 = T 1 f 1 ; f 3 = T 2 f 2 f 3 = T 1 T 2 f 1 TF 1 2 3 T G GY GY 1 T G 3 e 1 = T e 2 ; e 2 = G f 3 e 1 = TG f 3 f 2 = T f 1 ; e 3 = G f 2 e 3 = TG f 1 222 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 Initial bond graph: Table 5.4 Transforming symmetric junction elements through a transformer. Equivalent bond graph: Constitutive equations: Constitutive equations: TF 1 2 T 1 4 3 TF 3 T 1 1 TF 5 2 4 T e 1 =T(e 3 + e 4 ) e 1 = T e 3 + T e 4 f 3 = T f 1 ; f 4 = T f 1 f 3 = T f 1 ; f 4 = T f 1 TF 1 2 T 0 4 3 TF 3 1/T 1 0 TF 5 2 4 T e 1 = T e 4 ; e 3 = e 4 e 1 = T e 4 ; e 3 = (1/T)T e 4 f 4 = T f 1 - f 3 f 4 = T f 1 - T(1/T) f 3 For passive elements, the relation is quite different. If a passive element is linear, the parameter of the equivalent (transformed) element is related to the parameter of the original element through the square of the transformer coefficient. One way to remember this is to note that to obtain the equivalent relation for an active element, we had to "look through" the transformer once. In contrast, to obtain the equivalent relation for a passive element, we had to "look through" the transformer twice, and had to multiply (or divide) by the transformer coefficient on each pass. These relations are summarized in table 5.5 223 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 Initial bond graph: Table 5.5 Transforming passive elements through a transformer. Equivalent bond graph: Constitutive equations: Constitutive equations: Constitutive equations: Constitutive equations: TF C: C 1 2 T C: C/T 2 1 de 2 1 e 1 = T e 2 ; dt = C f 2 ; f 2 = T f 1 de 1 T 2 dt = C f 1 TF I : I 1 2 1/T 1 I : T 2 I dt = I 1 df 2 1 1 f 1 = T f 2 ; e 2 ; e 2 = T e 1 df 1 1 dt = T 2 I e 1 TF R : R 1 2 T R : T 2 R 1 e 1 = T e 2 ; e 2 = R f 2 ; f 2 = T f 1 e 1 = T 2 R f 1 TF R : G 1 2 1/T R : G/T 2 1 f 1 = T f 2 ; f 2 = G e 2 ; e 2 = T e 1 1 1 f 1 = T 2 e 1 G Transforming Through a Gyrator Any element or subsystem transformed through a gyrator takes on the dynamic character of the dual element, and in addition its parameter values may change. That is, a zero junction appears to behave as a one junction, a capacitor appears to behave as an inertia, a flow source appears to behave as an effort source, and so forth. This is true whether the elements are linear or nonlinear. 224 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 Initial bond graph: Table 5.6 Transforming active elements through a gyrator. Equivalent bond graph: Constitutive equations: Constitutive equations: Table 5.7 1 GY 2 G S f : f(t) S e : Gf(t) 1 e 1 = G f 2 ; f 2 = f(t) e 1 = Gf(t) 1 GY S : e(t) 2 e 1/G 1 S f : e(t)/G f 1 = (1/G) e 2 ; e 2 = e(t) f 1 = e(t)/G Transforming asymmetric junction elements through a gyrator. Initial bond graph: Equivalent bond graph: Constitutive equations: Constitutive equations: GY 1 2 3 G 1/T TF GY 1 G/T 3 e 1 = G f 2 ; f 2 = (1/T) f 3 e 1 = (G/T) f 3 e 2 = G f 1 ; e 3 = (1/T) e 2 f 3 = (G/T) f 1 GY 1 2 GY G 1 1/G 3 2 TF 1 G 1 /G 2 3 e 1 = G 1 f 2 ; f 2 = (1/G 2 ) e 3 e 1 = (G 1 /G 2 ) e 3 e 2 = G 1 f 1 ; f 3 = (1/G 2 ) e 2 f 3 = (G 1 /G 2 ) f 1 225 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 6 Initial bond graph: Table 5.8 Transforming symmetric junction elements through a gyrator. Equivalent bond graph: Constitutive equations: Constitutive equations: GY 1 2 G 1 4 3 GY 3 G 1 0 GY 5 2 4 1/G e 1 =G f 3 ; f 4 = f 3 e 1 = G f 3 ; f 4 = (1/G)G f 3 e 3 = G f 1 - e 4 e 3 = G(f 1 - (1/G)e 4 ) GY 1 2 G 0 4 3 GY 3 G 1 1 GY 5 2 4 G e 1 = G(f 3 + f 4 ) e 1 = G f 3 + G f 4 e 3 = G f 1 ; e 4 = G f 1 e 3 = G f 1 ; e 4 = G f 1 The equivalent (transformed) active source elements are related to the original source elements in proportion to the gyrator coefficient, as summarized in table 5.6. The relation for two-port asymmetric junctions (transformers and gyrators) is summarized in table 5.7. The other possible causal assignments are treated similarly. Multi-port symmetric junctions (one and zero junctions) may also be transformed as summarized in table 5.8. Representative causal assignments are shown. The other possible causal assignments are treated similarly. As with transforming through a transformer, if a linear passive element is transformed through a gyrator, the parameter of the equivalent element is related to the parameter of the original element through the square of the gyrator coefficient. These relations are summarized in table 5.9 226 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 7 Table 5.9 Transforming passive elements through a gyrator. Initial bond graph: Equivalent bond graph: Constitutive equations: Constitutive equations: Constitutive equations: Constitutive equations: GY I : I 1 2 G C: I /G 2 1 df 2 1 dt = I e 1 = G f 2 ; e 2 ; e 2 = G f 1 de 1 G 2 dt = I f 1 GY C: C 1 2 1/G 1 I : G 2 C 1 de 2 1 1 f 1 = G e 2 ; dt = C f 2 ; f 2 = G e 1 df 1 1 dt = G 2 C e 1 GY R : G r 1 2 G R : G 2 /G r 1 e 2 e 1 = G f 2 ; f 2 = G r ; e 2 = G f 1 e 1 = G r f 1 G 2 GY R : R 1 2 1/G R : G r /G 2 1 f 1 = G e 2 ; e 2 = R f 2 ; f 2 = G e 1 1 1 f 1 = G 2 e 1 R These tables are not intended to be memorized; they are presented primarily to demonstrate the various equivalences. Summarizing: transforming an element through a transformer may change its parameters but not its form. Transforming an element through a gyrator changes the element to its dual, and may also change its parameter values. The parameters of passive linear elements are scaled by the square of the transformer or gyrator parameter. The constitutive equations of other linear elements are changed in proportion to the first power of the parameter. These equivalences permit entire subsystems within a bond graph to be transformed through a transformer or gyrator, thereby enhancing insight into the consequences of energetic interaction and simplifying the graph and the subsequent process of deriving equations . Beware, however, that the physical meaning of the quantities represented on the graph may become obscured. The fact that transforming through a gyrator dualizes an element has some intriguing consequences. A transformer is, in a sense, a superfluous element, as it could always be represented by a pair of gyrators. By a similar argument, one of each pair of the other dual elements (0 and 1, C and I, S e 227 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 8 and S f ) could always be replaced by the other transformed through a gyrator. Thus any bond- graph composed of the nine primitive elements we have defined so far (R, S e , S f , 0, 1, C, I, GY, TF) could be constructed from a set of five primitive elements, for example: (R, S f , 0, C, GY). This possibility and its implications have been explored at some length in the bond-graph literature. However, most engineers are quite comfortable with the dual concepts of inertia and elasticity, inductance and capacitance, and so forth, and it is not clear that this further abstraction would enhance understanding of the physical system. 228 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Examples of Equivalences Electrical Inductor An electrical inductor is usually implemented as a coil of wire wrapped on a magnetically permeable core, e.g. as sketched in figure 5.13. Thus it embodies a gyrational transduction between the electrical and magnetic domains and a capacitive energy storage element in the magnetic domain, as depicted by the bond graph of figure 5.14. It is most often used as an electrical circuit component. Assuming magnetic linearity, the equivalent behavior in the electrical domain is obtained by combining equations 5.33, 5.34 and 5.35: N 2 ì = R i = L i (5.64) where L = N 2 /R is the inductance of the coil. Thus the behavior of the electrical inductor arises from the behavior of a capacitor in the magnetic domain transformed through a gyrator. Differentiating results in a form of the constitutive equation which may be more familiar. e = L di/dt (5.65) Electrical Transformer with Parasitic Energy Storage A model of an electrical transformer including energy storage in the magnetic field was presented in figure 5.17, consisting of a pair of gyrators between the electrical and magnetic domains and a capacitive energy storage element in the magnetic domain. Reflecting the magnetic capacitor through the gyrator on the right results in the equivalent network shown below. e 2 e 1 S f N 1 TF 0 S e i 2 i 1 i L2 N 2 N 2 2 I : R Figure 5.30: Bond graph of an equivalent network representing the transformer model of figure 5.17. The source elements in figure 5.30 represent an assumed current input on the the left and voltage input on the right. After causality has been assigned, an equation for the current output on the right may be read from the graph. N 1 i 2 = N 2 i 1 - i L2 (5.66) 229 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 where i L2 is the inductor current, determined from the inductor constitutive equation expressed in integral causal form. di L2 N 2 2 dt = R e 2 (5.67) Differentiating equation 5.66 and substituting equation 5.67 yields equation 5.45 as before. Alternatively, the magnetic capacitor could also be transformed through the gyrator on the left of figure 5.17. The resulting equivalent network is shown below. e 2 e 1 S f 0 N 1 TF S e i 2 i 1 i L1 N 2 N 1 2 I : R Figure 5.31: Bond graph of an alternative equivalent network representing the transformer model of figure 5.17. The same network could be obtained by transforming the inductor of figure 5.30 through the transformer. Consequently, the values of the two equivalent inductors should be related by the square of the transformer coefficient, and they are. L 1 = N 1 2 R = 2 N 2 2 R = 2 N 1 N 1 ¸ ( ( ¸ ¸ ( ( ¸ L 2 (5.68) N 2 N 2 230 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Hydraulic Ram Figure 5.32 shows a sketch of a hydraulic actuator of the kind commonly used in heavy earth- moving machinery. The difference in fluid pressure acting on the two sides of the piston exerts a force on the rod. As pressure is defined as force per unit area, the power transduction phenomenon connecting the hydraulic and translational domains is that of a transformer. In general, the piston area in the two chambers need not be the same 1 , and the rate of increase of the volume of one chamber, Q c1 , need not equal the rate of decrease of the volume of the other chamber, Q c2 , so two transformers may be needed to describe the piston. They are connected to a one-junction on the translational side (representing the unique piston velocity) as shown in figure 5.33(a). cylinder rod fluid lines (a) F v Q piston o-ring seals supply line return line P P P Q c1 c2 1 2 1 P 2 (b) Figure 5.32: Sketch of a hydraulic actuator (a). A cross-sectional view is shown in (b). 1 For example, the sketch shows a rod protruding from each side of the piston, but single-rod cylinders are also common; with a single rod, the area on one side of the piston exceeds that on the other side by the area of the rod. 231 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 If the piston area in each chamber is the same, then the two transformers may be replaced by a single equivalent transformer as shown in figure 5.33(b). Any increase in volume of one chamber equals the decrease in volume of the other and the transformer is connected to a one junction in the fluid domain (representing the common rate of volume change, Q c ). Constitutive equations are then as follows. F = A (P c1 - P c2 ) (5.69) Q c = A v (5.70) TF 1 1 TF 2 Q c1 1 1 Q c2 A A v TF 1 1 Q c A v (a) (b) Figure 5.33: Bond graph fragments representing power transduction in the hydraulic actuator. (a): general model for unequal piston areas. (b): special case of equal piston areas. Parasitic Dynamics Like other power transduction devices, a hydraulic ram also embodies energy storage and dissipation effects which may need to be modeled. For simplicity, we will assume the hydraulic fluid to be incompressible (though that is often one of the prominent "parasitic" dynamic phenomena) and model only the inertial effects due to acceleration of the fluid and the power dissipation due to its motion through the lines and orifices. On the mechanical side we will model the inertia of the piston and rods (considered to be a single rigid body) and the friction due to the sliding seals, described for simplicity as a single linear friction element (though in reality, a significant component of the dissipative behavior is likely to resemble dry friction). Friction due to flow in a pipe was considered in Chapter 4. A crude model of the inertial effects due to acceleration of the fluid may be derived as follows. Consider an incompressible fluid moving in a pipe of uniform cross-sectional area, A p . The mass, m, of the "slug" of fluid between two cross sections separated by a distance l is determined by the dimensions of the pipe and the density, . m = lA p (5.71) Considerable simplification is obtained by assuming that the fluid flow velocity has the same value, v, at all points. That velocity is determined by the volumetric flow rate, Q, divided by the cross sectional area. v = Q/A p (5.72) Given these assumptions, the kinetic co-energy of the moving fluid is: 232 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 * m lA p Q 2 l E k = 2 v 2 = 2 A p 2 = A p Q 2 (5.73) Thus, by analogy with a rigid body in translation, the energy stored in the moving fluid may be described by an ideal inertia, characterized by a linear relation between a flow, Q, and a generalized momentum, , termed pressure momentum. Q = (5.74) I The parameter I = l/A p is the inertance of the fluid, determined by the geometric and material properties of the pipe and the fluid. From the fundamental definition of generalized momentum, the pressure difference accelerating the fluid is the time derivative of the pressure momentum. P = d/dt (5.75) Thus pressure difference is proportional to the rate of change of fluid flow rate. l dQ P = A p dt (5.76) A bond graph of the complete system is shown in figure 5.34(a). For the purpose of this example, it has been assumed that the supply line is connected to a power source at constant pressure, P 1 , the return line to a sump at constant pressure, P 2 , (e.g. atmospheric pressure) and that the mechanical load is a constant force, F. We may analyze this system by assigning causality. After assigning the required causality to the source elements, we are free to assign integral causality to one of the energy storage elements (arbitrarily chosen in figure 5.34(b) to be the translational inertia). Propagating the causal consequences of that choice through the graph completes the causality assignment. Causality assignment reveals that two of the three inertias are dependent energy-storage elements. The velocity of the piston determines the flow rate in the fluid lines and a single variable is sufficient to characterize the kinetic energy stored in both the translating masses and the moving fluid. In effect, the piston and rods and the fluid in the lines have been described as a single "rigid body" even though the fluid in the lines would hardly be considered rigid in the usual sense of the word. 233 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 P : S e 1 TF 1 1 Q c A v 1 1 Q Q 2 1 R I R I R : b : m : 1 I R : 2 R : 1 S e : F P : S e 2 I : I 2 Hydraulic domain Translational domain (a) I S e TF 1 1 1 1 R R I R S e S e I (b) Figure 5.34: (a) Bond graph of a model of the hydraulic actuator including parasitic effects. (b) Bond graph with causality assigned. The parasitic fluid elements may be transformed into the translational domain. The fluid inertias are equivalent to translational masses. Combining equations 5.69, 5.70 and 5.76 the effective mass due to each fluid line is m eff = A 2 I = A 2 l/A p (5.77) The fluid resistors are equivalent to linear translational friction elements. The effective friction coefficient due to each fluid line is b eff = A 2 R (5.78) 234 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 I S e I I 1 TF 1 S e S e R R R (a) S e P 1 : I : m + A I + A I 2 1 2 2 TF 1 1 R S e S e F : P 2 A : : b + A R + A R 1 2 2 2 (b) Figure 5.35: Bond graph of figure 5.34 with (a) parasitic fluid elements transformed into the translational domain and (b) inertial and dissipative elements combined. When the parasitic fluid elements are transformed into the translational domain it becomes clear that all of the parasitic elements are connected to a single one-junction as shown in figure 5.35(a). The compatibility equation associated with the one-junction means that forces sum at the one-junction. As a result, the parasitic elements may be combined as shown in figure 5.35(b) into a single effective inertia with an apparent mass of m app = m + A 2 I 1 + A 2 I 2 (5.79) and a single effective friction element with an apparent friction coefficient of b app = b + A 2 R 1 + A 2 R 2 (5.80) 235 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Momentum-based fluid-mechanical power transduction. A domestic vacuum cleaner is a good example of a consumer product that fundamentally involves power transduction between the electrical, mechanical and fluid domains. How should the transduction mechanism of the pump be modeled? The best answer is based on understanding the basic fluid-mechanical transduction phenomenon in a turbine or centrifugal pump. This is due to a change of fluid momentum which requires a force or torque. Thus fluid flow is related to rotational effort; this transduction phenomenon may be modeled by a gyrator. Suppose you didn't see this; how could you deduce it? One way is to consider what is required to reproduce observable behavior of the vacuum cleaner. A key clue is the observation that when the air hole of a vacuum cleaner is blocked the vacuum cleaner speeds up. Let's try to model this phenomenon. Motor: Assume no significant electrical dynamics; model the electric motor as a torque source; include motor inertia, bearing friction. Se I R 1 to pump :B m :J m t m : Fluid flow: Include orifice resistance, neglect hose capacitance and fluid inertia. from pump R:R orifice First, try modeling the pump as a transformer. The blocked flow boundary condition may be represented as a flow source with zero flow as follows. 1 Se I R 1 TF R Sf:0 :J m :B m :R orifice t m : 236 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Propagating causality, it can be seen that this model would predict that blocking the air hole would stall the motor. Now try modeling the pump as a gyrator: 1 Se I R 1 GY R Sf:0 :J m :B m :R orifice t m : In this case, propagating causality, it can be seen that this model would predict that blocking the air hole will not stall the motor. If we remove all bonds with zero power flow (shown in the following as dashed lines) we can see that the effect of blocking the flow is to eliminate the influence of the fluid resistance on the motor; i.e. the load on the motor decreases. 1 Se I R 1 GY R Sf:0 :J m :B m :R orifice t m : Influence of leakage flow One might reasonably argue that this marked difference between the predictions of the two models of the pump is an artifact of neglecting the possibility of leakage flow through the pump. If the motor were immobilized, it would still be possible to move air through the pump, yet the transformer model above would predict otherwise. To rectify that error we might include a leakage flow in the fluid flow model as follows: from pump 0 R:R leakage R:R orifice 237 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 Now if we model the pump as a transformer and impose a blocked-flow boundary condition we find that the motor is not immobilized. Removing all bonds with zero flow (shown as dashed lines) we find that blocked flow eliminates the influence on the motor of orifice resistance but not leakage resistance. Se I R 1 TF 1 0 R Sf:0 R :J m :B m :R orifice t m : R leakage : Modeling the pump as a gyrator yields a superficially similar result. Se I R 1 GY 1 0 R Sf:0 R t m : :J m :B m However, the fundamental difference between these two models may be seen by considering the equivalent rotational systems. If the pump is modeled as a transformer and a blocked-flow boundary condition is imposed, the equivalent rotational system is as follows. Se I R 1 1 0 R Sf:0 R :: B eq., leakage eq., orifice B t m : :J m :B m This may be easier to understand and visualize when represented using more conventional graphic symbols for rotational systems. First, without the flow blocked: 238 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 J m B m eq., orifice B t m B eq., leakage Then with the flow blocked as follows: J m B m eq., orifice B t m B eq., leakage blocked flow is equivalent to immobilizing this component From this we may see that blocking the flow is equivalent to increasing the value of B eq., orifice without limit, which increases the load on the motor. If the pump is modeled as a gyrator and a blocked-flow boundary condition is imposed, the equivalent rotational system is as follows. Se I R 1 1 0 R Se:0 R :: B* eq., leakage eq., orifice B* t m : :J m :B m To denote the fact that the equivalent rotational dissipators have different values form the previous case, they have been denoted by B* eq., leakage and B* eq., orifice respectively. Again, it may be easier to understand and visualize this using the more conventional graphic symbols for rotational systems. First, without the flow blocked: 239 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 5 J m B m eq., orifice B* t m +B* eq., leakage Then with the flow blocked: J m B m eq., orifice B* t m +B* eq., leakage blocked flow is equivalent to disconnecting this component Note that in contrast to the previous case, blocking the flow is equivalent to decreasing the value of B eq., orifice without limit, which decreases the load on the motor. 240 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 STATE EQUATION DERIVATION Summary of Basic Bond Graph Elements A large class of physical systems may be described using the basic lumped parameter elements — ideal active and passive one-port elements connected by multiport junction elements. The variables and primitive elements of the energy-based formalism are summarized in the following tables. Table 6.1 Fundamental variables Energy and power: E - E o = ] (P dt Conjugate power variables: P = e f Conjugate energy variables: p - p o = ] (e dt q - q o = ] (f dt Table 6.2 Active one-port elements (sources or boundary elements) effort source flow source constitutive equation e = e(t) f = f(t) bond graph symbol S e S f Table 6.3 Passive one-port element constitutive equations capacitor inertia resistor general form e = u(q) f = +(p) e = I(f) linear form e = q/C f = p/I e = Rf bond graph symbol C I R 241 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Table 6.4 Symmetric multi-port junction elements common flow junction common effort junction constitutive equations f i = f j , j = 1,n _ j=1 n o j e j = 0 e i = e j , j = 1,n _ j=1 n o j f j = 0 bond graph symbol 1 1 2 n 3 0 1 2 n 3 where o j = ¹ ´ ¦+1 if power positive in –1 if power positive out Table 6.5 Asymmetric two-port junction elements transformer gyrator constitutive equations e i = T e j f j = T f i e i = G f j e j = G f i bond graph symbol TF GY Sign Convention Half arrows denote the direction of positive power flow. For passive elements power flow is positive inwards. For active elements there is no fixed convention, though it is common to denote power flow as positive outwards. Multiport junction elements may serve to define effort or flow differences (e.g. pressure difference, relative motion). In that case the following sign convention is recommended. 242 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Causality Assignment The nine primitive elements may be used to represent energy storage and power flow within an energetic system. The energetic interactions represented by the elements constrain (but do not uniquely determine) the mathematical operations which may be used to describe the system behavior. The causality assignment procedure identifies those constraints. The procedure is repeated below (this time including the gyrator and transformer juction elements). 1. All elements in a bond graph which are constrained to have an unique causality must have the appropriate causality assigned and denoted by a causal stroke at the appropriate end of the associated bond. 1a. Active one-port elements (e.g. power sources) are always constrained to an unique causality. 1b. Remember that in some cases passive one-port elements may be constrained to an unique causality because of their constitutive equations; for example, this is often true of resistors. 2. Examine the junction elements connected to all elements with causal assignments. Propagate causal assignments through the junction structure until no further causal assignments can be made. 2a. If a one-junction has its unique flow determined by one of its bonds, then all other bonds must be assigned flow output causality. 2b. If a zero-junction has its unique effort determined by one of its bonds, then all other bonds must be assigned effort output causality. 2c. If a transformer has a flow input determined on one of its bonds, then a flow output must be assigned to its other bond. Equivalently, if an effort input is determined on one of its bonds, then an effort output must be assigned to its other bond. 2d. If a gyrator has a flow input determined on one of its bonds, then an effort output must be assigned to its other bond. Conversely, if an effort input is determined on one of its bonds, then a flow output must be assigned to its other bond. 3. Choose one of the energy storage elements which does not already have a causal assignment (it doesn't matter which one) and assign it integral causality (effort input to an inertia, flow input to a capacitor). 3a. Return to step 2 and propagate this causal assignment through the junction structure until no further causal assignments can be made. 3b. Repeat steps 3 and 3a until all energy-storage elements have a causal assignment. 243 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 4. Any elements remaining which have not been given a causal assignment are dissipators which are indifferent to their causal assignment. Choose one or those dissipators (it doesn't matter which one), assign it an arbitrary causality 4a. Return to step 2 and propagate this causal assignment through the junction structure until no further causal assignments can be made. 4b. Repeat steps 4 and 4a until all elements have a causal assignment. If any of the steps in this procedure cannot be completed without violating the causal constraints imposed by the junction structure, the model is operationally inconsistent. At least one quantity has been determined by more than one relationship. This is an error which must be corrected. In addition to uncovering logical inconsistencies in the system model, causality assignment also serves to identify dependent and independent energy storage elements. If, in the process, any energy storing element is assigned derivative causality, then that is a dependent storage element. Its stored energy is determined by the variables associated with the element from which the causal propagation began. Derivative causality on an energy storage element is not an error but it can have undesirable consequences; it may lead to extensive algebra in deriving state equations; it may also lead to numerical difficulties when simulating system behavior on a computer. If one or more energy storage elements in a model have derivative causality, you may want to modify the model to eliminate the dependencies, but this is not essential; indeed it is not always possible. Causality Assignment and State Variables Models composed of assemblies of the nine primitive elements are state-determined. To derive state equations, state variables must be chosen. In general, state variables required to describe a system are not unique; many equivalent sets may be identified and the best choice depends on the problem under consideration. For an energetic system the state variables must at least uniquely define the energy stored in the system. The minimum number of state variables required is determined by the number of independent energy storage elements in the system model. Those energy-storage elements which have been assigned integral causality are independent. The energy stored in the element at a given time may be prescribed as an independent initial condition and gives rise to an independent state variable. 244 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Substitution Methods for Deriving System Equations A system model represented by a bond graph with a complete and unambiguous causal assignment may be used to derive system equations by a straightforward method of successive substitution as follows. Models with Linear Energy Storage Elements: Power Variables In the important special case in which all of the model elements are linear, a convenient choice of state variables is the efforts and flows associated with independent energy storage elements — known as power state variables or circuit state variables. That choice is the basis of the following efficient, systematic procedure for deriving state equations by successive substitution. 5. Choose power state variables. These are the output variables from the independent energy storage elements, and the power conjugate variables are the input variables. 5a. efforts associated with independent capacitors. 5b. flows associated with independent inertias. 6. Using the constitutive equation for the corresponding energy storage element, write the rate of change of each state variable as an explicit function of its power conjugate variable. de f 6a. independent capacitor: = dq 1 = dt dt C C df e 6b. independent inertia: = dp 1 = dt dt I I 7. Reading from the causal strokes on the graph, write an expression for the input variable for each independent energy storing element as a function of the output variables from one or more of the other elements in the system. 7a. This step proceeds by direct substitution using the constitutive equation(s) for each of the elements as needed, expressed in the input/output form specified by the causal strokes on the graph. 7b. Continue substitution until the input to each independent energy storing element is expressed as a function only of one or more of the state variables or of the independent variables associated with sources. This seven-step causality assignment and substitution procedure will result in state equations for the system model. With a little practice, in most cases state equations can be written by inspection from the causally augmented bond graph of the system model. 245 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Example: Electric Circuit An electric circuit schematic is shown in figure 6.1a. This schematic is already a model of the system and specifies the energetic elements to be used: an inertia, capacitor, resistor and effort source. It is evident from the schematic that all elements share the same flow and thus are connected by a single one-junction. A bond graph for the circuit, with sign convention included, is shown in figure 6.1b. Figure 6.1a Figure 6.1b Assume all elements have linear constitutive equations. The effort source must impress an effort on the rest of the system as shown in figure 6.1c. Assigning integral causality to the inertia and propagating that causal assignment through the one-junction completes the causal assignment as in figure 6.1d. Figure 6.1c Figure 6.1d From the causal graph, the capacitor and inductor are independent energy storage elements. Power (or circuit) state variables are the capacitor voltage, e C , and the inductor current, i L . From the constitutive equations for these elements: d dt e C = i C C (6.1) d dt i L = e L L (6.2) 246 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 Reading from the causal graph, the inductor determines the current associated with the one- junction, which in turn determines the current for the other connected elements. Thus, i C = i L (6.3) Substituting, one state equation is: d dt e C = i L C (6.4) Again reading from the causal graph, the inductor voltage is determined by the other three one- port elements. e L = e S – e C – e R (6.5) Note that the signs in this junction equation are read directly from the half-arrows on the graph. In words, voltage into the inductor equals voltage into the junction from the source minus the voltages out of the junction into the capacitor and the resistor. From the causal graph, the appropriate form of the constitutive equation for the dissipator is resistance causality. e R = R i R (6.6) Again reading from the causal graph, the resistor current is determined by the inductor current. i R = i L (6.7) Substituting, the second state equation is: d 1 dt i L = L (e S – e C – R i R ) (6.8) Note that the substitution process stops when the rate of change of a state variable has been expressed as a function of state and input variables. The state equations may be written in the conventional vector/matrix form as follows. ( ( d e C( dt ¸ i L¸ = ¸ 0 1/C –1/L –R/L ( ( ¸ ¸ ( ( e C( i L¸ + ¸ 0 1/L ( ( ¸ e S (6.9) 247 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 State Variables and Energy Causality assignment procedures may be used to identify a minimum set of variables necessary to define the energetic state of a system. However, it is important to recognize that everything of interest about a system will not always be determined by its energetic state. We will frequently need to choose state variables in addition to those associated with independent energy storage elements. This is illustrated in the following example. Example: A Trivial Mechanical System rigid body F applied force F : S e 1 v I : m x (a) (b) Figure 6.3 Consider a rigid body subject to an applied force as depicted in figure 6.3a. A corresponding bond graph with casual assignment is shown in figure 6.3b. There is a single independent energy storage element in this system and its energetic state is determined by a single variable, either the momentum of the rigid body which determines the kinetic energy p 2 E k = 2m (6.37) or the coresponding speed which determines the kinetic co-energy E * k = 1 2 mv 2 (6.38) A corresponding state equation is as follows. d dt v = F m (6.39) However, we will often be concerned with the location of the rigid body, not just with its speed or energy. In that case we will need an additional state variable and a suitable set of state equations is as follows. ( ( ¸ = ¸ 0 1 ¸ ( ( ¸ ( ( ¸ 0 1/m F (6.40) d dt ¸ ( ( ¸ + ¸ x x v 0 0 v 248 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 By a similar argument, in other cases we may need other state variables (e.g. the time integral of position, etc.). Thus the state variables identified by the causality assignment procedures are merely a minimum set required to define the system energy. No Fixed Rules! To summarize: there are no fixed rules for choosing state variables. Using the substitution methods reviewed above, the following guidelines are offered: generally, for systems with linear energy-storage elements, power variables are appropriate; for systems with nonlinear energy- storage elements, energy variables are recommended. For mechanical systems, Lagrangian variables are recommended. Most important, use variables that make the most physical sense to you. 249 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Models with Nonlinear Energy Storage Elements: Energy Variables If any of the energy storage elements in a model have nonlinear constitutive equations, then power or circuit variables may be a poor choice for the state variables associated with those elements. This is because differentiating a nonlinear constitutive equation (in step 6 above) will not necessarily result in a function only of power variables and their rates of change. For example, consider a nonlinear capacitor: de dt = ?F ?q (q) dq dt = ?F ?q (q) f (6.10) Now it is necessary to substitute for the variable q as well as the variable f. One might try to do so by inverting the capacitor constitutive equation. de dt = ?F ?q ( ) F -1 (e) f (6.11) However, there are two problems with this approach: First of all, the required inverse function may not exist. Secondly, even if it does, the required algebra may be quite tedious. A better alternative is to choose different state variables: the displacements and momenta associated with independent energy storage elements — known as energy state variables or Hamiltonian state variables. Steps 5 and 6 of the substitution procedure are changed as follows. 5. Choose energy state variables. These are the displacements associated with independent capacitors and the momenta associated with independent inertias. The rate of change of each state variable is equal to the input variable to the corresponding independent energy storage element. 5a. independent capacitor: dq/dt = f 5b. independent inertia: dp/dt = e 6. Using its constitutive equation, write the output variable for each independent energy storage element as a function of the corresponding state variable. 6a. independent capacitor: e = u(q) 6b. independent inertia: f = +(p) The rest of the substitution procedure remains unchanged. Example: Nonlinear Electric Circuit Consider the electric circuit of figure 6.1a but assume that the capacitor and inductor have the following nonlinear constitutive equations. For the capacitor: 250 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 e C = 1 C q C q s q s 2 – q C 2 (6.12) where q s is the saturation charge, (a constant) the maximum charge which may be stored in the capacitor. Note that for q C << q s the constitutive equation reduces to that of an ideal (linear) capacitor. lim e C q C A0 = q C C (6.13) For the inductor: i L = 1 L ì L ì s ì s 2 – ì L 2 (6.14) where ì s is the saturation flux linkage, (a constant) the maximum flux linkge which may stored in the inductor. For ì L << ì s the constitutive equation reduces to that of an ideal (linear) inductor. lim i L l L A0 = ì L L (6.15) If we were to choose the capacitor voltage as a state variable, differentiating the constitutive equation would result in the following relation. d dt e C = \ | . | | q s C(q s 2 – q C 2 ) 1/2 + q C 2 q s C(q s 2 – q C 2 ) 3/2 i C (6.16) The nonlinear capacitor equation may be inverted as follows. q C = q s e C C q s 2 + e C 2 C 2 (6.17) Therefore, in this case, the charge q C may be eliminated from equation 6.16 and the capacitor voltage could be used as a state variable. The resulting state equation is a little intimidating: d dt e C = \ | . | | q s C \ | . | | q s 2 – q s 2 e C 2 C 2 q s 2 + e C 2 C 2 1/2 + \ | . | | q s 2 e C 2 C 2 q s 2 + e C 2 C 2 q s C \ | . | | q s 2 – q s 2 e C 2 C 2 q s 2 + e C 2 C 2 3/2 i C (6.18) 251 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 By a similar argument, the inductor current could also be used as a state variable. However, it is far simpler to use the capacitor charge and the inductor flux linkages as state variables. d dt q C = i C (6.19) d dt ì L = e L (6.20) Reading the junction equation from the causal graph (figure 6.1d) i C = i L . Using the nonlinear inductor constitutive equation (6.14) we obtain one state equation. dq C dt = 1 L ì L ì s ì s 2 – ì L 2 (6.21) As before, reading from the causal graph, the inductor voltage is determined by the other three one-port elements e L = e S – e C – e R (6.22) Substitute for e C and e R in equation 6.22 using the constitutive equations of the nonlinear capacitor (6.12) and the resistor (6.6). d dt ì L = e S – 1 C q C q s q s 2 – q C 2 – R i R (6.23) Reading the junction equation from the causal graph, the resistor current is determined by the inductor current i R = i L . Substituting using the inductor constitutive equation, the second state equation is: d dt ì L = e S – 1 C q C q s q s 2 – q C 2 – R ì L ì s ì s 2 – ì L 2 (6.24) As before, the substitution process stops when the rate of change of a state variable has been expressed as a function of state and input variables. Note that in this nonlinear system, there is no clear way to express the state equations in vector/matrix form. Energy variables may also be used for systems composed exclusively of linear elements. For this reason, energy variables have been proposed as the exclusive choice of state variables for systems 1 represented by bond graphs. However, there are several reasons why this is not recommended: 1 Karnopp, Dean and Rosenberg, Ronald (1975) System Dynamics: A Unified Approach, John Wiley & Sons, New York. 252 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 4 • physical system behavior is fundamentally independent of the choice of variables, so we should expect no universal rule for choosing state variables. • energy variables may needlessly complicate the equation derivation process. • energy variables may be less familiar and less comprehensible than the corresponding power variables. This last point may seem trivial; in fact, it is probably the most important. One of the primary reasons for developing models is to enhance understanding. For most of us, the current in an inductor is more meaningful than the corresponding flux linkage; the speed of a mass is more readily visualized than its momentum. 253 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 1 Models of Mechanical Systems: Lagrangian Variables For (sub)system models of phenomena in the domain of mechanical translation or rotation, neither power variables nor energy variables are necessarily the best choice. Mechanical inertias have linear constitutive equations (except at speeds approaching light speed — a rare occurence for systems of interest to a practicing engineer). Thus it is convenient to work with flows (speeds) of independent inertias as state variables. On the other hand, mechanical capacitors — elastic elements — frequently have nonlinear constitutive equations. Even in the linear case, rather than work with efforts, it is generally easier to identify displacements and relate their time derivatives to the speeds of associated inertias. For mechanical systems the most convenient state variables are the flows of independent inertias and the corresponding displacements — which may be termed Lagrangian state variables. An obvious hybrid of the above two procedures can be used to derive state equations. It is illustrated by the following simple example. Example: Linear Mechanical System rigid body spring rigid body spring x x 1 2 F applied force ²x 2 stationary frame Figure 6.2a A mechanical system schematic is shown in figure 6.2a. This schematic represents a model of the system and specifies the lumped-parameter energetic elements to be used. A corresponding bond graph is shown in figure 6.2b. It is evident that a single displacement or speed is common to the leftmost spring and inertia which are therefore connected by a one junction. Similarly, a single displacement or speed is common to the rightmost inertia and the source of the applied force which are therefore also connected by a one junction. The force in the remaining spring is applied to both inertias and a zero junction connects these three elements as shown. Note that a sign convention is chosen so that the displacement of the rightmost spring is determined by the difference of the displacements of the two inertias. 0 C 1 1 S e v 1 v 2 1/k 1 : : F I : m 1 C : 1/k 2 I : m 2 254 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 2 Figure 6.2b Assume all elements have linear constitutive equations. Causality is assigned following the procedure described above with the result shown in figure 6.2c. 0 C 1 1 S e v 1 v 2 1/k 1 : : F I : m 1 C : 1/k 2 I : m 2 Figure 6.2c From the causal graph, all energy storage elements are independent. Choose as state variables the speeds of the two inertias, v 1 and v 2 , and the corresponding displacements, x 1 and x 2 , shown in figure 6.2a. The first two state equations are: d dt x 1 = v 1 (6.25) d dt x 2 = v 2 (6.26) Subscripting variables in an obvious way, from the constitutive equations of the inertias we may write: d F 1 dt v 1 = (6.27) m 1 d F 2 dt v 2 = (6.28) m 2 Reading from the causal graph, the force on the leftmost inertia is determined by the two springs. F 1 = F k2 – F k1 (6.29) The constitutive equations of the springs relate these forces to the corresponding displacements. F k1 = k 1 x 1 (6.30) F k2 = k 2 Ax 2 (6.31) The displacement of the rightmost spring may be related to the positions of the inertias by reading directly from the causal graph. 255 Integrated Modeling of Physical System Dynamics © Neville Hogan 1994 page 3 Ax 2 = x 2 – x 1 (6.32) Substituting yields a third state equation. d 1 dt v 1 = m 1 [ k 2 (x 2 – x 1 ) – k 1 x 1] (6.33) Reading from the causal graph, the force on the rightmost inertia is determined by the associated spring and the applied force. F 2 = F – F k2 (6.34) Substituting the constitutive equation of the spring yields the fourth state equation. d 1 dt v 2 = m 2 [ F – k 2 (x 2 – x 1 ) ] (6.35) The state equations may be written in standard vector/matrix form as follows. x 1 x 2 v 1 v 2 ( ( ( ¸ = ¸ 0 0 1 0 0 0 0 1 –(k 1 +k 2 )/m 1 k 2 /m 1 0 0 k 2 /m 2 –k 2 /m 2 0 0 ( ( ( ¸ x 1 x 2 v 1 v 2 0 d dt ¸ ¸ ( ( ( ¸ + ( ( ( ¸1/m 2 ¸ 0 F (6.36) 0 256 December 6, 1994 page 1 Neville Hogan Ideal asymmetric junction elements Relax the symmetry assumption and examine the resulting junction structure. For simplicity, consider two-port junction elements. As before, assume instantaneous power transmission between the ports without storage or dissipation of energy. Characterize the power flow in and out of a two- port junction structure using four real-valued wave-scattering variables. Using vector notation: u = ¸ ¸ ( ( u 1 u 2 (A.1) v = ¸ ¸ ( ( v 1 v 2 (A.2) The input and output power flows are the square of the length of these vectors, their inner products. P in = _ i=1 2 u i 2 = u t u (A.3) P out = _ i=1 2 v i 2 = v t v (A.4) The constitutive equations of the junction structure may be written as follows. v = f(u) (A.5) Geometrically, the requirement that power in equal power out means that the length of the vector v must equal the length of the vector u, i.e. their tips must lie on the perimeter of a circle (see figure A.1). For any two particular values of u and v, the algebraic relation f( . ) is equivalent to a rotation operator. v = S(u) u (A.6) where the square matrix S is known as a scattering matrix. 257 December 6, 1994 page 2 Neville Hogan v 1 2 v u 1 u 2 u v S need not be a constant matrix, but may in general depend on the power flux through the junction, hence the notation S(u). However, S is subject to important restrictions. In particular, v t v = u t S t Su = u t u (A.7) S is ortho-normal matrix: the vectors formed by each of its rows (or columns) are (i) orthogonal and (ii) have unit magnitude; its transpose is its inverse. S t S = 1 (A.8) This constrains the coefficients of the scattering matrix as follows. S = ¸ ¸ ( ( a b c d (A.9) a 2 + c 2 = 1 (A.10) ab + cd = 0 (A.11) b 2 + d 2 = 1 (A.12) 258 December 6, 1994 page 3 Neville Hogan As there are only three independent equations and four unknown quantities, we see that this junction is characterized by a single parameter. We may also write the orthogonality condition as SS t = 1 (A.13) which yields the following equations. a 2 + b 2 = 1 (A.14) ac + bd = 0 (A.15) c 2 + d 2 = 1 (A.16) There are four possible solutions to these equations. Combining A.10 and A.16, a 2 = 1 - c 2 = d 2 . Thus a = ± d. If a = d then b =c = ± 1 - a 2 . One solution Choosing the positive root yields one solution. Assuming the coefficient a to be the undetermined parameter, S = ¸ ¸ ( ( ( a 1 - a 2 - 1 - a 2 a (A.17) Rewrite in terms of effort and flow variables. e = ¸ ¸ ( ( e 1 e 2 (A.18) f = ¸ ¸ ( ( f 1 f 2 (A.19) The relation between efforts and wave-scattering variables is as follows. e = (u - v) c = c (1 - S) u (A.20) where c is a scaling constant. The relation between flows and wave-scattering variables is as follows. 259 December 6, 1994 page 4 Neville Hogan f = (u + v)/c = 1/c (1 + S) u (A.21) If |a| = 1 then 1 + S and 1 - S are nonsingular matrices and the input wave scattering variables u 1 and u 2 may be eliminated as follows. e = c 2 (1 - S) (1 + S) -1 f (A.22) ¸ ¸ ( ( ( e 1 e 2 = c 2 ¸ ¸ ( ( ( 0 - (1 - a)/(1 + a) (1 - a)/(1 + a) 0 ¸ ¸ ( ( ( f 1 f 2 (A.23) Writing G = c 2 (1 - a)/(1 + a) we obtain the equation for an ideal gyrator. ¸ ¸ ( ( ( e 1 e 2 = ¸ ¸ ( ( ( 0 -G G 0 ¸ ¸ ( ( ( f 1 f 2 (A.24) Note that equations A.20 and A.21 imply a sign convention in effort-flow coordinates such that power is positive inwards on both ports. P net inwards = e t f = u t u - v t v (A.25) To follow the more common sign convention we may simply change the sign of f 2 in equation A.24. If a = 1, e is identically zero for all values of f. No energy is exchanged between the ports and the junction structure behaves like a dissipator with zero resistance. If a = -1, f is identically zero for all values of e. No energy is exchanged between the ports and the junction structure behaves like a dissipator with infinite resistance (zero conductance). 260 December 6, 1994 page 5 Neville Hogan A second solution Choosing a = d and using the negative root yields another solution. Again assuming the coefficient a to be the undetermined parameter, S = ¸ ¸ ( ( ( a - 1 - a 2 1 - a 2 a (A.26) In this case the relation between efforts and flows is ¸ ¸ ( ( ( e 1 e 2 = c 2 ¸ ¸ ( ( ( 0 (1 - a)/(1 + a) - (1 - a)/(1 + a) 0 ¸ ¸ ( ( ( f 1 f 2 (A.27) Again we obtain the equation for an ideal gyrator. ¸ ¸ ( ( ( e 1 e 2 = ¸ ¸ ( ( ( 0 G -G 0 ¸ ¸ ( ( ( f 1 f 2 (A.28) 261 December 6, 1994 page 6 Neville Hogan A third solution If a = - d, b = c = ± 1 - a 2 . Using the positive root and assuming a to be the undetermined parameter S = ¸ ¸ ( ( ( a 1 - a 2 1 - a 2 -a (A.29) In this case the matrices 1 + S and 1 - S are singular for all values of the parameter a. However, equations A.20 and A.21 may be combined as follows: ¸ ¸ ( ( ( e 1 /c e 2 /c cf 1 cf 2 = ¸ ¸ ( ( ( 1 - S ----- 1 + S ¸ ¸ ( ( ( u 1 u 2 (A.30) ¸ ¸ ( ( ( ( e 1 /c e 2 /c cf 1 cf 2 = ¸ ¸ ( ( ( ( 1 - a - 1 - a 2 - 1 - a 2 1 + a 1 + a 1 - a 2 1 - a 2 1 - a ¸ ¸ ( ( ( u 1 u 2 (A.31) If |a| = 1, the 4 x 2 matrix relating efforts and flows to the input scattering variables contains two nonsingular 2 x 2 submatrices. ¸ ¸ ( ( ( e 2 /c cf 1 = ¸ ¸ ( ( ( - 1 - a 2 1 + a 1 + a 1 - a 2 ¸ ¸ ( ( ( u 1 u 2 (A.32) ¸ ¸ ( ( ( e 1 /c cf 2 = ¸ ¸ ( ( ( 1 - a - 1 - a 2 1 - a 2 1 - a ¸ ¸ ( ( ( u 1 u 2 (A.33) Solving the second of these for u and substituting into the first we obtain a relation between efforts and flows. 262 December 6, 1994 page 7 Neville Hogan ¸ ¸ ( ( ( e 2 f 1 = ¸ ¸ ( ( ( - (1 + a)/(1 - a) 0 0 (1 + a)/(1 - a) ¸ ¸ ( ( ( e 1 f 2 (A.34) Writing T = (1 + a)/(1 - a) we obtain the equation for an ideal transformer. ¸ ¸ ( ( ( e 2 f 1 = ¸ ¸ ( ( ( -T 0 0 T ¸ ¸ ( ( ( e 1 f 2 (A.35) To follow the more common sign convention we may change the sign of e 2 . If the parameter a = ± 1, an argument similar to that used above shows that a degenerate case results in which no energy is exchanged between the ports. 263 December 6, 1994 page 8 Neville Hogan Final solution Choosing a = d and using the negative root we obtain the fourth solution. S = ¸ ¸ ( ( ( a - 1 - a 2 - 1 - a 2 -a (A.36) Once again, the matrices 1 + S and 1 - S are singular for all values of the parameter a, but by rearranging equations A.20 and A.21 as before the corresponding relation between efforts and flows is ¸ ¸ ( ( ( e 2 f 1 = ¸ ¸ ( ( ( (1 + a)/(1 - a) 0 0 - (1 + a)/(1 - a) ¸ ¸ ( ( ( e 1 f 2 (A.37) Again we obtain the equation for an ideal transformer ¸ ¸ ( ( ( e 2 f 1 = ¸ ¸ ( ( ( T 0 0 -T ¸ ¸ ( ( ( e 1 f 2 (A.38) 264 December 6, 1994 page 9 Neville Hogan Two-port junction elements There are only two possible power-continuous, asymmetric two-port junction elements, the gyrator and the transformer. Unlike the ideal symmetric junction elements (0 and 1) the ideal asymmetric junction elements may be nonlinear. The relation between efforts and flows must have a multiplicative form. The general asymmetric junction elements are a modulated gyrator (MGY) and a modulated transformer (MTF) respectively. 265 Junction elements in network models. Classify by number of ports and examine the possible structures that result. Using only one-port elements, no more than two elements can be assembled. Combining two two-ports yields another two-port. At most two one-port elements could be connected. Thus three-port junction elements are necessary. They are also sufficient insofar as junctions with more than three ports may be constructed from assemblages of three-ports. An assembly of two three-port elements yields a four-port element. Using two- and three-port junctions, an arbitrary number of one-port (or n-port) elements may be assembled into a system. Three-port junction elements A junction structure or junction element is treated as a system with certain special properties which are derived from energetic considerations. December 6, 1994 page 1 Neville Hogan 266 An ideal junction structure is power-continuous: power is transmitted instantaneously without storing or dissipating energy. instantaneous power out equals instantaneous power in. Power flow in and out of a three-port may be characterized by six real-valued wave-scattering variables u 1 , u 2 , u 3 , define the input power flow v 1 , v 2 , v 3 , define the output power flow Using vector notation: u = ¸ ¸ ( ( u 1 u 2 u 3 (A.1) v = ¸ ¸ ( ( v 1 v 2 v 3 (A.2) The input and output power flows are the square of the length of these vectors, their inner products. P in = _ i=1 3 u i 2 = u t u (A.3) P out = _ i=1 3 v i 2 = v t v (A.4) December 6, 1994 page 2 Neville Hogan 267 Constitutive equations of the junction structure may be written in vector form as an algebraic relation between input and output scattering variables. v = f(u) (A.5) This algebraic relation is constrained by the requirement that power in equal power out. Geometrically, the length of the vector v must equal the length of the vector u. The input and output vectors are both confined to the surface of a sphere. v 1 3 v 2 v u 1 u 2 u 3 u v Consequently, for any two particular values of u and v, the algebraic relation f( . ) is equivalent to a rotation operator and equation A.5 may be re-written as follows. v = S(u) u (A.6) where the square matrix S is known as a scattering matrix. December 6, 1994 page 3 Neville Hogan 268 S need not be a constant matrix, but may in general depend on the power flux through the junction, hence the notation S(u). However, S is subject to important restrictions. In particular, v t v = u t S t Su = u t u (A.7) S is ortho-normal: the vectors formed by each of its rows (or columns) are (i) orthogonal and (ii) have unit magnitude; its transpose is its inverse. S t S = 1 (A.8) (This is simply a mathematical statement of the conclusion that S is a rotation operator.) December 6, 1994 page 4 Neville Hogan 269 Symmetric three-port junction elements To evaluate S, we assume an additional symmetry property in the mathematical sense of invariance under the action of a group operator. Specifically, assume the constitutive equation is invariant under the operation of permuting or exchanging the ports of the junction structure. In terms of the coefficients of the scattering matrix S: S = ¸ ¸ ( ( ( a b b b a b b b a (A.9) The power output along any port depends upon the power input along all three ports, itself and its two neighbors. Each of the neighboring ports contribute the same fraction, b, of their input power That fraction may be different from the portion, a, of the power input which is "reflected" back out the same port. Each of the three ports is the same in this regard. Using equation A.9 in equation A.8 yields three equations for the coefficients a and b, but only two are distinct. a 2 + 2b 2 = 1 (A.10) b 2 + 2ab = 0 (A.11) These equations admit only two solutions, a = 1/3; b = -2/3 (A.12) or a = -1/3; b = 2/3 (A.13) so the symmetry condition constrains the scattering matrix to be a constant. There are only two possible symmetric, power-continuous, three-port junctions. They are both linear. December 6, 1994 page 5 Neville Hogan 270 Transform back to effort and flow variables. e = ¸ ¸ ( ( e 1 e 2 e 3 (A.14) f = ¸ ¸ ( ( f 1 f 2 f 3 (A.15) The relation between efforts and wave-scattering variables is as follows. e = (u - v) c = c (1 - S) u (A.16) where c is a scaling constant. Using equation A.12 for the value of S S = ¸ ¸ ( ( ( 1/3 -2/3 -2/3 -2/3 1/3 -2/3 -2/3 -2/3 1/3 (A.17) it can be seen that e = c ¸ ¸ ( ( ( 2/3 2/3 2/3 2/3 2/3 2/3 2/3 2/3 2/3 u (A.18) The efforts on each of the ports are equal. This is a common effort or type zero junction. e 1 = e 2 = e 3 (A.19) December 6, 1994 page 6 Neville Hogan 271 The relation between flows and wave-scattering variables is as follows. f = (u + v)/c = 1/c (1 + S) u (A.20) Using the same value for S f = 1 c ¸ ¸ ( ( ( 4/3 -2/3 -2/3 -2/3 4/3 -2/3 -2/3 -2/3 4/3 u (A.21) Summing down the columns, it can be seen that the relation between flows on the ports is f 1 + f 2 + f 3 = 0 (A.22) This is a continuity equation 1 . It is a derived property of the zero junction. 1 Note that power on all ports of the junction has been defined to be positive inwards. December 6, 1994 page 7 Neville Hogan 272 Using the other of the two values of S (equation A.13). S = ¸ ¸ ( ( ( -1/3 2/3 2/3 2/3 -1/3 2/3 2/3 2/3 -1/3 (A.23) From equation A.20 we find f = 1 c ¸ ¸ ( ( ( 2/3 2/3 2/3 2/3 2/3 2/3 2/3 2/3 2/3 u (A.24) The flows on each of the ports of this junction are equal. This is a common flow or type one junction. f 1 = f 2 = f 3 (A.25) Using equation A.16 e = c ¸ ¸ ( ( ( 4/3 -2/3 -2/3 -2/3 4/3 -2/3 -2/3 -2/3 4/3 u (A.26) Summing down the columns, it can be seen that the relation between efforts is e 1 + e 2 + e 3 = 0 (A.27) This is a generalized compatibility equation. It is a derived property of the zero junction. Equating effort and voltage, flow with current, we have derived Kirchhoff's current and voltage laws from energetic "first principles". We have also shown that an exactly analogous pair of equations can be derived for any of the energetic media and that they must always be linear. December 6, 1994 page 8 Neville Hogan 273 Conservation principles and symmetric junctions In terms of efforts and flows, power continuity constrains the constitutive equations for a junction as follows. _ e i f i = 0 (A.28) The bond graph zero-junction is associated with a generalization of Kirchhoff's node (current) law, a statement of flow (current) continuity that constrains the constitutive equations for a zero junction as follows. _ f i = 0 (A.29) But these two constraints are not sufficient to define the zero junction . A junction element known as a circulator (used in microwave circuit theory) satisfies the requirements of power continuity (equation A.28) and current continuity (equation A.29), but it is quite distinct from a zero junction. The defining property of a zero junction is its symmetry — the effort is invariant under the operation of permuting or exchanging the ports of the junction structure. e i = e j ¬ i, j (A.30) Combined with this condition, either of the previous two constraints implies the other. Thus, suppressing the subscript on the common effort, the power continuity condition may be written as follows. e _ f i = 0 (A.31) If this relation is to be identically true for all values of the (common) effort, the flow continuity condition must be satisfied. A similar argument holds for the one-junction. Thus symmetry is the fundamental property, continuity or compatibility are not. December 6, 1994 page 9 Neville Hogan 274 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.151: Advanced System Dynamics and Control –Spring 2003 QUIZ 1 This quiz is closed-book. You have 1.5 hours to complete it. 275 Problem 1(30pts): For each of the following bond-graphs, determine the system’s minimum order. For the first three systems, determine whether the system could exhibit oscillatory (resonant) behavior. 1 I C 0 C 1 Se GY R R C C 0 I 1 Sf TF R C 1 0 I I C I 1 Sf R 1 0 I 0 C I 1 Se 0 GY TF TF 1 C Se 276 Problem 2(30pts): For each of the following systems, draw a corresponding bond-graph and determine the system’s minimum order. Assume all elements have linear constitutive relations. For the first three systems, choose appropriate state variables and write state equations. k b m k b m F(t) Assume the cylinder (with inertia I) rolls without slipping. F(t) R k I Assume no leaking, and incompressible fluid flow with density ρ. F(t) Ap d l Vs(t) + - R R R R R 277 Problem 3: Typical examples of common linear drives are described in the attached pages taken from the Warner Electric catalog. Consider the in-line configuration shown at the bottom of the page. A permanent-magnet DC motor is used to drive a recirculating-ball screw (a nut and screw with ball-bearings interposed between the nut and screw so that frictional energy losses are minimized). The motor shaft is connected directly to the screw. The nut is prevented from rotating but may slide as the screw rotates within it. The screw has 5 threads per inch with a pitch diameter of 0.5 inches. The screw may be considered to be a steel cylinder 18 inches long with diameter equal to the screw pitch diameter. The nut may be considered to be a steel cylinder 3 inches long, outside diameter 3 inches, with a hole though its center equal to the pitch diameter of the screw. To keep things simple, ignore all frictional losses due to the recirculating-ball screw and the sliding nut. Parameters of the motor are given in the table below. Excerpt from specification sheet of a direct-current permanent-magnet motor MOTOR CONSTANTS: INTRINSIC (AT 25 DEG C) SYMBOL UNITS TORQUE CONSTANT KT OZ IN/AMP 5.03 BACK EMF CONSTANT KE VOLTS/KRPM 3.72 TERMINAL RESISTANCE RT OHMS 1.400 ARMATURE RESISTANCE RA OHMS 1.120 VISCOUS DAMPING CONSTANT KD OZ IN/KRPM 0.59 MOMENT OF INERTIA JM OZ IN SEC-SEC 0.0028 ARMATURE INDUCTANCE L MICRO HENRY <100.0 TEMPERATURE COEFFICIENT OF KE C %/DEG C RISE -0.02 a. First assume the DC motor is driven by a current-controlled amplifier (i.e., the current applied to the electrical terminals of the motor may be specified independent of the voltage required). Assuming all model elements have linear constitutive equations, develop a model competent to describe the linear displacement of the rod end in response to motor current input. Express your model as a bond graph, clearly identifying the different energy domains. b. What is the order of this model? c. Next assume the DC motor is driven by a voltage-controlled amplifier (i.e., the voltage applied to the electrical terminals of the motor may be specified independent of the current required). Assuming all model elements have linear constitutive equations, develop a model competent to describe the linear displacement of the rod end in response to motor voltage input. Express your model as a bond graph, clearly identifying the different energy domains. d. What is the order of this model? e. Suppose an external force is applied to the rod end. Numerically evaluate the total apparent translational inertia opposing that force (i.e., the equivalent mass “seen at” the rod end due to everything that moves). <Pages from Warner Electric catalog referenced above have been removed due to copyright considerations.> 278 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.151: Advanced System Dynamics and Control –Spring 2003 QUIZ 1 Problem 1(30pts): Two possible bond graphs are shown below. The minimum order is two. Though this system has a capacitance and inertance in integral causality, the gyrator does not allow oscillations. 1 I C 0 C 1 Se GY R R 1 I C 0 C 1 Se GY R R Two possible bond graphs are shown below. This system has a minimum order of five. It can oscillate. C C 0 I 1 Sf TF R C 1 0 I I C C 0 I 1 Sf TF R C 1 0 I I This system has a minimum order of two. It can oscillate. C I 1 Sf R 1 0 I 0 279 This system has a minimum order of two. C I 1 Se 0 GY TF TF 1 C Se C I 1 Se 0 GY TF TF 1 C Se Problem 2(30pts): k b m k b m F(t) 1 I2 C2 0 C1 1 Se R R 1 I1 We see from the bond graph the system is at least of fourth order, with states associated with the two springs and two masses. Choosing the displacement across both springs and the momentum of the two masses, our state equations are, d dt     x 1 x 2 p 1 p 2     =     0 0 1/m −1/m 0 0 0 1/m −k 0 −b/m b/m k −k b/m −2b/m         x 1 x 2 p 1 p 2     +     0 0 1 0     F(t) (1) If the force across the springs and the velocity of the masses is our state, then we have, d dt     F 1 F 2 ˙ x 1 ˙ x 2     =     0 0 k −k 0 0 0 k −1/m 0 −b/m b/m 1/m −1/m b/m −2b/m         F 1 F 2 ˙ x 1 ˙ x 2     +     0 0 1/m 0     F(t) (2) 280 F(t) R k I 1 I C 1 0 Se TF R II For the bond graph, not that the velocity of the cylinder can be related to the velocity of the rack, v and the angular velocity, ω, as follows, v m = v −rω (3) This relation indicates the need for two distinct flows and a zero junction. Choosing the spring displacement, the cylinder velocity and angular velocity, the state equations are, d dt   x c v m ω   =   0 1 0 −k/m 0 0 0 0 0     x c v ω   +   0 1/m r/I   F(t) (4) F(t) Ap d l 1 R Se TF I The bond graph above is an adequate description of the system. The fluid resistance can be a sum of the resistance through the pipe and the resistance due to the exit restriction. Choosing the volume flow rate as our state we get the following, ˙ Q = −RQ + A p I F(t) where I = 4ρL πd 2 and R = 32µL πd 2 + R restriciton (5) 281 The bond graph below indicates the system is zeroth order. Vs(t) + - R R R R R 1 0 1 Se R R 1 0 0 R R R Problem 3(40pts): a), b) A possible bond graph for the system driven with a current source is shown in the figure below. Note that including damping in the screw’s motion, or resistance in the motor’s circuit does not affect the minimum order, which is one. However, in order to describe the displacement of the nut, we need to add a second variable in our system description, raising the order to two. c), d) A possible bond graph for the system driven with a voltage source is shown below. Again, including resistive type elements does not change the minimum order, which is now three when we include the displacement of the nut. e) The total apparent inertia, seen through the rod’s end, is a sum of the nut’s and extension tube’s mass with the inertia of the screw and the motor armature as “seen” through the transformer. This effective inertia is, I effective = m nut + m tube + 1 T 2 (J screw + J armature ) (6) where T is the transformer modulus for the screw-nut interface. T = 1in 5threads 1thread rev 1rev 2πradians = 1ft 120π (7) Then nut dimensions are given in the problem statement. To calculate the extension tube mass, we’ll assume a length of one foot, and radii of 3in and 2.9in. m nut = ρπ(r 2 o −r 2 i )L = (487lb/ft 3 )π(3 2 −.5 2 )in 2 ( ft 2 144in 2 )3in( 1in 12ft ) = 23.2lb (8) m tube = ρπ(r 2 o −r 2 i )L = (487lb/ft 3 )π(3 2 −2.9 2 )in 2 ( ft 2 144in 2 )1ft = 6.3lb (9) J screw = 1 2 mr 2 = 1 2 π(487lb/ft)18in( 1ft 12in )(.5in) 4 ( 1ft 12in ) 4 = 3.5 ×10 −3 lbft 2 (10) J armature = 0.0028ozins 2 ( 1lb 16oz )( ft 12in )( 32.2ft s 3 ) = 4.7 ×10 −4 lbft 2 (11) I effective = 23.2lb + 6.3lb + 497.4lb + 66.8lb = 593.6lb (12) Note that the inertia of the screw and extension tube combined is only 5% of the total inertia “seen” at the rod end. 282 1 I I e e I e 1 Sf GY R R TF 1 1 CTRIC I C IC R T TI I C IC TR S TI I 1 I I e e I e 1 Se GY R R TF 1 1 CTRIC I C IC R T TI I C IC TR S TI I 283