Model-Based Design, Analysis, & Control: Valve-Controlled Hydraulic System K. Craig 1

Transcription

1 Model-Based Design, Analysis, & Control: K. Craig 1

2 K. Craig

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6 Mission: It s All About Process Dynamic System Investigation K. Craig 6

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12 References Academic: Fluid Power Fundamentals of Fluid Power and Control, J. Watton, Cambridge, 009. Hydraulic Control Systems, H. Merritt, Wiley, Hydraulic Control Systems, N. Manring, Wiley, 005. Modeling, Monitoring, and Diagnostic Techniques for Fluid Power Systems, J. Watton, Springer, 007. Control of Fluid Power: Analysis and Design, D. McCloy and H.R. Martin, nd Edition, Ellis Horwood, K. Craig 1

13 Industry: Fluid Power Industrial Hydraulics Manual, Eaton Corp., 010. Electrohydraulic Proportional and Control Systems, Bosch Automation, Electrohydraulic Proportional Valves and Closed Loop Control Valves, Bosch Automation, Closed Loop Electrohydraulic Systems Manual, Vickers, Inc., Basic Electronics for Hydraulic Engineers, Eaton Corp., Hydraulics, Festo Didactic, Electrohydraulics, Festo Didactic, K. Craig 13

14 Academic: System Dynamics and Control Introduction to System Dynamics, J. Shearer, A. Murphy, and H. Richardson, Addison-Wesley, Dynamic Modeling and Control of Engineering Systems, J. Shearer, B, Kulakowski, and J. Gardner, nd Edition, Prentice Hall, Modeling, Analysis, and Control of Dynamic Systems, W.J. Palm, nd Edition, Wiley, Mechatronics, S. Cetinkunt, Wiley, 007. Introduction to Fluid Mechanics, R. Fox and A. McDonald, 3 rd Edition, Wiley, Sensors and Actuators, Clarence de Silva, CRC Press, 007. K. Craig 14

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17 Physical Modeling Assumptions Supply pressure p s is constant. Reservoir pressure is constant at 0 psig. Valve is zero lap. Actuator pressures p cl and p cr each come to p s / at the servo rest condition. Neglect inertia of the fluid. Cylinder and piston are rigid. Sensor dynamics are negligible. Parameters are constant. K. Craig 17

18 Compressibility effects are neglected in the orifice flow equations, but not in the cylinder equations as pressures can be high during acceleration and deceleration periods and oil compressibility can have a destabilizing effect. Both flow orifices are identical, i.e., the flow and pressure coefficients are identical for both. Load force is zero. Spring force is zero. Time delays are neglected. K. Craig 18

19 Physical Model Parameters x u p s C d 0 inches, zero lap condition 1000 psig (constant), supply pressure 0.6, orifice discharge coefficient w 0.5 in, valve port width 7.8E-5 lbf-s /in 4, fluid density A p.0 in, piston area 100,000 psi, bulk modulus of fluid M 0.03 lbf-s /in, mass K pl in 3 /s-psi, piston leakage coefficient B 100 lbf-s/in, viscous damping coefficient K 0 lbf/in, spring constant V l0 4.0 in 3, volume at operating point of left cylinder V r0 4.0 in 3, volume at operating point of right cylinder p cl0 500 psi, initial pressure of left cylinder p cr0 500 psi, initial pressure of right cylinder x C0 0 in, initial displacement of mass x C0 0 in/sec, initial velocity of mass F L 0 lbf, load force 0 psig, return pressure p r K. Craig 19

20 Nonlinear Mathematical Model Equations for the orifice volume flow rates Q cl and Q cr for the left and right ends of the cylinder Equations for conservation of mass (continuity equation) for the left and right ends of the cylinder Newton s nd Law applied to the moving mass Load position feedback control equation with simple proportional control. K. Craig 0

21 Orifice Flow-Rate Equations x u positive: valve underlap x u negative: valve overlap x u zero: valve zero lap x v is displacement of valve spool (ps p cl ) Qcl Cdw(x u x v ) valid when (x u + x v ) is 0. This is flow into the left cylinder. (pcl p r ) Qcl Cdw(x u x v ) valid when (x u x v ) is > 0. This is flow out of the left cylinder. (pcr p r ) Qcr Cdw(x u x v ) valid when (x u x v ) is 0. This is flow out of the right cylinder. (ps p cr ) Qcr Cdw(x u x v ) valid when (x u x v ) is > 0. This is flow into the right cylinder. K. Craig 1

22 Conservation of Mass Equations Conservation of Mass The net rate of mass efflux through the control surface plus the rate of change of mass inside the control volume equals zero. Velocity is measured relative to the control volume. 0 dv v da t CV CS 0 V V Q CV CV CV CV net V CV 0 VCV Qnet Here we assume that all of the densities of the system (inlet flow, outlet flow, and control volume) are the same and equal to. K. Craig

23 This assumption is justified for incompressible fluids and is quite accurate for compressible fluids if pressure variations are not too large and the temperature of flow into the control volume is almost equal to the temperature of the flow out of the control volume. The equation of state for fluid is: (P,T) P,T P P T T P T P,T P,T P P0 T T0 Conservation of Mass can be written as: V 0 V PCV Q net dp P dt 0 P K. Craig 3

24 Evaluating terms: Left cylinder Q Q K (p p ) net cl pl cl cr C V Ap dt V P l0 p C cl The resulting equations for the left and right cylinders are: (V A x ) dp dx Q K (p p ) A dt dt r0 p C cr C cr pl cl cr p (V A x ) dp dx Q K (p p ) A dt dt l0 p C cl C cl pl cl cr p CV dx (V A x ) dp dt K. Craig 4

25 Newton s nd Law The sum of all forces acting on a non-accelerating control volume equals the rate of change of momentum inside the control volume plus the net rate of efflux of momentum through the control surface. F FS FB vdv vv da t All velocities are measured relative to the control volume. All time derivatives are measured relative to the control volume. CV dxc d xc cl cr p C L (p p )A B Kx F M dt CS dt K. Craig 5

26 Simulink Block Diagrams ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM Qcl input Qcr Xv Step Command Xv Command Xc Xv - Xc Positive Xv - Xc Negative Controller Xv - Xc Positive Xv - Xc Negative Pcl Xv - Xc Pos Xv - Xc Neg Pcr Flow Qcl Qcr Qcl Qcr Xc Pcl Pcl Xc Pcr Xc dot Mass Xc Xc dot Xc Qcl Qcr Xc dot Cylinder Pcl Pcr Pcl Pcr Pcr K. Craig 6

27 CONTROLLER SUBSYSTEM 1 Xv Command Sum Xv - Xc Kc Control Gain 0 Zero Switch1 1 Xv - Xc Positive Xc Switch Xv - Xc Negative 1 MASS SUBSYSTEM Pcl Pcr Sum3 Ap Gain 0 Disturbance fu Sum 1/M Gain B Gain5 1/s Integrator Xc dot 1/s Integrator1 1 Xc Xc dot K. Craig 7

28 Ps Supply P FLOW SUBSYSTEM 1 Pcl Sum /rho Gain4 sqrt(u) Fcn Cd*w Gain Product 1 /rho sqrt(u) Cd*w Sum Qcl Sum3 Gain5 Fcn1 Gain1 Product1 Xv - Xc Pos 3 Xv - Xc Neg Sum4 /rho Gain6 sqrt(u) Fcn -Cd*w Gain Product Sum1 Qcr 4 Pcr Sum5 Pr Reservoir P /rho Gain7 sqrt(u) Fcn3 -Cd*w Gain3 Product3 K. Craig 8

29 Vlo Constant1 1/MB 1/u 1 Xc Ap Gain Sum Gain1 Fcn CYLINDER SUBSYSTEM 1/MB 1/u Vro Constant Sum1 Gain Fcn1 1 Pcl Qcl Kpl Sum Product 1/s Integrator Gain3 Sum4 3 Qcr Sum3 Product1 1/s Integrator1 4 Xc dot Ap Gain4 Pcr K. Craig 9

30 MatLab File of Constants and Expressions M = 0.03; B=100; Ap =.0; Kpl = 0.001; Vlo = 4.0; Vro =4.0; MB=100000; Pr = 0; Ps=1000; rho = 7.8e-5; Cd = 0.6; w = 0.5; Pclo = 500; Pcro = 500; Xcdoto = 0; Xco = 0; Vo = 4.0; Cx = ; Cp = 0; A=[-(Cp+Kpl)*MB/Vo Kpl* MB /Vo 0 -Ap* MB/Vo; Kpl*MB/Vo -(Cp+Kpl)* MB/Vo 0 Ap* MB/Vo; ;Ap/M -Ap/M 0 -B/M]; B1=[MB*Cx/Vo 0;- MB*Cx/Vo 0;0 0;0 1/M]; C=[ ; ; ; ]; D=[0 0;0 0;0 0;0 0]; K=(*Cx*Ap)/(*Ap^+B*(Cp+*Kpl)); omegan=sqrt((mb*(*ap^+b*(cp+*kpl)))/(m*vo)); zeta=(b+((*mb*m)/vo)*kpl+((mb*m)/vo)*cp)/(*sqrt(((mb*m)/vo) *(*Ap^+B*(Cp+*Kpl)))); XcXvNum=K*omegan^; XcXvDen=[1 *zeta*omegan omegan^ 0]; %[XcXvNumCL,XcXvDenCL]=cloop(XcXvNum,XcXvDen,-1); Kc=1; XcXvNumCL=Kc*K*omegan^; XcXvDenCL=[1 *zeta*omegan omegan^ Kc*K*omegan^]; K. Craig 30

31 Linear System Analysis Restrict the analysis to small perturbations around a chosen operating point. A linearized approximate model may be obtained that provides many useful results. Valve flow equations can be thought of as relations between a dependent variable (flow rate) and two independent variables (spool motion and cylinder pressure) and thus can be linearized about any desired operating point. K. Craig 31

32 Q Q Q Q x p v v v v,0 v,p c,p xv p operating point c operating point flow gain = C pressure coefficient = C x p Q x v v Q p operating point v c operating point Qv Qv,0 Cxxv,p Cppc,p K. Craig 3

33 Linearized Set of Equations V dp C x C p K p p A dt 0 cl,p x v,p p cl,p pl cl,p cr,p p dx dt C,p V dp C x C p K p p A dt 0 cr,p x v,p p cr,p pl cl,p cr,p p dx dt C,p dx C,p d xc,p cl,p cr,p p U,p p p A B f M dt dt K. Craig 33

34 Simulink Block Diagram of the Linear System Pcl_l Cp Sum5 Sum Sum3 MB/Vo Gain MB/Vo Gain3 1/s Pcl Integrator Pcr 1/s Integrator3 Gain6 Sum Cp Gain7 Ap Gain4 Kpl Gain5 500 Constant1 Sum7 Qcl_l Pcr_l Ap Sum6 Gain9 Sum8 input_l 0 1/M 1/s 1/s Xc Qcr_l Sum4 Cx Gain8 Disturbance fu Sum1 Gain B Integrator Integrator1 Xc_l Gain1 Xv Step Command ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) K. Craig 34

35 If we take the Laplace Transform of these equations, we can derive six useful transfer functions relating the two inputs, x v and F L, to the three outputs, p cl, p cr, and x C. V s K C K A s 0 pl p pl p C C C K x x x cl v V pl 0s Kpl Cp Aps p cr x v x x x x F p p C C C A A Ms Bs p x C L K. Craig 35

36 One of these transfer functions is: K n x p p pl 0 C A A B C K A B C K MV p p p pl M M B Kpl C V 0 V 0 M A p B Cp K pl V 0 x s K s 1 C x v s s n n p K. Craig 36

37 Simulation Results: Step Command x V = 0.0 in. applied at t = sec x V + x v x C - CONTROLLER K c PLANT G(s) x K G(s) K K x 1 K G(s) s s s K K C c c n 3 V c n n c n K. Craig 37

38 Nonlinear and Linear Simulation Results: x C vs. time x C (in) solid: nonlinear dashed: linear time (sec) K. Craig 38

39 Nonlinear and Linear Simulation Results: Q cl vs. time 5 0 Q cl (in 3 /sec) 15 solid: nonlinear dashed: linear time (sec) K. Craig 39

40 Nonlinear and Linear Simulation Results: p cl vs. time p cl (psig) solid: nonlinear dashed: linear time (sec) K. Craig 40

41 Phase (deg); Magnitude (db) Open-Loop Frequency Response Plots with K c = 1 Bode Diagrams Frequency (rad/sec) K. Craig 41

42 Phase (deg); Magnitude (db) GM = 16. db = 6.46 PM = 74.8 Bode Diagrams Gm=16. db (at rad/sec), Pm= deg. (at rad/sec) Frequency (rad/sec) K. Craig 4

43 Phase (deg); Magnitude (db) Closed-Loop Frequency Response Plots with K c = 1 Closed-Loop Bandwidth = 13 Hz = 774 rad/sec Bode Diagrams At 774 rad/sec: Mag = Phase = Frequency (rad/sec) K. Craig 43

44 Simulink Block Diagram: Nonlinear Control with Time Delay input_l_nl Cx Gain Cp Gain1 Xv Step Command Sum1 Qcl_l_nl Sum3 Pcl_l_nl Delay.005 Sum Sign Gain4 Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux 500 Constant Pcr_l_nl 0 Distrbance fu Xc_l_nl Sum4 Clock time_l_nl Cx Gain Sum Cp Gain3 Qcr_l_nl ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) with Nonlinear On-Off Controller and Time Delay K. Craig 44

45 Curve A: Gain = 0.01, Delay = sec Curve B: Gain = 0.005, Delay = sec Curve C: Gain = 0.005, Delay = 0 sec x c (in) A C 0.01 B time (sec) K. Craig 45

46 Simulink Block Diagram: Linear System with Proportional Control Cx Cp Clock time_l_p input_l_p Gain Gain1 Sum1 Qcl_l_p Sum3 Pcl_l_p 500 Xv Step Command Sum Kc Controller Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux Constant Pcr_l_p 0 Disturbance fu Xc_l_p Sum4 Cx Gain Sum Cp Gain3 Qcr_l_p ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) With Proportional Control K. Craig 46

47 Imag Axis Root Locus Plot Pt. #1: K c = 6.46 Pt. #: K c = 3.70 Pt. #3: K c = 1.36 Pt. #4: K c = Real Axis K. Craig 47

48 Closed-Loop Time Response (Step) Plots K c = x C (in) K c = 1.36 K c = time (sec) K. Craig 48

49 Phase (deg); Magnitude (db) Closed-Loop Frequency Response Plots Bode Diagrams K c = K c =1.0 K c = K c =1.0 K c =3.70 K c = Frequency (rad/sec) K. Craig 49

50 Imaginary Axis Nyquist Diagram: K c = 1.0 Nyquist Diagrams Real Axis K. Craig 50

51 Imaginary Axis Nyquist Diagram: K c = 6.46 Nyquist Diagrams Real Axis K. Craig 51

52 Imaginary Axis Nyquist Diagram: K c = 10.0 Nyquist Diagrams Real Axis K. Craig 5

53 Stability Considerations Closed-Loop Transfer Function x K G(s) K K x 1 K G(s) s s s K K C c c n 3 V c n n c n K n x p p pl 0 C A A B C K A B C K MV p p p pl M M B Kpl C V 0 V 0 M A p B Cp K pl V 0 p K. Craig 53

54 Neglect leakage (K pl = 0) and consider the load as mainly inertia (B = 0, friction is ignored). The closedloop transfer function becomes: KC c x x A C p xv V0 M 3 MCp KcCx s s s A A A p p p Since the bulk modulus of the fluid is defined as: P The combined stiffness k 0 V / V0 of the two columns of fluid is: k 0 A V 0 p K. Craig 54

55 The valve stiffness k v is defined as: C C x p Q x v v Q p operating point v c operating point k v C x Ap C p The closed-loop transfer function can now be written as: KC c x x A C p x M V 3 CxM KC c x s s s k A k A 0 p v p K. Craig 55

56 Applying the Routh Stability Criterion to the characteristic equation of the closed-loop transfer function gives the relationship for stability as: k k 0 v In other words, the stiffness of the oil column must be greater than the effective valve stiffness if stability is to be satisfactory. K. Craig 56