Mechatronics Exercise: Modeling, Analysis, & Control of an Electrohydraulic Valve-Controlled Servomechanism

Transcription

1 Mechatronics Exercise: Modeling, Analysis, & Control of an Electrohydraulic Valve-Controlled Servomechanism 11

2 Introduction Although a wide variety of detailed hydraulic control schemes are in use, a useful classification is that of pump-controlled versus valve-controlled systems. Pump-controlled systems are usually relatively high power (above 1 or 2 hp) applications, where efficiency is economically significant and response speed requirements are modest (less than 1 Hz frequency response). 111

3 Valve-controlled systems are faster but are generally quite inefficient. For a low-power system, inefficiency has little economic impact. For fast high-power systems where speed specifications can only be met by valve control, the economic cost of low efficiency must be accepted. In pump-controlled systems, the fluid power supply must be included in the system model, while the analysis of valve-controlled systems can proceed without consideration of power supply details if one assumes the existence of a power source of constant supply pressure, p s, irrespective of flow demand. Power supplies that approximate this behavior are available in several different forms that trade off complexity, cost, efficiency, and static/dynamic pressure regulation accuracy. 112

4 For example, the spring-loaded relief valve is completely shut until pressure reaches the low end of the regulating range, whereupon it opens sufficiently to bypass any pump flow not required by the servovalve. The fluid power of the bypassed flow is completely lost and converted to heat. pump flow 113

5 When the servo system requires no flow, all the pumpgenerated power is converted to heat, giving zero efficiency. Supply pressure p s varies by about ± 3% over the regulating range for steady conditions and response to transient flow requirements is quite fast. Pump size must be chosen to match the largest anticipated servo-system demand. Thus the standard assumption of constant p s used in servo-system analysis is usually reasonable. 114

6 The figure below shows a flight simulator, a sophisticated application of valve-controlled servos where the motions of six actuators are coordinated to provide roll, pitch, and yaw rotary motions plus x, y, z translation. 115

7 Physical System: Valve-Controlled Servo 116

8 Real valves always exhibit either underlap (x u > ) or overlap (x u < ) behavior. Underlap / Overlap effects are usually very small. spool valve 117

9 Physical Modeling Assumptions supply pressure is constant at 1 psig reservoir pressure is constant at psig valve is zero lap actuator pressures p cl and p cr each come to p s /2 at the servo rest condition neglect inertia of the fluid cylinder and piston are rigid sensor dynamics are negligible parameters are constant 118

10 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 disturbance to the mass is zero 119

11 f U p r Physical Model Parameters x u inches, zero lap condition p s 1 psig (constant), supply pressure C d.6, orifice discharge coefficient w.5 in, valve port width ρ 7.8E-5 lbf-s 2 /in 4, fluid density A p 2. in 2, piston area β 1, psi, bulk modulus of fluid M.3 lbf-s 2 /in, mass K pl.1 in 3 /s-psi, piston leakage coefficient B 1 lbf-s/in, viscous damping coefficient V l 4. in 3, volume at operating point of left cylinder V r 4. in 3, volume at operating point of right cylinder p cl 5 psi, initial pressure of left cylinder p cr 5 psi, initial pressure of right cylinder x C in, initial displacement of mass ẋ C in/sec, initial velocity of mass lbf, disturbance psig, return pressure 12

12 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 2 nd Law applied to the moving mass Kinematic relation representing the mechanical feedback 121

13 The Variable Orifice The variable orifice is at the heart of most fluid power systems and is the most popular device for controlling fluid flow. It is the fluid equivalent of the electrical resistor and like the resistor its use leads to energy dissipation. Overriding advantages include simplicity, reliability, and ease of manufacture. The orifice can be used in analog (infinite number of positions) or discontinuous (fully open or fully closed) modes, depending on the application. 122

14 Knowledge of the orifice equations for incompressible and compressible flow is essential. Here we consider incompressible flow. The orifice equation for the volume flow rate Q of an incompressible fluid, assuming that the upstream area is much greater than the orifice area A, is: Q = CA d ( P ) 2P u ρ d C d = orifice discharge coefficient In a given system the dominating variables are usually the pressure drop and the orifice area. 123

15 The predominant nonlinearity is the square root term, but C d depends on the Reynolds number and cavitation. Cavitation refers to the formation and collapse of cavities, containing air or gas, in the liquid. If the pressure is reduced far enough hydraulic oil will vaporize and vapor cavities will form. The pressure at which vaporization commences is called the vapor pressure of the liquid and is very dependent on the temperature of the liquid. As the temperature increases, the vapor pressure increases. 124

16 The phenomenon of cavitation damage in hydraulic machinery, turbines, pumps, and propellers is well known. It has been shown both analytically and experimentally that when cavities collapse as a result of increased hydraulic pressure, very large pressures can be developed. However, there is still controversy about the exact mechanism of the damaging process. 125

17 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 2(ps p cl) Qcl = Cw(x d u + x) v valid when (x u+ x v) is >. ρ This is flow into the left cylinder. 2(pcl p) r Qcl = Cw(x d u x) v valid when (xu x) v is >. ρ This is flow out of the left cylinder. 2(pcr p) r Qcr = Cw(x d u + x) v valid when (xu + x) v is >. ρ This is flow out of the right cylinder. 2(ps p cr) Qcr = Cw(x d u x) v valid when (xu x) v is >. ρ This is flow into the right cylinder. 126

18 Conservation of Mass Equations Conservation of Mass = ρ dv+ ρvda i t CV CS =ρ V + V ρ +ρq CV CV CV CV net V ρ CV = VCV + ρ+ 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 ρ. 127

19 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 isothermal bulk modulus is given by: Therefore: ρ dp ρ= = Pdt ρ β P P β=ρ = ρ P,T Conservation of Mass can be written as: V = V + P CV + Q β net P ρ/ ρ P,T 128

20 Evaluating terms: Left cylinder Q = Q + K (p p ) net cl pl cl cr C V= Ap dt V P β CV = dx (V + Ax ) dp β dt l p C cl The resulting equations for the left and right cylinders are: (V Ax ) dp dx Q K (p p ) A β dt dt r p C cr C cr + pl cl cr = p (V + Ax ) dp dx Q K (p p ) A β dt dt l p C cl C cl pl cl cr = p 129

21 V l and V r are the volumes at the operating point of the left and right cylinders, respectively. β is the bulk modulus (isothermal) of the fluid defined by the expression: dp β= V dv (V l + A p x C ) and (V r - A p x C ) represent the compressed volumes of the left and right sides of the cylinder, respectively, which include the lines from the valve to the actuator plus the ends of the cylinder. 13

22 Newton s Second Law & Feedback Equations 2 dxc dxc cl cr p + U = 2 (p p )A B f M dt dt xv = xv xc 131

23 Simulink Block Diagrams Xv Step Command ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM input To Workspace6 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 To Workspace2 Qcr To Workspace3 Qcl Qcr Xc To Workspace1 Pcl To Workspace4 Pcl Xc Pcr Xc dot Mass Xc Xc dot Xc Qcl Qcr Xc dot Cylinder Pcl Pcr Pcl Pcr Pcr To Workspace5 Clock time To Workspace 132

24 CONTROLLER SUBSYSTEM 1 Xv Command Sum Xv - Xc Switch1 1 Xv - Xc Positive Zero 2 Xc Switch 2 Xv - Xc Negative 1 MASS SUBSYSTEM Pcl 2 Pcr Sum3 Ap Gain2 Disturbance fu Sum 1/M Gain B Gain5 Xc dot 1/s 1/s Integrator Integrator1 1 Xc 2 Xc dot 133

25 Ps Supply P FLOW SUBSYSTEM 1 Pcl Sum2 2/rho Gain4 sqrt(u) Fcn Cd*w Gain Product 1 2/rho sqrt(u) Cd*w Sum Qcl 2 Sum3 Gain5 Fcn1 Gain1 Product1 Xv - Xc Pos 3 Xv - Xc Neg Sum4 2/rho Gain6 sqrt(u) Fcn2 -Cd*w Gain2 Product2 Sum1 2 Qcr 4 Pcr Sum5 Pr Reservoir P 2/rho Gain7 sqrt(u) Fcn3 -Cd*w Gain3 Product3 134

26 Vlo Constant1 1/MB 1/u 1 Xc Ap Gain Sum Gain1 Fcn CYLINDER SUBSYSTEM 1/MB 1/u Vro Constant2 Sum1 Gain2 Fcn1 1 Pcl 2 Qcl Kpl Sum2 Product 1/s Integrator Pcl Gain3 Sum4 3 Qcr Sum3 Product1 1/s Integrator1 Pcr 4 Xc dot Ap Gain4 2 Pcr 135

27 MatLab File of Constants and Expressions M=.3; B=1; Ap=2.; Kpl=.1; Vlo=4.; Vro=4.; MB=1; Pr=; Ps=1; rho=7.8e-5; Cd=.6; w=.5; Pclo=5; Pcro=5; Xcdoto=; Xco=; Vo=4.; Cx= ; Cp=; A=[-(Cp+Kpl)*MB/Vo Kpl* MB /Vo -Ap* MB/Vo; Kpl*MB/Vo -(Cp+Kpl)* MB/Vo Ap* MB/Vo; 1;Ap/M -Ap/M -B/M]; B1=[MB*Cx/Vo ;- MB*Cx/Vo ; ; 1/M]; C=[1 ; 1 ; 1 ; 1]; D=[ ; ; ; ]; K=(2*Cx*Ap)/(2*Ap^2+B*(Cp+2*Kpl)); omegan=sqrt((mb*(2*ap^2+b*(cp+2*kpl)))/(m*vo)); zeta=(b+((2*mb*m)/vo)*kpl+((mb*m)/vo)*cp)/(2*sqrt(((mb*m)/vo) *(2*Ap^2+B*(Cp+2*Kpl)))); XcXvNum=K*omegan^2; XcXvDen=[1 2*zeta*omegan omegan^2 ]; %[XcXvNumCL,XcXvDenCL]=cloop(XcXvNum,XcXvDen,-1); Kc=1; XcXvNumCL=Kc*K*omegan^2; XcXvDenCL=[1 2*zeta*omegan omegan^2 Kc*K*omegan^2]; 136

28 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. 137

29 Q Q Q Q + x + p v v v v, v,p c,p xv p operating point c operating point flow gain = C pressure coefficient = C x p Q = x v Q = p v operating point v c operating point Qv Qv, + Cx x v,p Cp p c,p 138

30 Assume that Q v, = and that the numerical values of C x and C p are equal for the Q cl and Q cr equations (correct assumptions for commonly used operating points). Qcl Cx x v,p Cp p cl,p Qcr Cx x v,p Cp p cr,p Take the volumes (V l + A p x C ) and (V r - A p x C ) to be constant at V l = V r = V, a good approximation for small changes in x C. 139

31 Linearized Set of Equations: V dp Cx Cp K p p = A β dt ( ) cl,p ( ) x v,p p cl,p pl cl,p cr,p p dx dt C,p V dp Cx Cp + K p p = A β dt ( ) cr,p ( ) x v,p p cr,p pl cl,p cr,p p dx dt C,p 2 dxc,p dxc,p cl,p cr,p p + U,p = 2 ( ) p p A B f M dt dt 14

32 Simulink Block Diagram of the Linear System Pcl_l Cp Sum5 To Workspace1 MB/Vo 1/s Pcl Gain6 Ap Sum Sum3 Gain2 MB/Vo Gain3 Integrator2 1/s Pcr Integrator3 Sum2 Cp Gain7 Gain4 Kpl Gain5 5 Constant1 Sum7 Qcl_l To Workspace4 Pcr_l Clock time_l To Workspace Ap Gain9 Sum6 To Workspace2 Sum8 input_l To Workspace6 Sum4 Cx Gain8 Disturbance fu Sum1 1/M Gain 1/s Integrator B Gain1 1/s Xc Integrator1 Qcr_l To Workspace5 Xc_l To Workspace3 Xv Step Command ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) 141

33 If we take the Laplace Transform of these equations, we can derive six useful transfer functions relating the two inputs, x v and f U, to the three outputs, p cl, p cr, and x C. ( ) Vs +β Kpl + Cp Kpl As p Cxβ Cx Cx Vs β K + C ( ) p x cl v Kpl pl p As p p cr = x v Cx Cxβ C x x 2 C fu Ap Ap Ms + Bs 142

34 One of these transfer functions is: x ( s) K = s C 2 x v s 2s ζ 2 ωn ωn where K = ω = n ζ= x p ( ) 2 p p pl 2C A 2A + BC + 2K ( ) β 2A + BC + 2K MV 2 p p pl 2βM βm B+ Kpl C V + V βm 2 2 2Ap + BCp + 2Kpl V p ( ) 143

35 The state variables for the linearized equations are: The state-variable equations are: q q 1 p p cl 2 cr = q3 xc q x 4 C ( ) β Cp + Kpl Kplβ Apβ βcx V V V q 1 q 1 V q Kpl ( Cp K β β + pl ) Apβ 2 q 2 βcx = V V V + x q q V q 4 q 4 Ap Ap B M M M [ v ] 144

36 Simulink Block Diagram: Linear System in State-Variable Form Cx Cp Clock time_lss To Workspace1 input_lss To Workspace6 Gain Gain1 Sum1 Qcl_lss To Workspace4 Sum3 Pcl_lss To Workspace2 5 Xv Step Command Sum Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux Constant Pcr_lss Distrbance fu Xc_lss To Workspace Sum4 To Workspace3 Cx Gain2 Sum2 Cp Gain3 Qcr_lss To Workspace5 ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR - STATE SPACE) 145

37 Simulation Results: Step Command x V =.2 in. applied at t =.3 sec x V + x v x C - CONTROLLER K c PLANT G(s) x KG(s) KKω x 1 KG(s) s 2 s s KK 2 C c c n = = V + c + ζω n +ω n + c ωn 146

38 Nonlinear and Linear Simulation Results: x C vs. time x C (in) solid: nonlinear dashed: linear time (sec) 147

39 Nonlinear and Linear Simulation Results: Q cl vs. time 25 2 Q cl (in 3 /sec) 15 solid: nonlinear dashed: linear time (sec) 148

40 Nonlinear and Linear Simulation Results: Q cr vs. time -5 Q cr (in 3 /sec) -1 solid: nonlinear dashed: linear time (sec) 149

41 Nonlinear and Linear Simulation Results: p cl time 8 75 p cl (psig) 7 65 solid: nonlinear dashed: linear time (sec) 15

42 Nonlinear and Linear Simulation Results: p cr vs. time p cr (psig) 4 35 solid: nonlinear dashed: linear time (sec) 151

43 Open-Loop Frequency Response Plots with K c = 1 Bode Diagrams Phase (deg); Magnitude (db) Frequency (rad/sec) 152

44 GM = 16.2 db = 6.46 PM = 74.8 Bode Diagrams Gm=16.2 db (at rad/sec), Pm= deg. (at rad/sec) 2 Phase (deg); Magnitude (db) Frequency (rad/sec) 153

45 Closed-Loop Frequency Response Plots with K c = 1 Closed-Loop Bandwidth = 123 Hz = 774 rad/sec Bode Diagrams At 774 rad/sec: Mag =.77 Phase = Phase (deg); Magnitude (db) Frequency (rad/sec) 154

46 Simulink Block Diagram: Nonlinear Control with Time Delay input_l_nl Cx Cp To Workspace6 Gain Gain1 Xv Step Command Sum1 Qcl_l_nl Sum3 Pcl_l_nl To Workspace2 Delay To Workspace4.1 Sum Sign Gain4 Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux 5 Constant Pcr_l_nl Distrbance fu Xc_l_nl To Workspace Sum4 To Workspace3 Clock time_l_nl To Workspace1 Cx Gain2 Sum2 Cp Gain3 Qcr_l_nl To Workspace5 ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) with Nonlinear On-Off Controller and Time Delay 155

47 Curve A: Gain =.1, Delay =.1 sec Curve B: Gain =.5, Delay =.1 sec Curve C: Gain =.5, Delay = sec x c (in).15 A C.1 B time (sec) 156

48 Simulink Block Diagram: Linear System with Proportional Control Cx Cp time_l_p Gain Gain1 Clock To Workspace1 Sum1 Pcl_l_p input_l_p Qcl_l_p Sum3 To Workspace2 To Workspace6 To Workspace4 5 Xv Step Command Sum Kc Controller Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux Constant Pcr_l_p Disturbance fu Xc_l_p To Workspace Sum4 To Workspace3 Cx Gain2 Sum2 Cp Gain3 Qcr_l_p To Workspace5 ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) With Proportional Control 157

49 Root Locus Plot Pt. #1: K c = 6.46 Pt. #2: K c = 3.7 Pt. #3: K c = 1.36 Pt. #4: K c = Imag Axis Real Axis 158

50 Open-Loop Poles:, ± 1993i K c =6.46 Closed-Loop Poles: -3384, ± 2614i K c = 3.7 Closed-loop Poles: -2677, -353 ± 2195i K c = 1.36 Closed-Loop Poles: -1133, ± 1737i K c = 1. Closed-Loop Poles: -732, ± 1771i 159

51 Closed-Loop Time Response (Step) Plots.3.25 K c = x C (in) K c = 1.36 K c = time (sec) 16

52 Closed-Loop Frequency Response Plots Bode Diagrams K c = K c =1. K c =1.36 Phase (deg); Magnitude (db) K c =1. K c =3.7 K c = Frequency (rad/sec) 161

53 Nyquist Diagram: K c = 1. Nyquist Diagrams Imaginary Axis Real Axis 162

54 Nyquist Diagram: K c = 6.46 Nyquist Diagrams Imaginary Axis Real Axis 163

55 Nyquist Diagram: K c = 1. Nyquist Diagrams Imaginary Axis Real Axis 164

56 Stability Considerations Closed-Loop Transfer Function x KG(s) KKω x 1 KG(s) s 2 s s KK 2 C c c n = = V + c + ζω n +ω n + c ωn K n = ω = ζ= x p ( ) 2 p p pl 2C A 2A + BC + 2K ( ) β 2A + BC + 2K MV 2 p p pl 2βM βm B+ Kpl C V + V βm 2 2 2Ap + BCp + 2Kpl V ( ) 165 p

57 Neglect leakage (K pl = ) and consider the load as mainly inertia (B =, friction is ignored). The closed-loop transfer function becomes: x x C V = KC c x Ap VM MC KC s + s + s+ 2βA 2A A 3 p 2 c x 2 2 p p p Since the bulk modulus β of the fluid is defined as: P β= V/V The combined stiffness k of the two columns of fluid is: k = 2βA V 2 p 166

58 The valve stiffness k v is defined as: C C x p Q = x v Q = p v operating point v c operating point k v = C x 2A p C p The closed-loop transfer function can now be written as: KC c x x x C V = Ap M CM KC s + s + s+ k Ak A 3 x 2 c x p v p 167

59 Applying the Routh Stability Criterion to the characteristic equation of the closed-loop transfer function gives the relationship for stability as: k > k v In other words, the stiffness of the oil column must be greater than the effective valve stiffness if stability is to be satisfactory. 168