Model-Based Design, Analysis, & Control: Valve … Design, Analysis, & Control: Valve-Controlled...

Valve-ControlledHydraulic System K. Craig 1

Model-Based Design, Analysis, & Control:Valve-Controlled Hydraulic System






Mission: It’s All About Process

Dynamic System Investigation







References

• Academic: Fluid Power– Fundamentals of Fluid Power and Control, J.

Watton, Cambridge, 2009.

– Hydraulic Control Systems, H. Merritt, Wiley, 1967.

– Hydraulic Control Systems, N. Manring, Wiley, 2005.

– Modeling, Monitoring, and Diagnostic Techniques for

Fluid Power Systems, J. Watton, Springer, 2007.

– Control of Fluid Power: Analysis and Design, D.

McCloy and H.R. Martin, 2nd Edition, Ellis Horwood,

1980.


• Industry: Fluid Power– Industrial Hydraulics Manual, Eaton Corp., 2010.

– Electrohydraulic Proportional and Control

Systems, Bosch Automation, 1999.

– Electrohydraulic Proportional Valves and Closed

Loop Control Valves, Bosch Automation, 1989.

– Closed Loop Electrohydraulic Systems Manual,

Vickers, Inc., 1998.

– Basic Electronics for Hydraulic Engineers, Eaton

Corp., 1988.

– Hydraulics, Festo Didactic, https://festo.my-e-

campus.com

– Electrohydraulics, Festo Didactic, https://festo.my-

e-campus.com


• Academic: System Dynamics and Control– Introduction to System Dynamics, J. Shearer, A.

Murphy, and H. Richardson, Addison-Wesley, 1967.

– Dynamic Modeling and Control of Engineering

Systems, J. Shearer, B, Kulakowski, and J. Gardner,

2nd Edition, Prentice Hall, 1997.

– Modeling, Analysis, and Control of Dynamic Systems,

W.J. Palm, 2nd Edition, Wiley, 1999.

– Mechatronics, S. Cetinkunt, Wiley, 2007.

– Introduction to Fluid Mechanics, R. Fox and A.

McDonald, 3rd Edition, Wiley, 1985.

– Sensors and Actuators, Clarence de Silva, CRC

Press, 2007.




Physical Modeling Assumptions

• Supply pressure ps is constant.

• Reservoir pressure is constant at 0 psig.

• Valve is zero lap.

• Actuator pressures pcl and pcr each come to ps/2 at

the servo rest condition.

• Neglect inertia of the fluid.

• Cylinder and piston are rigid.

• Sensor dynamics are negligible.

• Parameters are constant.


• 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.


Physical Model Parameters xu 0 inches, zero lap condition

ps 1000 psig (constant), supply pressure

Cd 0.6, orifice discharge coefficient

w 0.5 in, valve port width

7.8E-5 lbf-s2/in

4, fluid density

Ap 2.0 in2, piston area

100,000 psi, bulk modulus of fluid

M 0.03 lbf-s2/in, mass

Kpl 0.001 in3/s-psi, piston leakage coefficient

B 100 lbf-s/in, viscous damping coefficient

K 0 lbf/in, spring constant

Vl0 4.0 in3, volume at operating point of left cylinder

Vr0 4.0 in3, volume at operating point of right cylinder

pcl0 500 psi, initial pressure of left cylinder

pcr0 500 psi, initial pressure of right cylinder

xC0 0 in, initial displacement of mass

C0x 0 in/sec, initial velocity of mass

FL 0 lbf, load force

pr 0 psig, return pressure


Nonlinear Mathematical Model

• Equations for the orifice volume flow rates Qcl

and Qcr 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 2nd Law applied to the moving mass

• Load position feedback control equation with

simple proportional control.


Orifice Flow-Rate Equations

s clcl d u v u v

cl rcl d u v u v

2(p p )Q C w(x x ) valid when (x + x ) is 0.

This is flow into the left cylinder.

2(p p )Q C w(x x ) valid when (x x ) is > 0.

This is flow out of the left cyli

cr rcr d u v u v

s crcr d u v u v

nder.

2(p p )Q C w(x x ) valid when (x x ) is 0.

This is flow out of the right cylinder.

2(p p )Q C w(x x ) valid when (x x ) is > 0.

This is flow into the rig

ht cylinder.

xu positive: valve underlap

xu negative: valve overlap

xu zero: valve zero lap

xv is displacement of valve spool


Conservation of Mass Equations

• Conservation of Mass

• Here we assume that all of the densities of the

system (inlet flow, outlet flow, and control volume)

are the same and equal to .

CV CS

CV CV CV CV net

CVCV net

0 dV v dAt

0 V V Q

V0 V Q

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.


– 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:

• Conservation of Mass can be written as:

0dPP

P dt

CV net

V0 V P Q

0 0 0 0

0 0 0 0

P ,T P ,T

0 0 0

(P,T) P ,T P P T TP T

11 P P T T


• Evaluating terms:

• The resulting equations for the left and right cylinders

are:

r0 p C cr Ccr pl cl cr p

l0 p C cl Ccl pl cl cr p

(V A x ) dp dxQ K (p p ) A

dt dt

(V A x ) dp dxQ K (p p ) A

dt dt

net cl pl cl cr

Cp

l0 p C clCV

Q Q K (p p )

dxV A

dt

(V A x ) dpVP

dt

Left cylinder


Newton’s 2nd 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.

– All velocities are measured relative to the control

volume.

– All time derivatives are measured relative to the

control volume.

S B

CV CS

F F F v dV v v dAt

2

C Ccl cr p C L 2

dx d x(p p )A B Kx F M

dt dt


Simulink Block Diagrams

Xc dot Pcr

PclXc

Qcr

Qcl

ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM

Xv - Xc Positive

Xv - Xc Negative

Xv Step Command

input

Pcr

Pcl

Qcr

Qcl

Xc

Pcl

Pcr

Xc

Xc dot

Mass

Pcl

Xv - Xc Pos

Xv - Xc Neg

Pcr

Qcl

Qcr

Flow

Xc

Qcl

Qcr

Xc dot

Pcl

Pcr

Cylinder

Xv Command

Xc

Xv - Xc Positive

Xv - Xc Negative

Controller


Xc dot

MASS SUBSYSTEM

2

Xc dot

1

XcSum3

Sum

1/s

Integrator1

1/s

Integrator

B

Gain5

Ap

Gain2

1/M

Gain

0

Disturbance

fu

2

Pcr

1

Pcl

Xv - Xc

CONTROLLER SUBSYSTEM

2

Xv - Xc Negative

1

Xv - Xc Positive

0

Zero

Switch1

Switch

Sum

Kc

Control

Gain

2

Xc

1

Xv Command


FLOW SUBSYSTEM

2

Qcr

1

Qcl

Ps Supply P

Sum5

Sum4

Sum3

Sum2

Sum1

Sum

Pr Reservoir P

Product3

Product2

Product1

Product

2/rho

Gain7

2/rho

Gain6

2/rho

Gain5

2/rho

Gain4

-Cd*w

Gain3

-Cd*w

Gain2

Cd*w

Gain1

Cd*w

Gain

sqrt(u)

Fcn3

sqrt(u)

Fcn2

sqrt(u)

Fcn1

sqrt(u)

Fcn

4

Pcr

3

Xv - Xc Neg

2

Xv - Xc Pos

1

Pcl


CYLINDER SUBSYSTEM

2

Pcr

1

Pcl

Sum4

Sum3

Sum2

Sum1

Sum

Product1

Product

1/s

Integrator1

1/s

Integrator

Ap

Gain4

Kpl

Gain3

1/MB

Gain2

1/MB

Gain1Ap

Gain

1/u

Fcn1

1/u

Fcn

Vro

Constant2

Vlo

Constant1

4

Xc dot

3

Qcr

2

Qcl

1

Xc


M = 0.03;

B=100;

Ap = 2.0;

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 = 1074.172;

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;

0 0 0 1;Ap/M -Ap/M 0 -B/M];

B1=[MB*Cx/Vo 0;- MB*Cx/Vo 0;0 0;0 1/M];

C=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1];

D=[0 0;0 0;0 0;0 0];

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 0];

%[XcXvNumCL,XcXvDenCL]=cloop(XcXvNum,XcXvDen,-1);

Kc=1;

XcXvNumCL=Kc*K*omegan^2;

XcXvDenCL=[1 2*zeta*omegan omegan^2 Kc*K*omegan^2];

MatLab File of Constants and Expressions


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.


v vv v,0 v,p c,p

v coperating point operating point

Q QQ Q x p

x p

v v,0 x v,p p c,pQ Q C x C p

vx

v operating point

vp

c operating point

Qflow gain = C

x

Qpressure coefficient = C

p


Linearized Set of Equations

cl,p C,p0x v,p p cl,p pl cl,p cr,p p

dp dxVC x C p K p p A

dt dt

cr,p C,p0x v,p p cr,p pl cl,p cr,p p

dp dxVC x C p K p p A

dt dt

2

C,p C,p

cl,p cr,p p U,p 2

dx d xp p A B f M

dt dt


Simulink Block Diagram of the Linear System

Pcl

Pcr

Xc

ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR)Xv Step Command

input_l

Qcr_l

Qcl_l

Xc_l

Pcr_l

Pcl_l

Sum8

Sum7

Sum6

Sum5

Sum4

Sum3

Sum2

Sum1

Sum

1/s

Integrator3

1/s

Integrator2

1/s

Integrator1

1/s

Integrator

Ap

Gain9

Cx

Gain8

Cp

Gain7

Cp

Gain6

Kpl

Gain5

Ap

Gain4

MB/Vo

Gain3

MB/Vo

Gain2

B

Gain1

1/M

Gain

0

Disturbance

fu

500

Constant1


• If we take the Laplace Transform of these equations,

we can derive six useful transfer functions relating

the two inputs, xv and FL, to the three outputs, pcl, pcr,

and xC.

L

0 pl p pl p

x x x

cl v

0 pl ppl p

cr v

x x x

C2

p p

V s K C K A s

C C Cp x

V s K CK A s p x

C C Cx F

A A Ms Bs


C

2v

2

n n

x Ks

x s 2 ss 1

x p

2

p p pl

2

p p pl

n

0

pl p

0 0

2

p p pl

0

2C AK

2A B C 2K

2A B C 2K

MV

2 M MB K C

V V

M2 2A B C 2K

V

One of these transfer functions is:


Simulation Results:Step Command xV = 0.02 in. applied at t = 0.003 sec

PLANT G(s)CONTROLLER

Kc

+

-

xV

xv

xC

2

C c c n

3 2 2 2

V c n n c n

x K G(s) K K

x 1 K G(s) s 2 s s K K


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.020

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

time (sec)

xC (in)

solid: nonlinear

dashed: linear

Nonlinear and Linear Simulation Results: xC vs. time


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.020

5

10

15

20

25

time (sec)

Nonlinear and Linear Simulation Results: Qcl vs. time

Qcl (in3/sec)

solid: nonlinear

dashed: linear


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02500

550

600

650

700

750

800

Nonlinear and Linear Simulation Results: pcl vs. time

time (sec)

solid: nonlinear

dashed: linearpcl (psig)


Frequency (rad/sec)

Ph

ase

(d

eg

); M

ag

nitu

de

(d

B)

Bode Diagrams

-40

-20

0

20

40

60

10-1

100

101

102

103

104

-250

-200

-150

-100

Open-Loop Frequency Response Plots with Kc = 1


Frequency (rad/sec)

Ph

ase

(d

eg

); M

ag

nitu

de

(d

B)

Bode Diagrams

-40

-20

0

20

Gm=16.2 dB (at 2614.1 rad/sec), Pm=74.783 deg. (at 527.04 rad/sec)

102

103

-250

-200

-150

-100

GM = 16.2 dB = 6.46

PM = 74.8


Frequency (rad/sec)

Ph

ase

(d

eg

); M

ag

nitu

de

(d

B)

Bode Diagrams

-50

-40

-30

-20

-10

0

102

103

104

-250

-200

-150

-100

-50

Closed-Loop Frequency Response Plots with Kc = 1

Closed-Loop Bandwidth = 123 Hz = 774 rad/sec

At 774 rad/sec:

Mag = 0.707

Phase = -72.2


Simulink Block Diagram: Nonlinear Control with Time Delay

ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR)

with Nonlinear On-Off Controller and Time Delay

Xv Step Command

input_l_nl

Qcr_l_nl

Qcl_l_nl

Pcr_l_nl

Pcl_l_nl

time_l_nl

Xc_l_nl

x' = Ax+Bu

y = Cx+Du

System

Sum4

Sum3

Sum2

Sum1

Sum Sign Mux

Mux

.005

Gain4

Cp

Gain3

Cx

Gain2

Cp

Gain1

Cx

Gain

0 Distrbance fu

Demux

Demux

Delay

500

Constant

Clock


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.020

0.005

0.01

0.015

0.02

0.025

0.03

xc (in)

time (sec)

A

B

C

Curve A: Gain = 0.01, Delay = 0.001 sec

Curve B: Gain = 0.005, Delay = 0.001 sec

Curve C: Gain = 0.005, Delay = 0 sec


Simulink Block Diagram: Linear System with Proportional Control

ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR)

With Proportional Control

Xv Step Command

input_l_p

Qcr_l_p

Qcl_l_p

Pcr_l_p

Pcl_l_p

time_l_p

Xc_l_p

x' = Ax+Bu

y = Cx+Du

System

Sum4

Sum3

Sum2

Sum1

Sum

Mux

Mux

Cp

Gain3

Cx

Gain2

Cp

Gain1

Cx

Gain

0 Disturbance fu

Demux

Demux

Kc

Controller

500

Constant

Clock


-4000 -3000 -2000 -1000 0 1000 2000 3000 4000-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

Real Axis

Ima

g A

xis

Root Locus Plot

1

1

1

2

2

2

3

3

3

4

4

4

Pt. #1: Kc = 6.46 Pt. #2: Kc = 3.70

Pt. #3: Kc = 1.36 Pt. #4: Kc = 1


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.020

0.005

0.01

0.015

0.02

0.025

0.03

Closed-Loop Time Response (Step) Plots

time (sec)

xC (in)

Kc = 3.70

Kc = 1.36

Kc = 1.0


Closed-Loop Frequency Response Plots

Frequency (rad/sec)

Ph

ase

(d

eg

); M

ag

nitu

de

(d

B)

Bode Diagrams

-100

-80

-60

-40

-20

0

101

102

103

104

105

-250

-200

-150

-100

-50

0

Kc =3.70

Kc =3.70

Kc =1.36

Kc =1.36

Kc =1.0

Kc =1.0


Real Axis

Ima

gin

ary

Axis

Nyquist Diagrams

-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Nyquist Diagram: Kc = 1.0


Real Axis

Ima

gin

ary

Axis

Nyquist Diagrams

-1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2

-3

-2

-1

0

1

2

3



Real Axis

Ima

gin

ary

Axis

Nyquist Diagrams

-2.5 -2 -1.5 -1 -0.5-5

-4

-3

-2

-1

0

1

2

3

4

5



Stability Considerations

2

C c c n

3 2 2 2

V c n n c n

x K G(s) K K

x 1 K G(s) s 2 s s K K

x p

2

p p pl

2

p p pl

n

0

pl p

0 0

2

p p pl

0

2C AK

2A B C 2K

2A B C 2K

MV

2 M MB K C

V V

M2 2A B C 2K

V

Closed-LoopTransfer Function


• Neglect leakage (Kpl = 0) and consider the load as

mainly inertia (B = 0, friction is ignored). The closed-

loop transfer function becomes:

• Since the bulk modulus of the fluid is defined as:

• The combined stiffness k0

of the two columns of fluid is:

c x

pC

p3 20 c xV

2 2

p p p

K C

Ax

MCV M K Cxs s s

2 A 2A A

2

p

0

0

2 Ak

V

0

P

V / V


• The valve stiffness kv is defined as:

• The closed-loop transfer function can now be written

as:

xv p

p

Ck 2A

C

c x

pC

3 2 c xxV

0 p v p

K C

Ax

K CM C Mxs s s

k A k A

vx

v operating point

vp

c operating point

QC

x

QC

p


• Applying the Routh Stability Criterion to the

characteristic equation of the closed-loop transfer

function gives the relationship for stability as:

• In other words, the stiffness of the oil column must be

greater than the effective valve stiffness if stability is

to be satisfactory.

0 vk k

Date post:	12-Jun-2018
Category:	Documents
Upload:	dangdang
View:	234 times
Download:	4 times

Model-Based Design, Analysis, & Control: Valve … Design, Analysis, & Control: Valve-Controlled...

Documents