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ME in actuator technology

Course title: Principles designing hydraulic servoactuator systems

Code: 521

Teacher: Prof. Veljko Potkonjak

Abstract.

Principles of hydraulic systems.

Actuators. Hydraulic cylinder with piston. Rotary actuator. Mathematical models ofactuator dynamics.

Electrohydraulic servovalves principles and mathematics. Permanent-magnet torque

motor. Single-stage electrohydraulic servovalve. Two-stage electrohydraulic servovalve

with direct feedback. Two-stage electrohydraulic servovalve with force feedback.Specification, selection and use of servovalves.

Mathematical modeling. Mathematical model of the complete system. Linearization of

the 5-th order model. Reduction of the system (to 3-rd order form). Linearization of the3-rd order model.

Nonlinearities. Saturation. Deadband. Backlash and hysteresis. Friction. etc.

Closed-loop control of electrohydraulic system.

Simulation. Simulation model. Simulation in system design.

Literature:

H. E. Merit, Hydraulic Control Systems, John Wiley & Sons, New York

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1. INTRODUCTION

Advantages and Disadvantages of Hydraulic Systems

ADVANTAGES:

- No heating problems ... the fluid carries away the

heat ...

- Lubrification ...

- No saturation ...

- Fast response ... fast start/stop ... high torque-to-

inertia ratio => high accelerations ...- All working modes ... continuous, intermittent,

reversing, ...

- High stiffness ... little drop in speed as loads are

applied ...

- Open and closed loop control ...

- Other aspects ...

DISADVANTAGES:

- Power not so readily available ...- High costs for small tolerances ...

- Upper temperature limit ... fire danger ; messy due

to leakage

- Fluid contamination ... dirt in fluid (contamination)

is chief source of hydraulic control failure ...

- Complex modeling ... very often the design is not

based on a sophisticated mathematical model ...

- Inappropriate for low power ...

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2. HYDRAULIC FLUIDS (LIQUIDS, OIL)

NOT GAS!

2.1. Density

)(

)()(

volumeV

weightGdensityweight = ,

typically 33 /......../03.0 mNinlb ==

)(

)()(

volumeV

massmdensitymass = ,

typically 3424 /......../sec1078.0 mkginlb ==

g= )/81.9( 2smg= (2.1)

2.2. Equation of State

Expression that relates density (or volume V), pressure P , and

temperature T .

Volume (and density) changes little. So, a linear approximation isjustified:

)()( 000 TTT

PPP PT

+

+=

(2.2)

or

))()(1

1( 000 TTPP +=

(2.3)

where

TT

V

PV

P

=

= 00

,PP T

V

VT

=

=00

11

(2.4), (2.5)

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isotermal bulk modulus (compressibility). IMPORTANT!

- It relates to the stiffness of the liquid (a kind of a sping effect).

- It have in important influence to the precision of hydraulicactuator.

- It is desired to be as high as possible.-Presence of air (gas) in the liquid, even small, decreases sharply

the bulk modulus.

anddepend on the temperature:

2.3. Viscosity

It expresses the internal friction of the liquid and its resistance

to shear.

Necessary for lubrification.

If too low leakage!

If too large power loss due to friction (lower efficiency)!

ln

T T

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Friction force is proporional to the contact area A and to the velocityx , and inversly proportional to the film thickness rC :

xC

DL

C

xAF

rr

== , absolute viscosity (coeff. of visc.) (2.7)

=v kinematic viscosity (2.8)

depends on the temperature:

=0e (T - T0) (2.9)

leakage

leakage

of liquidmotionx ,velocity

Cr radialclearance

F

LD

Piston in a cylinder

resistive

friction

force

Fig. 2.2

T

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2.4. Thermal Properties

Specific heatis the amount of energy (heat) needed to raise the

temperature by 10.

Thermal conductivity is the measure of the rate of heat flow

through an area for a temperature gradient in the direction of

heat flow.

2.5. Effective Bulk Modulus

Interaction of the spring effect of a liquid and the masses of

mechanical parts gives a resonance in nearly all hydrauilic

components.

The bulk modulus can be loweredby intruducing

- mechanical compliance and/or- air compliance.

For instance:

- the container can be flexible (mechanical

compliance), and/or

- bubbles or pocket of gas are present inside (gas

compliance).

(see Fig. 2.4)

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The expression for the effective (total) bulk moduluse can be

found in the form:

)11

(111

lgt

g

lce V

V

++= + (2.20)

where:

c the bulk modulus for the container, lfor the liquid,gfor

the gas; Vg the volume of the gas, and Vt the total volume.

Since gl >> , (2.20) becomes:

)1

(111

gt

g

lce V

V

++= (2.21)

If there is no gas (so, only mechanical compliance), one obtains:

lce

111+= (2.22)

2.6. Chemical and Related Properties

Vt

Vc

liquid, volumeVl

gas pocket,

volume Vg

liquid

Vg

gas

Fig. 2.4

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- Lubricity

- Thermal stability

- Oxidative stability

- Hydrolytic stability

- Compatibility- Foaming

- Flash point, fire point, autogenous ignition temperature

- Pour point

- handling properties (toxity, color, odor, ...)

2.7. Types of Hydraulic Fluids

Petroleum based fluids, and

Synthetic fluids

Characteristics

2.8. Selection of the Hydraulic Fluid

3. FLUID (LIQUID) FLOW FUNDAMENTALS

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It is assumed that the general theory of fluid flow is elaborated in

theprevious courses.

Among numerous problems, we highlight here the topic:

3.4. Flow Through Orifices Turbulent Flow

0

2

A

ACc = contraction coefficient (3.28)

Let: u fluid velocity, P pressure . We apply:

- Bernullis equation )(2

21

2

1

2

2 PPuu = (3.29)

- Equation of incompressibility 332211 uAuAuA == (3.30)

- Volumetric flow rate (theflow) 22uAQ =

- Contaction coefficient (3.28) 02 / AACc =

- velocity coefficient98.0vC

(sometimes adopted1vC

)(velocity is slightly smaller due to friction)and we obtain

)(2

210 PPACQ d =

(3.33)

where

A2, jetarea is minimum

jet area A0

vena contracta the

jet area is minimimum

Fig. 3.10.1 2 3

3

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2

10

2)/(1 AAC

CCC

c

cvd

= (3.34)

is the discharge coefficient. Since 1vC and 10 AA

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our primary interest

hydrodynamic machines(turbines, etc.)

Hydraulic machines

positive displacement mach.!

limited travel machines

continuous travel machines

rotary machines

piston machines (translation)

Piston actuator (cylinder with a piston) limited travel mach.

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The piston moves due to the pressure force created by the

different pressures on the two sides of the piston: P1 in the

forward chamber andP2 in the backward chamber.

cylinder

xp

piston

position

fluid IN fluid OUT

fluid IN :pressure P

1

fluid OUT :pressureP

2

Single rod

actuator

Double rod

actuator

forwardchamber

backwardchamber

motion

motion

pressure force

load

force

piston

Fig. 4.1

piston parameters:

Mtmass of the

piston plus

refered masses

Ap

effective piston

area

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When the piston moves to the right, the fluid enters the forward

chamber (fluid IN), and leaves the backward chamber (fluid

OUT).

Mathematical description:

Differential pressure PL (difference between the twopressures):

21 PPPL =

Pressure force (generated force) isLpg

PAF =

Load force oroutput forceis FL

There is a spring effectassociated with the piston:Kxp , whereKis the gradient (stiffness).

There is a viscous damping effect associated with the piston:

pp

xB

, whereBp is the viscous damping coefficient.

Dynamics of the motor (i.e. dynamics of the piston)

Newtons law gives:

LpppptLp FKxxBxMPA +++= (A.1)

Vane rotary actuator limited travel mnachine

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Pressure torque force (generated torque) is rPA Lpg =

Load torque oroutput torqueis L

There is a torsionspring effectassociated with the rotor:K,whereKis the gradient (torsion stiffness).

There is a viscous damping effect associated with the piston:B , whereB is the viscous damping coefficient.

Dynamics of the motor (i.e. dynamics of the rotor)

rotor

backward

chambre

forward

chambre

Vane

Rotation

angle

Pressuremakes

a resultant

force andconsequently

a torque

fluid IN :

pressure P1

fluid OUT :

pressure P2

r

rotor

parameters:

It

moment

of inertia

Ap

effectivevane area

housing

(stator)

14

Fig. 4.2 a

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mechanic energy

OUTPUT:mechanic

energy

Newtons law for rotation gives:

LtLp KBIrPA +++= (A.2)

Double vane rotary actuator is shown in

Spur gear rotary machine (actuator or pump) is shown in

It allows continuous rotation.

show different types (examples) of hydraulic

machines.

In this course, we are primarily interested in

actuators. The ususl example will be a

piston actator or a vane rotary motor

The pumps are used just as a source of

hydraulic energy.

5. HYDRAULIC CONTROL VAVES

Valves are are the interface between the the sorce of hydraulicenergy and the actuator.

15

hydraulic energy

Figs. 4.4

PUMP(source of hydro

energy): convertsmechanical

energy into

hydraulic energy

HYDRO

ACTUATOR:converts hydro

energy into

mechanicalenergy

ELECTRIC

MOTOR(source of

mechanic energy):

converts electricener. into

mechanical ener.

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Actuator (motor) is e.g. a cylinder with a piston or a vane

rotary motor.

Energy source is a pump (of any type).

Valve is a devices that uses mechanical motion to control thedelivery of power to the actuator.

5.1. Valve Configurations

sliding type(a, b, c, d in Fig. 5.1)

Config. classification seating type (e in Fig. 5.1)

16

control the

delivery ofenergy

controlled source of energy(controlled by means of mechanical motion)

source of

hydrauluicenergy

Oil supply

(pressure

supplay)VALVE Actuator

oil flow oil flow

Unit which creates the

mechanical motion

that controls

the valve

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flow deviding type (f in Fig. 5.1)

Sliding valves are classified according to:

- number of ways - the number of input/output oil

lines;- number of lands,

- type of center when spool is in neutral position.

(a) two-land-four-way spool valve:

(b) three-land-four-way spool valve:

17

flow to

source

return

supply

flow to

actuator

mechanical motion

that controls the valve

spool stroke xv

Fig. 5.1 (a)Fig. 5.1 (a)

mechanical motion thatcontrols the valve

spool stroke xv

flow to

source

return

supply

flow to

actuator

Fig. 5.1 (b)Fig. 5.1 (b)

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(c) four-land-four-way spool valve:

(d) two-land-three-way spool valve:

mechanical motion

that controls the valve

spool stroke xv

flow to

source

return

supply

flow toactuator

Fig. 5.1 (c)Fig. 5.1 (c)

18

mechanical motion that

controls the valve

spool stroke xv

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(e) two-jet flapper valve:

(f) jet pipe valve:

flow to

source

return

supplyflow to

actuator

Fig. 5.1 (d)Fig. 5.1 (d)

flapper

supply

pivot

Fig. 5.1 (e)Fig. 5.1 (e)

flow to

actuator

motion of

the flapper

controls the

valve

return

to source

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Spool valves:matching tolerances are required =>

- expensive and

- sensitive top oil contamination

Flapper valves:

leakage =>- for low power or

- as a first stage in a two-stage systems. Jet pipe valves:

- large null flow,

- characteristics are not easy to predict,

- slow response.

supply

pivot

rotation of the

jet controls

the valve

Fig. 5.1 (f)Fig. 5.1 (f)

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For further discussion spool valves.

Number of lands:

- two , in primitive valves;

- three or four , in a usual case- up to six , for special valves.

Ratio between the land width and the port:

If land width < port : open center or underlapped valve

If land width = port : critical center orzero lapped valve

If land width > port : closed center or overlapped valve

width

port

width port

widthport

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open center valve : large power loss ion neutral position; only

for some special systems

critical center valve : our choice; linear characteristics

closed center valve : deadband near null causes steady state error

and stability problems.

flow

Q

spool strokexv

critical

center

closed

center

overlap region

underlap region

flow gain doubles

near null

Fig. 5.2

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5.2. General Valve Analysis

General Flow Equations

Neglecting the compressibility, continuity request yields:

- to actuator: 41 QQQL = (5.1)

- from actuator: 23 QQQL = (5.2)

The differential pressure is21 PPPL = (5.3)

L2

L1

L2

L1

spool stroke

P2

P1

3

2

1

4Supply:- flow Q

s

- pressure Ps

Return:

- flow Qs

- pressure P0 0

To actuator:

- flow QL

- pressureP1

From actuator:

- flow QL

- pressureP2

PL= P

1 P

2

ForceFi

Fig. 5.3.

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4,3,2,1,)0( 0 == jAAj

So, only one orifice area need to be described. If the orifice area

is linear with the valve stroke (as is usually tha case), only one

defining parameter is needed:

w the width of the slot (hole) in the valve sleeve (cover) .

w For linear valves (like with rectangular ports), this is thearea gradient for each orifice (and so for the whole

valve).

For matched and symmetrical orifices, it holds that

4231 , QQQQ == (5.15), (5.16)

Substituting (5.4), (5.5) and (5.6) into (5.15) one obtains:

21 PPPs += (5.17)

Relation (5.16) may give the same result.

Equations (5.3) and (5.17) can be combined to produce:

21

Ls PPP+

= (5.18)

22

Ls PPP

= (5.19)

From Fig. 5.3, it follows that the total supply flow can be written

as21 QQQ s += (5.20)

and as21 QQQ s += (5.21)

In summary, for a matched and symmatrical valve, relations(5.15), (5.16) and (5.18), (5.19) applies and equations (5.1) and

(5.2) both become

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- Flow-pressure coef. 0>

L

Lc

P

QK (5.26)

- Pressure sensitivityc

q

v

Lp

K

K

x

PK =

(5.27), (5.28)

Flow gainaffects the open-loop gain constant and thus has

a direct influience on the system stability.

Flow-pressure coeficient directly affects the damping

ratio of valve-motor combination.

Pressure sensitivity of valves is quite large which shows

the ability of valve-motor combination to breakaway

large friction loads with little error.

Now, (5.24) becomes

LcvqLPKxKQ = (5.29)

The most important operating point is the origin: 0,0,0 111 === LLv QPx .

- In this case, qK is largest (thus, high system gain)

and cK is smallest (thus, low damping), and

accordingly this operating point is most critical from

a stability viewpoint.

- If we achieve stability for this point, the system will

be stable for all other operating points.- Valve coefficinets calculated for thgis point are

called null valve coefficients.

For this operating point ( 0,0,0 111 === LLv QPx ), it holds that:

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0,2

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NOTE a difference regarding previous figures. The forward

flow (to the actuator: Q 1) is not equal to the return flow

(from the actuator: Q2). Previously, it was equal: Q1=Q2= Q L

This is due to some effects that have been neglected in the

previous discussions and now we take care of them. Theseeffects are:

- Leakage,

- Compression.

Valve controlled flowLinear analysis

Starting from relation (A.3) ( LcvqL PKxKQ = ), one may writeexperessions forQ1and Q2:

11 2 PKxKQ cvq = (6.1)22 2 PKxKQ cvq += (6.2)

xp

Returnline:

Q2,P

2

Forwardline:

P1

, Q1

Cylinder with

a Piston

VALVE

SupplyPs

Fig. A.1

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- If the valve is matched and symmetrical, the pressures in the

lines will rise above and below 2/sP by equal amounts so that

the pressure drops across the two valve orifices are identical .

Hance the valve coefficients qK for forward and return flows

are the same.- The flow-pressure coefficient cK is twice that for the whole

valve since qK was defined with respect to PL and the change

inPL is twice that which occurs across a port.

Adding tha above two equations, it follows that

LcvqL PKxKQ = (6.3)

So, the same form was obtained like expression (A.3).However, here, the load flow is the average :

2

21 QQQL+

= . (6.4)

and it is not equal to the flow in each line ( 21 QQQL ).

The load pressure (diffrerencial pressure) is still 21 PPPL = . Valve controlled flowNon-linear analysis

Instead of (6.3) , the nonlinear expression for the flow (eq.

(5.33)), can be applied (like in later Section 6.7.)

== L

v

vsvdLvLL P

x

xPxwCPxQQ

1),( (5.33)

Flow through the actuator continuity relations .

Let us turn to the actuator chambers and look at Fig. 6.6.

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Analyzing the flow, we take care of

Piston motion. The corresponding flow is the rate of volume

change: dV/dt.

Leakage (internal and external). Flow due to leakage is

proportional to the pressure drop.

Compression (effective due to air and mech. compliance;

oil itself might be considered noncompressible or

compressible). Flow due to compression is derived starting

from eq. (2.4) the definition of the bulk modulus:

(2.4):

==V

PV0 =>

==

dtdV

dtdPV

/

/0 =>

dt

dPV

dt

dV

0=

Applying the equation of continuity for chambers 1 and 2, one

obtains

ForceFiand

motionxv

(to control the valve)

PL

=P1

P2

P2

,V2P

1,V

1

xp

VALVE

Fig. 6.6

Load.

- ForceFL

- spring effect

- dampingeffect

Piston parameters:M

tmassof the

piston plus

refered masses

Ap

effective piston

area

Forward

line:

P1

, Q1

External leakage External leakage

Return

line:

Q2,P

2

Internal leakage

Supply

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dt

dPV

dt

dVPCPPCQ

e

epip111

1211 )(

+= (6.27)

dt

dPV

dt

dVQPCPPC

e

epip222

2221 )(

+= (6.28)

where

V1 volume of the chamber 1 of the actuator plus relatedvolumes: connecting line, and the refered volume in the

valve)

V2 volume of the chamber 2 plus related volumes

Cip internal leakage coefficient

Cep external leakage coefficient

The volumes of the chambers may be writted as

ppxAVV += 011 (6.29)ppxAVV = 022 (6.30)

where V01 and V02 are the initial volumes (for the null position of

the piston,xp= 0). The piston is usually centered, and then:V01= V02 = V0 .

Now, from (29) and (6.30), the derivatives are

dt

dxA

dt

dV

dt

dxA

dt

dV pp

p

p ==11 ; ;

=

dt

dV

dt

dV21

(A.4)

The sum of the two volumes is contant and independent of piston

motion:

0020121 2VVVVVVt =+=+= (6.32)

Vt is the total volume of fluid under compression in bothchambers.

We now combine (6.29), (6.30), (A.4) and (6.27), (6.28) toobtain

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dt

PPdxA

dt

PPdVPP

CC

dt

dxA

QQQ

e

pp

e

ep

ip

p

p

L

)(

2

)(

2))(

2(

2

2121021

21

++

+++

=+

=

If

0VxA pp

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Dynamics of the motor (i.e. dynamics of the piston)

Newtons law gives:

LpppptLp FKxxBxMPA +++=

(A.1)=(6.34)

6.A. Mathematical Model of the Valve-Controlled

Actuator

Actuator controlled by the valve strokeAs mentioned several times, the velve control the actuator by the

spool stroke xv .

(I) Dynamics of the piston motion is desribed by (6.34):LpppptLp FKxxBxMPA +++= (6.34)

(II) Load flow is described by continuity equation (6.33):

L

e

tLtpppL P

VPCxAQ

4

++=(6.33)

(III) Valve control the flow by relation

- (6.3) in the case of linear analysis, or

- (5.33) in the case of non-linear analysis:LcvqL PKxKQ = (6.3)

or

== L

v

vsvdLvLL P

x

xPxwCPxQQ

1),( (5.33)

Eqs. (I)(III), i.e. - (6.34), (6.33) and (6.3) (for lin. case) or

` - (6.34), (6.33) and (5.33) (for non-lin. case),

define the mathematical model.

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Important notes about the load.

The model derived (eqs. (I) (IV)) includes the load force FL. Itis not a known force but it depends on the dynamics of the load.

In a general case, the load is a dynamic system (that may have its

own degrees of freedom). So, the load force FL represents theinteraction between the two systems (actuator and load see Fig.

A.2).

According to the law of action and reaction, the force that actsfrom the actuator to the load (action) is equal and oposite to the

force that acts from the load to the actuator (reaction).

So, the load forceFL is unknown and has to be expressed fromthe mathematical model of the load dynamics.

Hence, in order to complete the system of equations (i.e. to make

it solveble), it will be necessary to specify the load and formulate

its mathematical model.

Canonic form of the mathematical model

- For the analysis of system: dynamic characteristics, control

syntehis, stability analysis, and finally simulation, it is desired

to put thge mathematical model in the canonic form.

actionFL

Load

reactionFL

Actuator

Fig. A.2

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(I) Lptp FKzzBzMzA +++= 1223

(II) 3324

zV

zCzAQ

e

t

tppL

++=

(III) 3zKuKQ cqL = (for linear analysis), or

= 3

1z

u

uPuwCQ sdL

(for nonlinear analysis)

From (A.9), it follows that 21 zxz p == .

By combining the above relations, for the linear case one

gets:

21 zz =

L

tt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 += (A.11)

uV

KzV

CKzV

Azt

eq

t

etpc

t

ep

44)(

4323 ++=

i.e.. in a matrix form (A.8) it is:

L

t

t

eq

t

etpc

t

ep

t

p

t

p

t

FM

u

VKz

zz

VCK

VA

M

A

M

B

M

K

z

zz

HED

+

+

+

=

0

10

400

4)(

40

010

3

2

1

3

2

1

(A.12)

For the nonlinear case one gets the form (A.7):

21 zz =

L

tt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 +=

+= 3323

1444z

u

uPuwC

Vz

VCz

VAz sd

t

e

t

e

tp

t

e

p

(A.13)

44

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For a linear analysis, load force is substituted into (A.11) (or,

may be it is simpler to substitute into (I)). In any case, one

gets:

21 zz =

3212)3/2()3/2()3/2(

zmM

Az

mM

Bz

mM

Kz

t

p

t

p

t ++

+

+=

uV

KzV

CKzV

Azt

eq

t

etpc

t

ep

44)(

4323 ++=

or in a matrix form

u

VKz

z

z

VCK

VA

mM

A

mM

B

mM

K

z

z

z

ED

t

e

q

t

etpc

t

ep

t

p

t

p

t

+

+

++

+=

40

0

4)(

40

)3/2()3/2()3/2(

010

3

2

1

3

2

1

which is the final form (A.6).

For a nonlinear analysis, load force is substituted into (A.13)(or, into (I)), to get:

21zz =

3212)3/2()3/2()3/2(

zmM

Az

mM

Bz

mM

Kz

t

p

t

p

t ++

+

+=

+= 3323

1444z

u

uPuwC

Vz

VCz

VAz sd

t

e

t

e

tp

t

e

p

which is the final form (A.5).

EXAMPLE 2

Form the linear mathematical model for the system of Fig. A.4 !

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The control input is the valve stroke.

NOTE: The load introduces one additional degree of freedom

(x2r) and accordingly two additional state variabls ),( 22 rr xx .

The actuator is modeled by (A.11) for a linear analysis or

(A.13) for a nonlinear analysis.

The model includes the load forceFL .

We now look for the mathematical model of the load in order to

express the load forceFL .

NOTE:FL

is

in reverse

direction

(negative)

FL

x2r

Load

Body: massm2

Cylinder:

mass m1

radius r

Fig. A.4

Actuator

xp= x

1

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Besides the old state variables (comming from the actuator),

i.e. z1, z2, z3, we have introduced two new state variables

(due to the new degree of freedom of the load,x2r):rxz 24 = , rxz 25 = .

In this case the above equations of dynamics become

LFzmzm = 51212

1

gmzmmzm 251222 )2

1( =+

with54 zz = ,

or, after additional transformation,

gmm

mmz

mm

mmmF

L

12

212

12

2

121

)2/1(

)2/1(

)2/1(

)2/1()2/3(

+

=

2

12

2

12

2

5)2/1()2/1(

zmm

mg

mm

mz

= (*)

54 zz =

For a linear analysis, (A.11) is combined with the above three

relations. First,FL from the first relation is substituted into thesecond equation from (A.11) (note that the sign ofFL has

changed due to the oposite action of the force). Then, from this

modified second equation of (A.11), 2z is substituted into the

second relation of the above set (*). Now, this modified

second relation form (*), and the third relation from (*) are

supplemented to the set (A.11). In this way, five state

equations are obtained:

21 zz =

23232221212 GzDzDzDz +++= uEzDzDz

33332323 ++=

54 zz =

53532521515 GzDzDzDz +++=

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Control input is the force to valve spool,i

Fu = . (A.15)

Let us rewrite (I)-(IV) acoording to notation (A.14) and

(A.15):

(I) Lptp FKzzBzMzA +++= 1223

(II) 3324

zV

zCzAQe

t

tppL

++=

(III) 34 zKzKQ cqL = (for linear analysis), or

= 3

4

44

1z

z

zPzwCQ sdL

(for nonlinear analysis)

(IV) 455 zKzBzMu ffs ++=

From (A.14), it follows that21 zxz p == and 54 zxz v ==

By combining the above relations, for the linear case one

gets:21 zz =

L

tt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 +=

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

++= (A.16)

54 zz =

uM

zM

Bz

M

Kz

ss

f

s

f 1545+=

i.e.. in a matrix form (A.8) it is

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L

t

s

t

eq

t

etpc

t

ep

t

p

t

p

t

FM

u

Mz

z

z

z

z

V

K

V

CK

V

A

M

A

M

B

M

K

z

z

z

z

z

HED

+

+

+

=

0

0

0

10

10

0

0

0

10000

044

)(4

0

00

00010

5

4

3

2

1

5

4

3

2

1

(A.17)

For the nonlinear case one gets the form (A.7):

21zz =

L

tt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 +=

+= 3

4

44323

1444z

z

zPzwC

Vz

VCz

VAz

sd

t

e

t

etp

t

ep

54 zz =

uM

zM

Bz

M

Kz

ss

f

s

f 1545+=

(A.18)

How to handle the load ?We could explain this by examples ! The examples would be

done completely analogously like Examples 1 and 2, so like

it was done for the spool-stroke controlled actuator.

6.3. Three-Way Valve Controlled Piston

6.4. Pump Controlled MotorNOT DISCUSSED FOR THE MOMENT

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6.5. Valve Controlled Motor with Load Having Many

Degrees of Freedom

Let the load be in the form of n masses connected by means ofsprings (stiffness) and dampers, as shown in Fig. 6.8. A

combination of a spring and a damper will be called simply

spring (a real spring actually involves stiffness and damping).

m1, m2, ... , mn masses

k1, k2, ... , kn stiffnesses

b1, b2, ... , bn damping constants

Position coordinates (degrees of freedom) for the entire system:

- xp ,xv(for the acruator and valve) plus- x1,x2, ... ,xn (for the load)

Dynamics of the actuator and the valve is described by

eqs. (I) (IV) .

QL

kn

xn

x2

x1

b1

k2

m1

k1

xp

Fi

xv

QL

Valve

m2

b2 m

n

bn

Fig. 6.8

loadFL

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This model includes the load forceFL.

Dynamics of the load can be described by the following set of nequations:

2springinforce)(1springinforce

)]()([)]()([ 212212111111 xxbxxkxxbxxkxm

LF

pp ++=

=

3springinforce2springinforce

)]()([)]()([32332321221222xxbxxkxxbxxkxm ++=

. . .

. . .

. . .

n

nnnnnnnn xxbxxkxmspringinforce

)]()([ 11 +=

(A.19)

The complete mathematical model (actuator plus load) includes:

- eqs. (I) (IV) , fot the acatuator and valve, plus

- set ofn equations (A.19).

Force FL in (I)(IV) can be eliminated since it is the force inspring 1 and it is

)()( 1111 xxbxxkF ppL += ,as given in the first equation of the set (A.19).

The load has intruduced additional degrees of freedom and

accordingly additional state variables. The entire set of state

variables (vectorz) is :

z=( px , px , LP , vx , vx , (from the actuator) nn xxxxxx ,,,,,, 2211 (from the load)).

The model can be put in a canonical form.

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6.6. Pressure Transients in Power ElementsNOT DISCUSSED FOR THE MOMENT

6.7. Non-linear Analysis of Valve Controlled ActuatorsWe, in our course (and this text), discussed

nonlinear analysis in Section 6.2. Equation (5.33),

used in Sec. 6.2., concides with (6.93) being

crucial in the current section 6.7.

7. ELECTROHYDRAULIC SERVOVALVES

As we have mentioned, the valve and the actuator were controlled

by

- spoll stroke xv , or- forceFi imposed on the valve spool.

In any case, there is a question:

Question: How to generate the appropriate stroke or force ? ?

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ANSWER: Some kind of motor is needed to create the force (or

torque) and consequently the stroke ! It is called the torque

motor.

So, servovalve means the valve (one or two stages)

plus the torque motor .

7.1. Types of Electrohydraulic Servovalves

Single-stage servovalve

The torque motor is directly connected to the spool valve.

Torque motors have limited power capabilities. This- limits the torque/force that can be generated,

- limits the flow capacity of the valve, and

- may lead to stability problems in some applications.

Two-stage servovalve

Force/torqueTorque motor

Spool of the

valve

Torque motor

Stage 1

Valve ofdifferent type

Stage 2

Spool valve

force/torqueamplified

force/torque

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Stage 1 is a hydraulic preamplifier. It augments the

force/torque generated by the motor to the level that can

overcome all the problems: flow forces, stiction, acceleration,

vibrations, etc.

Stage 1 can be:

- spool valve,

- jet pipe valve, and

- flapper valve.

Stage 2, the main spool, is alvays a spool valve.

Types of feedback between the two stages (most common types):

- direct feedback,

- force deedback, and

- spring centered spool.

With direct feedback, the main spool follows the first stage in

a one-to-one relation. We talk about hydraulic follower.

With force feedback, there is a deformable element, a spring,

between the two stages.

7.2. Permanemnt Magnet Torque Motor

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