Prof. Veljko Potkonjak

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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 of actuator 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 the 3-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

)()()(

volumeVweightGdensityweight =γ ,

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

)()()(

volumeVmassmdensitymass =ρ ,

typically 3424 /......../sec1078.0 mkginlb =×= −ρ

g⋅= ργ )/81.9( 2smg = (2.1) 2.2. Equation of State ● Expression that relates density ρ (or volume V), pressure P , and

temperature T . Volume (and density) changes little. So, a linear approximation is justified:

)()( 000 TTT

PPP PT

−⎟⎠⎞

⎜⎝⎛∂∂

+−⎟⎠⎞

⎜⎝⎛∂∂

+=ρρρρ (2.2)

or

))()(11( 000 TTPP −−−+= αβ

ρρ (2.3)

where

4

TT V

PVP⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= 00 ρρβ ,

PP TV

VT⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

−=00

11 ρρ

α (2.4), (2.5)

β – 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 hydraulic

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

the bulk modulus.

▪ ρ and β depend 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)!

ρ lnβ

T T

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♦ Friction force is proporional to the contact area A and to the velocity

x , and inversly proportional to the film thickness rC :

xCDL

CxAF

rr

µπµ == , µ – absolute viscosity (coeff. of visc.) (2.7)

ρµ

=v – kinematic viscosity (2.8)

▪ µ depends on the temperature:

µ = µ0 e – λ (T - T0) (2.9)

µ

T

leakage

leakage of liquid motion x ,

velocity x

Cr – radial clearance

F

L

D

Piston in a cylinder

resistive friction force

Fig. 2.2

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2.4. Thermal Properties ► Specific heat is 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 lowered by 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 modulus βe can be

found in the form:

)11(111lgt

g

lce VV

βββββ−++= + (2.20)

where: βc – the bulk modulus for the container, βl – for the liquid, βg – for the gas; Vg – the volume of the gas, and Vt – the total volume. Since gl ββ >> , (2.20) becomes:

)1(111gt

g

lce VV

ββββ++= (2.21)

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

lce βββ

111+= (2.22)

∆Vt ∆Vc

liquid, volume Vl

gas pocket, volume Vg

liquid

∆Vg

gas

Fig. 2.4

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2.6. Chemical and Related Properties - 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

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3. FLUID (LIQUID) FLOW FUNDAMENTALS It is assumed that the general theory of fluid flow is elaborated in the previous courses. Among numerous problems, we highlight here the topic: 3.4. Flow Through Orifices – Turbulent Flow

0

2

AACc = – contraction coefficient (3.28)

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

- Bernulli’s equation )(221

21

22 PPuu −=−

ρ (3.29)

- Equation of incompressibility 332211 uAuAuA == (3.30)

- Volumetric flow rate (the flow) 22uAQ = - Contaction coefficient (3.28) 02 / AACc = - velocity coefficient 98.0≈vC (sometimes adopted 1≈vC ) (velocity is slightly smaller due to friction)

A2 , jet area is minimum jet area A0

vena contracta – the jet area is minimimum

Fig. 3.10.1 2 3

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and we obtain

)(2210 PPACQ d −=

ρ (3.33)

where

210

2 )/(1 AACCCC

c

cvd

−= (3.34)

is the discharge coefficient. Since 1≈vC and 10 AA << it follows that cd CC ≈ . If 10 AA << , the theoretical value for the the discharge coefficient for all sharp-edged orifices, regardless of the geometry, is 6.0611.0)2/( ≈=+= ππcC .

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4. HYDRAULIC PUMPS and MOTORS ♦ Conversion of energy:

◦ Pump: mechanical energy hydraulic energy

◦ Motor (actuator): hydraulic energy mechanical energy

our primary interest hydrodynamic machines (turbines, etc.) ♦ Hydraulic machines positive displacement mach.! limited travel machines ♦ continuous travel machines rotary machines ♦ piston machines (translation)

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

cylinder

xp

piston position

fluid IN fluid OUT

fluid IN : pressure P1

fluid OUT : pressure P2

Single rod actuator

Double rod actuator

forward chamber

backward chamber

motion

motion

pressure force

load force

piston

Fig. 4.1

piston parameters: Mt – mass of the

piston plus refered masses

Ap – effective piston area

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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 and P2 in the backward chamber. 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 two

pressures):

21 PPPL −= ▫ Pressure force (generated force) is

Lpg PAF = ▫ Load force or output force is FL

▫ There is a spring effect associated with the piston: Kxp , where

K is the gradient (stiffness). ▫ There is a viscous damping effect associated with the piston:

pp xB , where Bp is the viscous damping coefficient. ▪ Dynamics of the motor (i.e. dynamics of the piston)

Newton’s law gives:

LpppptLp FKxxBxMPA +++= (A.1)

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♦ Vane rotary actuator – limited travel mnachine

▫ Pressure torque force (generated torque) is rPA Lpg =τ ▫ Load torque or output torque is τL ▫ There is a torsion spring effect associated with the rotor: Kφ ,

where K is the gradient (torsion stiffness).

rotor

backward chambre

forward chambre

Vane

Rotation angle φ

Pressure makes

a resultant force and

consequently a torque

fluid IN : pressure P1

fluid OUT :pressure P2

r

rotor parameters: It –

moment of inertia

Ap – effective vane area

housing(stator)

Fig. 4.2 a

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mechanic energy

OUTPUT:mechanic

energy

▫ There is a viscous damping effect associated with the piston: ϕB , where B is the viscous damping coefficient.

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

Newton’s law for rotation gives: LtLp KBIrPA τϕϕϕ +++= (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.

hydraulic energy

Fig. 4.2 (b) .

Fig. 4.3 .

Figs. 4.4 – 4.12

PUMP (source of hydro energy): converts

mechanical energy into

hydraulic energy

HYDRO ACTUATOR: converts hydro

energy into mechanical

energy

ELECTRIC MOTOR (source of

mechanic energy): converts electric

ener. into mechanical ener.

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5. HYDRAULIC CONTROL VAVES

♦ Valves are are the interface between the the sorce of hydraulic

energy and the actuator.

▪ 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 the

delivery of power to the actuator.

control the delivery of

energy

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

source of hydrauluic

energy

Oil supply (pressure supplay)

VALVE Actuator

oil flow oil flow

Unit which creates the mechanical motion

that controls the valve

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

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

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:

flow to source

return

supply

flow to actuator

mechanical motion that controls the valve – spool stroke xv

Fig. 5.1 (a)

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

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


flow to source

return

supply

flow to actuator

Fig. 5.1 (c)


flow to source

return

supply flow to actuator

Fig. 5.1 (b)

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

(e) two-jet flapper valve:


flow to source

return

supply flow to actuator

Fig. 5.1 (d)

flapper

supply

pivot

Fig. 5.1 (e)

flow to actuator

motion of the flapper controls the valve

return

to source

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(f) jet pipe valve:

♦ 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)

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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 or zero lapped valve

▪ If land width > port : closed center or overlapped valve

width port

width port

width port

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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 stroke xv

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 is

21 PPPL −= (5.3)

L2

L1

L2

L1

spool stroke vx

P2

P1

3

2

1

4Supply: - flow Qs - pressure Ps

Return: - flow Qs - pressure P0 ≈ 0

To actuator: - flow QL - pressure P1

From actuator:- flow QL - pressure P2

PL= P1 – P2

Force Fi

Fig. 5.3.

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▪ According to equation (3.33), the flows through the valving

orifices is:

)(2111 PPACQ sd −=

ρ (5.4)

)(2222 PPACQ sd −=

ρ (5.5)

2332 PACQ d ρ

= (5.6)

1442 PACQ d ρ

= (5.7)

▪ The orifices areas depend on the valve geometry and the valve

displacement (spool stroke) xv :

)(,)(,)(,)( 44332211 vvvv xAAxAAxAAxAA −==−== (5.8) ▪ The set (5.1) – (5.8) copntains 11 equations that can be combined

to give the load flow as a function of the spool stroke xv and the diffeerential pressure PL:

),( LvLL PxQQ = (5.9)

The plot of (5.9) is known as as the pressure-flow curves for the valve and is a complete description of stady state valve performance. All of the performance parameters, such as valve coefficients, can be obtained from such curves.

▪ In the vast majority of cases, the valving orifices are matched and

symmetrical. Matched orifices require 4231 , AAAA == (5.10), (5.11)

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and symmetrical orifices require

)()(,)()( 4321 vvvv xAxAxAxA −=−= (5.12), (5.13) Therefore, in the neutral position of the spool, all four areas are aqual: 4,3,2,1,)0( 0 == jAA j 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 the area 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:

21Ls PPP +

= (5.18)

22Ls PPP −

= (5.19)

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

as 21 QQQ s += (5.20)

and as

21 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

)(1)(121 LsdLsdL PPACPPACQ +−−=

ρρ (5.22) and similar treatment yields (using (5.20) and (5.21)):

)(1)(121 LsdLsds PPACPPACQ ++−=

ρρ (5.23)

● Linearization – Valve Coefficients ♦ Sometimes, a nonlinear form of the matyhematical model causes

problems and linearization is needed. ▪ Equation (5.9), describing the load flow, can be expanded in the

Taylor’s series about a particular operating point 1: 111 ),( LLv QPx →

producing

+∆⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∆⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+= LL

Lv

v

LLL P

PQx

xQQQ

111

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▪ If the working mode is such that ),( Lv Px are kept in the vicinity

of the operating point 1, i.e. close to ),( 11 Lv Px , then ),( 11 Lv Px ∆∆ will be small and it is jusrtified to keep only the

linear terms in the Taylor’s expansion. Thus:

LL

Lv

v

LLLL P

PQx

xQQQQ ∆⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∆⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∆=−11

1 (5.24)

The partial derivatives are obtained analytically or numerically.

▪ Valve coefficients (!!!)

- Flow gain : 0>∂∂

≡v

Lq x

QK (5.25)

- Flow-pressure coef. 0>∂∂

−≡L

Lc P

QK (5.26)

- Pressure sensitivity c

q

v

Lp K

KxPK =

∂∂

≡ (5.27), (5.28)

► Flow gain affects 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.

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▪ Now, (5.24) becomes

LcvqL PKxKQ ∆−∆=∆ (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:

,

,

,

1

1

11

LLLL

LLLL

vvvv

QQQQ

PPPP

xxxx

=−=∆

=−=∆

=−=∆

and accordingly, (5.29) becomes

LcvqL PKxKQ −= (A.3)

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5.3. Critical-Center Spool Valve Analysis ● Pressure-Flow Curves We are going to derive the exact form of the relation (5.9)

),( LvLL PxQQ =

for the case of a critical-center valve. ▪ We assume the ideal valve geometry, and hence, leakage iz zero:

0,0,0 42 >== vxforQQ , (so, (5.1) becomes QL= Q1 ) and

0,0,0 31 <== vxforQQ , (so, (5.2) becomes QL= – Q2 = – Q4) ▪ Substituting (5.18), (5.4) into (5.1), one obtains

0,2

21 >⎟

⎠⎞

⎜⎝⎛ −

= vLs

dL xforPPACQρ (5.30)

▪ For negativevalve displacements, (5.18), (5.7), substituting into

(5.2), yield

0,2

22 <⎟

⎠⎞

⎜⎝⎛ +

−= vLs

dL xforPPACQρ (5.31)

▪ For symmetrical valve, eq. (5.12) holds and (5.30) and (5.31) can

be written as a single relation:

⎟⎟⎠

⎞⎜⎜⎝

⎛−== L

v

vs

v

vdLvLL P

xxP

xxACPxQQ

ρ1),( 1 (5.32)

30

QL

▪ If rectangular ports are used with an area gradient w, one obtains

⎟⎟⎠

⎞⎜⎜⎝

⎛−== L

v

vsvdLvLL P

xxPxwCPxQQ

ρ1),( (5.33)

This is the pressure-flow curve mentioned earlier as eq. (5.9). Family of curves, for different xv is shown in Fig. 5.4.

PL

Ps

– Ps

xv increasing in positive sense

xv increasing in negative sense

Fig. 5.4

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● Valve Coefficients We recall the linearized form (A.3),

LcvqL PKxKQ −= (A.3) and look for the coefficients.

▪ Differentiation of (5.33) gives

► )(1Lsd

v

Lq PPwC

xQK −=∂∂

≡ρ (5.35)

► )(2))(/1(

Ls

Lsvd

L

Lc PP

PPxwCPQK

−−

=∂∂

−≡ρ

(5.36)

► v

Ls

c

qp x

PPKK

K )(2 −== (5.37)

▪ For the null operating point (being the most important) i.e. for

0,0,0 === LLv QPx , the null coefficients for the ideal critical-center valve are:

► ρs

dqPwCK =0 (5.38)

► 00 =cK (5.39)

► ∞=0pK (5.40) ▪ The computed value for 0qK is close to a realistic value (obtained

by tests). However, the computed values for 0cK and 0pK are far from the values obtained by testing a realistic valve. => So, we have to consider leakage !!!

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● Leakage Characteristics of Practical Critical-Center Valves

– just some comments – ▪ Ideal valve ↔ ideal geometry => no leakage ▪ Real valve ↔ radial clearance => leakage ▪ Example: Realistic pressure sensitivity curve for blocked lines

(so, only leakege flolw exists)

Ps

– Ps

load pressure difference PL

valve stroke xv

the slope is not infinite, i.e., Kp ≠ ∞

Fig. 5.5

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● Stroking Forces – Dynamics of the Valve (Spool)

Analysis is based on the Figure 5.3. ▪ Mathematical description:

▫ Force Fi is imposed to control the spool motion (stroke) i.e.

to control the valve ▫ There are flow forces that oppose the spool motion. These

forces are derived from eqs. (5.90) and (5.93) in Section 5.6. and from (5.48) and (5.49) in Section 5.3. The result is:

◦ There is a spring effect associated with the spool motion (like a centerung spring). It is the steady-state flow force: Kf xp, where )(cos2 Lsvdf PPwCCK −= θ is the gradient (like a stiffness). ◦ There is a viscous damping effect associated with the spool

motion. It is the transient flow force: vf xB , where )()( 12 Lsdf PPwCLLB −−= ρ is the damping coefficient.

▫ Mass Ms defines the inertia: vs xM .


vfvfvsi xKxBxMF ++= (5.50)

34

5.4. Open-Center Spool Valve Analysis HOMEWORK 1a - Ramadan 5.5. Three-Way Spool Valve Analysis HOMEWORK 1b - Mohamad 5.6. Flow Forces on Spool Valves HOMEWORK 1c - Ismail 5.7. Lateral Forces on Spool Valves 5.8. Spool Valve Design NOT DISCUSSED FOR THE MOMENT 5.9. Flapper Valve Analysis and Design

Single-jet, Double-Jet, Flow Forces HOMEWORK 1d - Abdulhalim

35

6. HYDRAULIC POWER ELEMENTS 6.1. Valve Controlled Rotary Motor NOT DISCUSSED FOR THE MOMENT 6.2. Valve Controlled Piston

NOTE a difference regarding previous figures. The forward flow (to the actuator: Q 1) is not equal to the return flow (from the actuator: Q 2). Previously, it was equal: Q 1= Q 2= Q L This is due to some effects that have been neglected in the previous discussions and now we take care of them. These effects are:

- Leakage, - Compression.

xp Return line: Q 2 , P2

Forward line:

P1 , Q 1

Cylinder with a Piston

VALVE

Supply Ps

Fig. A.1

36

♦ Valve controlled flow – Linear analysis ▪ Starting from relation (A.3) ( LcvqL PKxKQ −= ), one may

write experessions for Q 1 and Q 2 : 11 2 PKxKQ cvq −= (6.1)

22 2 PKxKQ cvq += (6.2)

- 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 in PL 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 :

221 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 −= .

37

♦ Valve controlled flow – Non-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

xxPxwCPxQQ

ρ1),( (5.33)

♦ Flow through the actuator – continuity relations .

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

Force Fi and motion xv (to control the valve)

PL = P1 – P2

P2 , V 2 P1 , V 1

xp

VALVE

Fig. 6.6

Load. - Force FL

- spring effect - damping effect

Piston parameters: Mt – mass of the

piston plus refered masses

Ap – effective piston area

Forward line:

P1 , Q 1

External leakage External leakage

Return line: Q 2 , P2

Internal leakage

Supply

38

▪ 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) : ⎟⎠⎞

⎜⎝⎛∂∂

−==VPV0β => ⎟

⎠⎞

⎜⎝⎛−==

dtdVdtdPV

//

0β => dtdPV

dtdV

β0−=

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

dtdPV

dtdVPCPPCQ

eepip

1111211 )(

β+=−−− (6.27)

dtdPV

dtdVQPCPPC

eepip

2222221 )(

β+=−−− (6.28)

where

V1 –volume of the chamber 1 of the actuator plus related volumes: 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

pp xAVV += 011 (6.29)

pp xAVV −= 022 (6.30)

39

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

dtdx

Adt

dVdt

dxA

dtdV p

pp

p −== 11 ; ; ⎟⎠⎞

⎜⎝⎛ −=

dtdV

dtdV 21 (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 both chambers.

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

obtain

dtPPdxA

dtPPdVPP

CC

dtdx

A

QQQ

e

pp

e

epip

pp

L

)(2

)(2

))(2

(

221210

21

21

++

−+−++

=+

=

ββ

▪ If 0VxA pp << , the last term may be neglected ▪ ▪ So, we finaly come to

Le

tLtpppL PVPCxAQ

β4++= (6.33)

where 2/epiptp CCC += is the total leakage coefficient.

40

♦ Mathematical description of the piston dynamics (this has been already discussed in Ch. 4 – we repeat here): ▫ Differential pressure PL (difference between the two

pressures):

21 PPPL −=

▫ Pressure force (generated force) is

Lpg PAF =

▫ Load force or output force is FL ▫ There is a spring effect associated with the piston: Kxp , where

K is the gradient (stiffness). ▫ There is a viscous damping effect associated with the piston:

pp xB , where Bp is the viscous damping coefficient. ▪ ▪ Dynamics of the motor (i.e. dynamics of the piston)


LpppptLp FKxxBxMPA +++= (A.1)=(6.34)

41

6.A. Mathematical Model of the Valve-Controlled Actuator

♦ Actuator controlled by the valve stroke

As 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):

Le

tLtpppL PVPCxAQ

β4++= (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

xxPxwCPxQQ

ρ1),( (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.

▪ State variables are piston position, its velocity, and load

pressure: px , px , LP .

▪ Control input is the valve spool stroke, vx .

♠ Question: If the spool stroke controls the actuator, how to generate the appropriate spool stroke ?

42

♠ ANSWER: We use a force to move the spool ! =>

♦ Actuator and valve controlled by the force imposed to the spool

Figure 5.3 showed that the spool stroke is generated by the force Fi imposed to the spool.

(IV) ► We relate the force Fi with the spool motion xv by

dynamic equation (5.50):

vfvfvsi xKxBxMF ++= (5.50) ● Eqs. (I)–(IV), i.e. - (6.34), (6.33), (6.3), (5.50) (linear case) or

` - (6.34), (6.33), (5.33), (5.50) (non-lin.), define the mathematical model.

▪ State variables are piston position and velocity, load

pressure, spool position (stroke) and velocity: px , px , LP ,

vx , vx , ▪ Control input is the force imposed to valve spool: Fi.

♠ Question: If the force imposed to the spool controls the valve and the actuator, how to generate the appropriate force ? ?

♠ ANSWER requires a more detailed analysis of the valve. Some

kind of motor will be needed to create the force ! This will be elasborated in Chapter 7.

43

♦ Important notes about the load.

▪ The model derived (eqs. (I) – (IV)) includes the load force FL. It is 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 the interaction between the two systems (actuator and load – see Fig. A.2). According to the law of action and reaction, the force that acts from 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 force FL is unknown and has to be expressed from

the 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.

- Let ),( ,21 zzz = be state vector and let u be the input control signal.

action FL

Load

reaction FL

Actuator

Fig. A.2

44

▪ The canonic form is then:

),( uzfz = for nonlinear systems, (A.5)

and

uEzDz += for linear systems, (A.6)

where D and E are system matrices. ▪ The model that we discuss includes the load force FL , and it

may introduce additional state variables. So, with the force FL (A.5) and (A.6) become

),,( LFuzfz = for nonlin. case, (A.7)

and

LHFuEzDz ++= for linear case. (A.8) ♦ Actuator controlled by the valve stroke

The model involves (I) – (III) .

▪ The state variables and state vactor are:

pxz =1 , pxz =2 , LPz =3 , ),,( Lpp Pxxz = (A.9) ▪ Control input is the valve spool stroke,

vxu = . (A.10)

45

▪ Let us rewrite (I)-(III) acoording to notation (A.9) and (A.10):

(I) Lptp FKzzBzMzA +++= 1223

(II) 332 4zVzCzAQ

e

ttppL β

++=

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

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 3

1 zuuPuwCQ sdL ρ (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 =

Ltt

p

t

p

t

FM

zMA

zMB

zMKz 1

3212 −+−−= (A.11)

uV

KzV

CKzV

Azt

eq

t

etpc

t

ep

βββ 44)(4323 ++−−=

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

Lt

t

eq

t

etpc

t

ep

t

p

t

p

t

FM

u

VKz

zz

VCK

VA

MA

MB

MK

zzz

HED

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+−−

−−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0

10

400

4)(40

010

3

2

1

3

2

1

βββ

(A.12)

46

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

21 zz =

Ltt

p

t

p

t

FM

zMA

zMB

zMKz 1

3212 −+−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−−= 3323

1444 zuuPuwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep ρ

βββ

(A.13) ◘ How to handle the load ? Let us explain this by examples !

♠ EXAMPLE 1 Form the complete mathematical model for the system of Fig. A.3! The control input is the valve stroke. NOTE: The load does not introduce any new state variable.

Load

Rolling without sliding

Cylinder:mass m radius r

Fig. A.3

Actuator

47

○ The actuator is modeled by (A.11) for a linear analysis or (A.13) for a nonlinear analysis.

○ The model includes the load force FL .

● We now look for the mathematical model of the load in order to express the load force FL .

▫ Eqations of load dynamics:

frL FFam −= , for translation

rFI fr=α , for rotation (about the center) where a is the acceleration, α is the angular acceleration, and I is the moment of inertia. Note that there is no sliding and accordingly NFfr µ≠ (thus

friction frF is unknown).

▫ Having in mind: ra /=α and 2

21 rmI = , the above equations

yields:

maFL 32

= , maFfr 31

= .

FL Load force i.e. actuator output force FL

Friction force (dry) Ffr

48

▫ The motion of the wheel center equals the the piston motion xp, and so:

2zxa p == => 232 zmFL = .

One can see that FL does not introduce new state variables but depends on the existing one. ▪ 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(z

mMA

zmM

Bz

mMKz

t

p

t

p

t ++

+−

+−=

uV

KzV

CKzV

Azt

eq

t

etpc

t

ep

βββ 44)(4323 ++−−=

or in a matrix form

u

VKz

zz

VCK

VA

mMA

mMB

mMK

zzz

ED

t

eq

t

etpc

t

ep

t

p

t

p

t

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+−−

++−

+−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

βββ

400

4)(40

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

010

3

2

1

3

2

1

which is the final form (A.6).

49

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

21 zz =

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

mMA

zmM

Bz

mMKz

t

p

t

p

t ++

+−

+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−−= 3323

1444 zuuPuwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep ρ

βββ

which is the final form (A.5).

♠ EXAMPLE 2 Form the linear mathematical model for the system of Fig. A.4 ! 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 .

NOTE: FL is in reverse direction (negative)

FL

x2r

Load

Body: mass m2

Cylinder: mass m1

radius r

Fig. A.4

Actuator

xp= x1

50

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

(A.13) for a nonlinear analysis. ○ The model includes the load force FL .

● We now look for the mathematical model of the load in order to express the load force FL .

▫ Eqations of load dynamics:

- for the wheel: translation and rotation LFFam −= 111

rFI 11 =α - for the body (translation only)

1222 Fgmam −=

▫ Accelerations are:

pxa =1 , rpr xxaaxa 22122 +=+== , rr xa 22 =

x2= =xp+ x2r

m1g

x2rF1

F1

xp= x1

FL

FL

m1

m2

51

▫ Having in mind: ra r /2=α and 211 2

1 rmI = , the equations of

load dynamics, after some transformations, become

Lrp Fxmxm −=− 211 21

gmxmmxm rp 22122 )21( =−+

▫ 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 −=− 5121 21

gmzmmzm 251222 )21( =−+

with 54 zz = , or, after additional transformation,

gmm

mmzmm

mmmFL12

212

12

2121

)2/1()2/1(

)2/1()2/1()2/3(

−+

−−

−=

212

2

12

25 )2/1()2/1(

zmm

mgmm

mz−

−−

= (*)

54 zz =

52

▪ For a linear analysis, (A.11) is combined with the above three relations. First, FL from the first relation is substituted into the second equation from (A.11) (note that the sign of FL 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 +++=

where

1)2/1(2

21)2/1(21)2/3(

21

mm

mmmtM

KD

−

−+

−= ;

1)2/1(2

21)2/1(21)2/3(

22

mm

mmmtM

pBD

−

−+

−=

1)2/1(2

21)2/1(21)2/3(

23

mm

mmmtM

pAD

−

−+

= ;

gG

mm

mmmtM

mm

mm

1)2/1(2

21)2/1(21)2/3(

1)2/1(2

21)2/1(

2

−

−+

−=

t

ep V

AD β432 −= ;

t

etpc V

CKD β4)(33 +−= ; t

eq V

KE β43 =

53

2112

251 )2/1(

Dmm

mD−

−= ; 2212

252 )2/1(

Dmm

mD−

−=

2312

253 )2/1(

Dmm

mD−

−= ; 212

2

12

25 )2/1()2/1(

Gmm

mgmm

mG−

−−

=

The obtained model describes the dynamics of the entire system. The model is in a linear canonical form, like (A.6). The matrix form is

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

5

2

3

5

4

3

2

1

535251

3332

232221

5

4

3

2

1

00

0

00

00

00100000000000010

G

Gu

E

E

zzzzz

DDDD

DDDDD

zzzzz

♦ Actuator and valve controlled by the force on the spool

The model involves (I) – (IV) .

▪ The state variables and state vactor are:

pxz =1 , pxz =2 , LPz =3 , vxz =4 , vxz =5 , ),,,,( vvLpp xxPxxz = (A.14)

▪ Control input is the force to valve spool,

iFu = . (A.15)

54

▪ Let us rewrite (I)-(IV) acoording to notation (A.14) and (A.15):

(I) Lptp FKzzBzMzA +++= 1223

(II) 332 4zVzCzAQ

e

ttppL β

++=

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

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 3

4

44

1 zzzPzwCQ sdL ρ (for nonlinear analysis)

(IV) 455 zKzBzMu ffs ++= From (A.14), it follows that

21 zxz p == and 54 zxz v ==

◘ By combining the above relations, for the linear case one gets:

21 zz =

Ltt

p

t

p

t

FM

zMA

zMB

zMKz 1

3212 −+−−=

432344)(4 zV

KzV

CKzV

Azt

eq

t

etpc

t

ep

βββ++−−= (A.16)

54 zz =

uM

zMB

zMK

zss

f

s

f 1545 +−−=

55

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

Lt

s

t

eq

t

etpc

t

ep

t

p

t

p

t

FM

u

Mzzzzz

VK

VCK

VA

MA

MB

MK

zzzzz

HED

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡−

+

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+−−

−−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

000

10

10000

10000

044)(40

00

00010

5

4

3

2

1

5

4

3

2

1

βββ

(A.17)

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

21 zz =

Ltt

p

t

p

t

FM

zMA

zMB

zMKz 1

3212 −+−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−−= 3

4

44323

1444 zzzPzwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep ρ

βββ

54 zz =

uM

zMB

zMK

zss

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.

56

6.3. Three-Way Valve Controlled Piston 6.4. Pump Controlled Motor NOT DISCUSSED FOR THE MOMENT 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 of springs (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

QL

kn

xnx2x1

b1

k2

m1

k1 xp

Fi

xv

QL

Valve

m2b2 mn

bn

Fig. 6.8

load FL

57

▪ 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) . This model includes the load force FL. ▪▪ Dynamics of the load can be described by the following set of n

equations:

2springinforce)(1springinforce

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

LFpp −+−−−+−=

=

3springinforce2springinforce

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

. . .

. . .

. . .

nnnnnnnnn xxbxxkxm

springinforce

)]()([ 11 −+−= −−

(A.19) ▪▪ The complete mathematical model (actuator plus load) includes:

- eqs. (I) – (IV) , fot the acatuator and valve, plus - set of n equations (A.19).

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

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

58

▪▪ The load has intruduced additional degrees of freedom and

accordingly additional state variables. The entire set of state variables (vector z) 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. 6.6. Pressure Transients in Power Elements NOT DISCUSSED FOR THE MOMENT 6.7. Non-linear Analysis of Valve Controlled Actuators

We, 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.

59

7. ELECTROHYDRAULIC SERVOVALVES As we have mentioned, the valve and the actuator were controlled by

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

In any case, there is a question: ♠ Question: How to generate the appropriate stroke or force ? ? ♠ 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.

Force/torque Torque motor Spool of the

valve

60

♦ Two-stage servovalve

▫ 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.

Torque motor Stage 1 Valve of

different type

Stage 2 Spool valve

force/torque amplified force/torque

61

7.2. Permanemnt Magnet Torque Motor

62

Documents

Prof. Veljko Potkonjak