Prof. Veljko Potkonjak
description
Transcript of 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
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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
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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:
mechanical motion that controls the valve – spool stroke xv
flow to source
return
supply
flow to actuator
Fig. 5.1 (c)
mechanical motion that controls the valve – spool stroke xv
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:
mechanical motion that controls the valve – spool stroke xv
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)
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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 .
Newton’s law gives:
vfvfvsi xKxBxMF ++= (5.50)
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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
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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
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♦ 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 −= .
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♦ 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
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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) : ⎟⎠⎞
⎜⎝⎛∂∂
−==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)
Newton’s law gives:
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