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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)
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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.
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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)
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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:
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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)
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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)
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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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