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Dynamic SystemInvestigation
Process
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K. Craig 5Typical Fluid Power System
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Advantages / Disadvantages / Challenges• Fluid Power is the transmission of forces and motions
using a confined, pressurized fluid.
– In hydraulic fluid power systems the fluid is oil, or,
less commonly, water.
– In pneumatic fluid power systems the fluid is air.
• Fluid Power System Advantages – High Power Density
• Fluid power is ideal for high speed, high force, high
power applications. Compared to all otheractuation technologies, including electric motors
which are limited by magnetic saturation, fluid
power is unsurpassed for force and power density.
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– Responsiveness and Bandwidth of Operation
• Fluid power systems have a higher bandwidth thanelectric motors and can be used in applications
that require fast starts, stops, and reversals, or that
require high-frequency oscillations.
– High Accuracy and Precision
• Oil has a high bulk modulus; hydraulic systems
can be accurately and precisely controlled.
• Advances in pneumatic components and controltheory have opened up new opportunities for
pneumatic control applications.
– Heat Dissipation and Lubrication Ensures Reliability• Fluid circulating to and from an actuator removes
heat generated by the actuator doing work, and
also lubricates moving parts of the components.
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• Heat is the predominant damaging mechanism
in electric and electronic systems andelectromechanical actuators and motors have
limited ability to dissipate heat generated.
– Compactness, Light Weight, and Flexibility
• Fluid power cylinders and motors are relatively
small and light weight. Flexible hoses allow
compact packaging.
– Stiffness• Hydraulic drives are stiff with respect to load
disturbances; stiffness is the slope of the speed
– torque (force) curve. Control gains required
in a high-power hydraulic control system would
be significantly less than the gains required in a
comparable electromagnetic control system.
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• Fluid Power System Disadvantages
– Electric power is more readily available, cleaner and
quieter, and easier to transmit. Hydraulic systems
require pumps.
– Oil leakage, flammability, and fluid contamination – Fluid cavitation and entrained air
– Challenging physics leads to more difficult modeling and
control.
• Fluid Power System Challenges
– Increase Efficiency
– Compact Energy Storage and Compact Power Sources
– Noise, Vibration, Leakage, Safety, and Ease of Use
– Create Portable, Untethered, Human-Scale Applications
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Pump-Controlled vs. Valve-Controlled Systems
• Hydraulic actuation devices may be linear (piston / cylinder)
or rotary (motor) and these may be controlled by a pump or
a valve, giving four basic hydraulic system combinations.
• The pump-controlled system consists of a variable-delivery
pump supplying fluid to an actuation device. The fluid flow
is controlled by the stroke of the pump to vary output speed,and the pressure generated matches the load.
• The valve-controlled system consists of a servovalve /
proportional valve controlling the flow from a hydraulic
power supply to an actuation device. The hydraulic powersupply is usually a constant-pressure type with two basic
configurations.
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– a constant delivery pump with a pressure relief
valve and
– a variable-delivery pump with a stroke control to
regulate pressure.
• Pump-Controlled vs. Valve-Controlled Comparison – Response Speed
• A pump-controlled system has slow response
because pressures must be built up, containedvolumes are large, and the stroke servo has
comparatively slow response.
• A valve-controlled system has fast response to
valve and load inputs because containedvolumes are small and the supply pressure is
constant.
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– Efficiency
• A pump-controlled system is much more efficientsince both pressure and flow are closely matched to
load requirements.
• A valve-controlled system is much less efficient (≈ 2/3)
because of the constant supply pressure regardless ofload, the large pressure drop across the control valve,
and significant leakage.
– Size• A pump-controlled system has a bulky power element
which makes the application difficult if the pump is
close coupled to the actuator.
• A valve-controlled system has a small and light powerelement but a bulky hydraulic power supply is
required.
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– Heat Dissipation
• A pump-controlled system requires an auxiliarypump and valving to provide oil for
replenishment and cooling.
• A valve-controlled system has a build-up of oiltemperature because of inefficiency which
necessitates heat exchangers to dissipate the
wasted energy.
– Cost and Complexity• A pump-controlled system generally requires
an electrohydraulic servo to stroke the pump
which increases system cost and complexity.
• Several valve-controlled systems can be
powered by a single hydraulic power supply.
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– A valve-controlled system has a build-up of oil
temperature because of inefficiency whichnecessitates heat exchangers.
• Cost and Complexity
– A pump-controlled system generally requires anelectrohydraulic servo to stroke the pump which
increases system cost and complexity.
– Several valve-controlled systems can be fed from
a single hydraulic power supply.
• Here, we focus on a valve-controlled, linear actuation
system, with a 4-way, 3-position valve, a single-rod,
double-acting cylinder, and a constant-displacementpump with a pressure-relief valve.
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Physical System
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Physical System Components• Single-Rod, Double-Acting Hydraulic Cylinder
• 4-Way, 3-Position, Solenoid-Operated,Spring-Return, Proportional Control Valve
• Check Valve
• Pilot-Operated Pressure Relief Valve• Fixed-Displacement Hydraulic Pump (Gear
Pump) with Motor
• Transmission Lines• Option: Accumulator
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System Description• This system is one of the most basic hydraulic control
systems.
– It uses a standard four-way, three-position valve tocontrol the output characteristics of a single-rod,
double-acting linear actuator by controlling the
volumetric flow of hydraulic fluid into and out of the
actuator.
– The load to be moved by the actuator is shown as a
single mass-spring-damper system with a load-
disturbance force. – A fixed-displacement pump provides a constant
volumetric flow rate into the supply line to the valve.
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– The pump is sized according to its volumetric
displacement and is driven by an external powersource at an angular velocity ω.
– The pressure in the discharge line of the pump is
controlled using a pilot-operated pressure relief valve set at the desired supply pressure.
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System Diagram Used For Analysis
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Analysis• Force Analysis
– F = load force = dynamic term that represents the
force of the load on the actuator
– m = inertia of the load; we often neglect the inertia
of the actuator itself because it is much smaller
than the actual forces that are generated by theactuator. This high force-to-inertia ratio is one of
the principle advantages of using a hydraulic
system as opposed to an electric system.
– c = equivalent viscous damping constantrepresenting energy dissipation effects associated
with the load and moving parts of the actuator.
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– k = equivalent spring constant representingcompliance / flexibility effects associated with the
load and moving parts of the actuator
– Apply Newton’s Second Law to obtain the
equation of motion:
– P A and PB are the fluid pressures on the A and Bsides of the actuator.
– ηaf is the force efficiency of the actuator.
– F0 is the nominal spring or bias load that is appliedto the actuator when y equals zero.
2
af A A B B 02
d y dym c ky (A P A P ) F F
dt dt
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– Nominal Steady-State Operating Conditions of the
System:
• Note that we assume that the actuator pressures P A and PB each
come to Ps/2 at the servo rest condition.
– The nominal force exerted on the actuator by the load isthen given as:
– The equation of motion then becomes:
– We see that the required inputs for adjusting the position
of the load are P A and PB. These pressures result fromthe changing flow and volume conditions within the
actuator itself.
sA B
Py 0 P P F 02
saf A B 0
P(A A ) F
2
2
s saf A A B B2
P Pd y dym c ky A P A P F
dt dt 2 2
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• Background on Hydraulic Cylinders
– The major function is to convert hydraulic power
into linear mechanical force to perform work or
transmit power.
– Hydraulic power (P AQ A) is delivered to the systemthrough port A. As the piston moves to the right,
power (PBQB) is expelled from the actuator
through port B. Heat is lost to the atmosphere
resulting from viscous shear and Coulomb friction.The useful output power of the linear actuator is
Fv.
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– The overall efficiency η of the linear actuator is
defined as the ratio of the useful output power to thesupplied input power.
– The overall efficiency η can be separated into twocomponents: the volumetric efficiency ηv and the
force efficiency ηf .
– In general, the volumetric efficiency will be less than
unity owing to fluid compression and leakage past thepiston. The force efficiency will be less than unity
owing to Coulomb friction and viscous shear.
A A
Fv
P Q
Av f v f
A A A
A v FQ P A
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• Pressure Analysis
– We assume that the pressure transients that result fromfluid compressibility are negligible.
• This assumption is especially valid for a system
design in which the transmission lines between the
valve and the actuator are very short, i.e., small
volumes of fluid exist on either side of the actuator.
• The omission of pressure transient effects is also valid
for systems in which the load dynamics are muchslower than the pressure dynamics themselves.
• Since the load dynamics typically occur over a range
of seconds and the pressure dynamics typically occur
over a range of milliseconds, it is usually safe to
neglect the time variation of the pressure in favor of
the time variation of the overall systems.
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• For design and operating conditions in which long
transmission lines between the valve and theactuator are used or in which a large amount of
entrained air is captured within the fluid causing
the fluid bulk modulus to be reduced, it may be
necessary to conduct a transient analysis of thepressure conditions on both sides of the actuator.
This also may be the case if the actuator dynamics
are very fast.
– Under these assumptions:
– Q A and QB are the volumetric flow rates into and out
of the actuator, respectively, and ηav is the volumetric
efficiency of the actuator.
A B A B
av av
A dy A dyQ Q
dt dt
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• Review: Conservation of Mass
– Here we assume that all of the densities of the system (inlet
flow, outlet flow, and control volume) are the same and equal
to . – This assumption is justified for incompressible fluids and is
quite accurate for compressible fluids if pressure variations
are not too large and the temperature of flow into the control
volume is almost equal to the temperature of the flow out ofthe control volume.
CV CS
CV CV CV CV net
CVCV net
0 dV v dAt
0 V V Q
V0 V Q
The net rate of mass effluxthrough the control surface plus
the rate of change of mass
inside the control volume
equals zero. Velocity ismeasured relative to the control
volume.
CV net
V0 V P Q
0dP P
P dt
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– From the flow rates shown at the four-way spool
valve below, we see that:
A 2 1 B 4 3Q Q Q Q Q Q
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– The linearized flow equations for fluid passing across
the metering lands of the four-way spool valve aregiven by:
– Kq and Kc are the flow gain and pressure-flow
coefficients for the valve, respectively. Kp is the
pressure sensitivity.
s1 c q c A r
s2 c q c s A
s3 c q c s B
s4 c q c B r
PQ K K x K P P
2P
Q K K x K P P2
PQ K K x K P P
2P
Q K K x K P P2
d
q d
0 0
dc
0 0
q
p
0 0 0c
2Q AC P
Q A 2 AK C P
A x x
ACQK
P 2 P
KP Q / x 2P AK
x Q / P K A x
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– The volumetric flow rates into and out of the
actuator may be expressed as:
– The return pressure Pr = 0.
– The equations for the operating pressures on both
sides of the linear actuator then are:
s A q c A
sB q c B
PQ 2K x 2K P
2
PQ 2K x 2K P2
s A A p
c av
s BB p
c av
P A dyP K x
2 2K dt
P A dyP K x
2 2K dt
A A
av
BB
av
A dyQ
dt
A dyQ
dt
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– We see that the fluid pressure on side A of the
linear actuator is increased by moving the spool
valve in the positive x direction and that the fluid
pressure on side B of the actuator is decreased bythe same spool motion.
– This adjustment in fluid pressure by motion of the
spool valve is the mechanism for adjusting the
output motion of the load.
– The equation also shows a linear velocity
dependence for the fluid pressure as well, which
will result in favorable damping characteristics forthe system.
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• Analysis Summary
s A A p
c av
s BB p
c av
P A dyP K x
2 2K dt
P A dyP K x2 2K dt
2
s s
af A A B B2
P Pd y dy
m c ky A P A P Fdt dt 2 2
Substitution into
Result is:
2 2 2A B
af A B p2
c
d y A A dym c ky A A K x Fdt 2K dt
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– In this result we have assumed that ηaf
/ηav ≈ 1.
– The mechanical design of the linear actuator and
the four-way spool has a decisive impact on the
overall dynamics of the hydraulic control system.
– These design parameters help to shape theeffective damping of the system and provide an
adequate gain relationship between the input
motion of the spool valve and the output motion of
the load.
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Design• Actuator Design
– We start by sizing the linear actuator in accordance with
the expected load requirements. – Usually for a given application the working load force is
known and can be used to determine the required
pressurized areas that are needed within the actuator to
develop this load for a specified working pressure.
2
s saf A A B B2
P Pd y dym c ky A P A P F
dt dt 2 2
s saf A A B B
P PF A P A P
2 2
0 =
steady state steadyworkingforce
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– At the working force that is specified for the
application, it is common to use fluid pressures inthe linear actuator that are less than the full supply
pressure so as to provide a margin of excessive
force capability for the system.
– Typical design specifications at the working force
conditions are:
– Fw is the working force of the hydraulic control
system
– Ps is the supply pressure to the hydraulic controlvalve.
w A s B s
3 1F F P P P P
4 4
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– Substitution:
– Result is:
– Simultaneous solution yields:
w A s B s
3 1F F P P P P
4 4
s s
af A A B B
P PF A P A P
2 2
w 0A B A B
af s af s
F FA A 4 A A 2
P P
s
af A B 0
P(A A ) F
2
into
and
w 0 w 0A B
af s af s
2F F 2F FA A
P P
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– These results may be used to design or select a
linear actuator that will provide a sufficient workingforce for a given supply pressure.
– Other design considerations must also be taken
into account when designing the linear actuator.
• The stroke of the actuator or distance of piston
travel must be sufficient for the application.
• Pressure vessel stresses within the actuator
must be acceptable.• Sealing mechanisms must be used to minimize
both internal and external leakage.
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• Valve Design
– Once the actuator has been designed to generatethe necessary working force for the control
application, the next step is to design a control
valve that will provide sufficient flow and pressure
characteristics for the linear actuator.
– These characteristics are designed by specifying
the appropriate flow gain Kq and pressure-flow
coefficient Kc for the valve.
qs A A p p
c av c
s Ac A q
av
KP A dyP K x and K
2 2K dt K
P A dyK P K x
2 2 dt
Combine:
Result is:
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– For specifying the appropriate valve coefficients,
usually two operating conditions are considered:
• the steady working force condition
• the steady no-load velocity of the actuator
– For both these conditions, the displacement of the
open-centered valve is usually taken to be ¼ the
under-lapped valve dimension u. A specified
working valve displacement of this magnitude ishelpful because this valve keeps the valve
operating near the vicinity of the null position
about which the valve flow equations have been
linearized, and it provides a margin of excessivecapability for the system design.
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– Substitution:
– Result is:
– Similarly, we use the following parameters to
identify the no-load velocity conditions of thehydraulic control system:
– Where v0 is the no-load velocity that is specified
when the load disturbance force F is zero.
A s
1 3 dyx u P P 0
4 4 dt s Ac A q
av
P A dyK P K x
2 2 dt
c s qK P K u
A s 0
1 1 dyx u P P v
4 2 dt
s Ac A q
av
P A dyK P K x
2 2 dt
Aq 0
av
2A0 K u v
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– Simultaneous solution:
– Results in the following specifications for the
valve-flow gain, the pressure-flow coefficient, andthe pressure sensitivity:
– The valve coefficients are heavily dependent on
the shape of the flow passages that are used to
design the valve.
c s qK P K u A
q 0
av
2A0 K u v
and
q A 0 A 0 sq c p
av s av c
K2A v 2A v PK K K
u P K u
dq d c
0 0 0 0
ACQ A 2 A QK C P K
A x x P 2 P
If rectangular flow passages are used in the
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– If rectangular flow passages are used in the
design, we can show that the flow gain and
pressure-flow coefficients are given by:
– Here h is the height of the rectangular flow
passage, Cd is the discharge coefficient for the
flow passage, and ρ is the fluid density.
– Equate the following two expressions:
– Result is:
s dq d c
s
P uhCK hC K
P
A 0 A 0q c
av s av
2A v 2A vK K
u P
s d
q d c
s
P uhCK hC K
P
A 0
sd
av
2A vuh
PC
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– If it is assumed that the under-lapped dimension of
the open-centered valve is ¼ of the flow passageheight, the following design characteristics of the
valve may be written:
– It can be seen that as the actuator area A A and theno-load velocity requirement v0 become large, the
valve size also must grow accordingly. The valve
must be matched with the load requirements in
order to satisfy the overall design objectives of thehydraulic control system.
A 0
sd
av
A v
u h 4uP2C
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• Pump Design
– Now that the linear actuator and control valvehave been designed, it is time to specify the
hydraulic pump that will be used to power the
hydraulic control system.
– The pump is a fixed-displacement pump that
produces a volumetric flow rate that is proportional
to the angular input speed of the pump shaft ω.
– Within the pump itself there is internal leakage thatresults in a reduction of the pump volumetric
efficiency ηpv.
– Let’s review some details on pump efficiency (seeslides 52-56 .)
If we assume that the relief valve is closed the
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– If we assume that the relief valve is closed, the
supply flow to the four-way control valve is given
by:
– Where Vp is the volumetric displacement of the
pump per unit of rotation, and ηpv is the volumetricefficiency of the pump.
– The supply flow must equal the sum of the
volumetric flow rates that are crossing metering
lands 2 and 3 on the four-way spool valve.
s pv pQ V
s2 c q c s A
s3 c q c s B
s 2 3 c s c A B
PQ K K x K P P
2
PQ K K x K P P2
Q Q Q 3K P K P P
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– Using the following equations:
– Substitution:
– Result is:
s A A p
c av
s BB p
c av
A B
A B s
c av
P A dyP K x2 2K dt
P A dyP K x
2 2K dt A A dy
P P P2K dt
A B A B s
c av
A A dyP P P
2K dt
s c s c A BQ 3K P K P P
A Bs c s
av
A A dyQ 2K P
2 dt
supplied volumetricflow rate to the valve
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– Substitution:
– Result is the expression for the requiredvolumetric flow displacement for the supply pump,
where dy/dt has been replaced with the no-load
velocity requirement v0 for the system.
– The physical construction of the pump now needsto be addressed, e.g., gear pump or axial-piston
pump.
A Bs c s
av
A A dyQ 2K P
2 dt
s pv pQ V
A Bc sp 0
pv pv av
A A2K PV v
2
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• Input Power Design
– The input power that is required to operate the
hydraulic control system is given by:
– Where η is the overall pump efficiency.
p s
V PPower
v t d
v
d
Q
V
d d
t
V P
T
Eff
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Review: Pump Efficiency
• The task of the hydraulic pump is to convert rotating
mechanical shaft power into fluid power that may be
used downstream of the pump.
• None of the hydraulic pumps is 100% efficient. They
all lose power in the process of converting power.
– Fluid leaks away from the main path of powertransmission.
– Friction exists within the machine.
• The diagram on the next slide shows the power that
flows in and out of a typical hydraulic pump,
regardless of type.
Power Flowing In and Out of The Pump
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Power Flowing In and Out of The Pump
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– Power is supplied to the pump through the rotating
shaft by an external drive device (not shown).
– As the shaft rotates, the pump draws fluid into the
inlet side and pushes fluid out of the discharge side.
– The input power to the shaft is torque times the shaftangular velocity , i.e., Tω.
– Power is also delivered to the pump on the inlet side
by any pressure that may exist at the intake port of
the pump. This hydraulic power is equal to the
pressure times the volumetric flow rate, i.e., PiQi.
– The discharge power of the pump is equal to the
discharge pressure times the discharge volumetricflow rate, i.e., PdQd.
– Power also leaks away from the pump in the form of
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Power also leaks away from the pump in the form of
internal leakage. This power loss is calculated as
PℓQℓ, where Pℓ is the pressure drop across the leakpath, and Qℓ is the leakage volumetric flow rate.
– Finally, power also leaves the pump in the form of
dissipating heat.• The overall pump efficiency is defined as the useful
output power divided by the supplied input power:
• We can use the volumetric displacement of the pump Vdto separate the overall efficiency into two components:
the volumetric efficiency ηv and the torque efficiency ηt.
d dP QT
v t
Th l t i ffi i i i b Q
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Dynamic System Investigation K. Craig 55
• The volumetric efficiency is given by:
– It is used for describing power loses that result
from internal leakage and fluid compressibility.
• The torque efficiency is given by:
– It is used for describing power loses that resultfrom fluid shear and internal friction.
• The volumetric displacement Vd is given in units of
volume per radian. The volumetric displacement per
revolution is given by 2πVd.
d dt
V P
T
dv
d
Q
V
C St d
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Dynamic System Investigation K. Craig 56
Case Study
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Dynamic System Investigation K. Craig 57
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Control
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Control
• Position Control
– One control objective is to accurately position the load at
a prescribed location within the trajectory range of theactuator.
– Typically, this control function is carried out under slowly
moving conditions of the actuator ; therefore, the plant
description for this control problem may neglect safely
any transient contributions that normally would be
significant during high-speed operations.
2 2 2
A Baf A B p2
c
d y A A dym c ky A A K x F
dt 2K dt
Neglect
Using a standard PI controller the control law for
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Dynamic System Investigation K. Craig 61
– Using a standard PI controller, the control law for
the displacement of the four-way spool valve is:
– yd is the desired position of the load. – The most practical way to enforce the control law
is to use an electrohydraulic position control of the
four-way spool valve coupled with a
microprocessor that is capable of readingfeedback information and generating the
appropriate output signal for the valve actuator.
See diagram on the next slide.
– A block diagram for the position control system is
also shown on the slide after that.
e d i dx K y y K y y dt
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Dynamic System Investigation K. Craig 62
Solenoid-Actuated Two-Stage Electrohydraulic Valve
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Position Control of theFour-Way Valve-Controlled Linear Actuator
– If we assume that the position control of the system is a
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Dynamic System Investigation K. Craig 64
regulation control problem, and that the load force and
the desired position of the load yd are constants, the
equation of motion for the closed-loop system is:
– This is a first-order dynamic system.
– The design objective for the position control problem is
to shape the first-order response by designing theappropriate time constant for the system. This is
accomplished by proper selection of the proportional and
integral control gains.
– NOTE: This response is based on a slowly-movingdevice. If the settling time becomes too short, higher-
order dynamics will become significant.
• Velocity Control
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Dynamic System Investigation K. Craig 65
y
– Another common control objective is velocitycontrol. The objective is to establish a specific
output velocity for the load based on a desired
velocity that is prescribed by the application. This
objective typically is carried out for load systemsthat do not include a load spring.
– The equation of motion (with k = 0) for the
velocity-controlled system is:
– v is the instantaneous velocity of the load.
2 2
A Baf A B p
c
dv A Am c v A A K x F
dt 2K
– For the velocity control objective, the PID
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Dynamic System Investigation K. Craig 66
controller will be used for the spool valve:
– vd is the desired velocity of the load.
– Again, the most practical way to enforce the
control law is to use an electrohydraulic position
control of the four-way spool valve coupled with amicroprocessor that is capable of reading
feedback information and generating the
appropriate output signal for the valve actuator.
– The system block diagram for the velocity control
is shown on the next slide.
e d i d d dd
x K v v K v v dt K v vdt
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Velocity Control of theFour-Way Valve-Controlled Linear Actuator
– If it is assumed that the velocity control of the
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Dynamic System Investigation K. Craig 68
y
system is a regulation control problem and that theload force F and the desired velocity of the load vdare constants, the equation of motion for the
closed-loop system may be written as:
22 2
n n n d2
d v dv2 v v
dt dt
– This is a second-order dynamic system.
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Dynamic System Investigation K. Craig 69
– The design objective for the velocity-controlproblem is to shape the second-order response by
designing the appropriate undamped natural
frequency and damping ratio for the system.
– By making the proper selection of control gains,the undamped natural frequency and damping
ratio for the control system may be adjusted.
• Force Control
Another control objective that is used for a hydraulic
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– Another control objective that is used for a hydraulic
control system is force control.
– Force-controlled systems usually are configured very
much like a velocity-controlled system without a load
spring, and when they are used, they often switchbetween velocity and force control depending on the
immediate needs of the application.
– Furthermore, force-controlled systems typically are
operated in slow motion so as to gradually apply the loadforce to whatever the application is trying to resist.
– The equation of motion is:
2 2 2
A Baf A B p2
c
d y A A dym c ky A A K x F
dt 2K dt
Neglect
– Use a standard PI controller for the displacement
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Dynamic System Investigation K. Craig 71
of the four-way spool valve.
– Fd
is the desired force that is to be exerted on the
load. In most applications, the instantaneous load
force F is measured by sensing the fluid pressures
in sides A and B of the linear actuator and
calculating the load force according to thefollowing equation:
e d i dx K F F K F F dt
s saf A A B B
P PF A P A P
2 2
– Again, the most practical way to enforce the control law
is to use an electrohydraulic position control of the four-
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Dynamic System Investigation K. Craig 72
is to use an electrohydraulic position control of the four-
way spool valve coupled with a microprocessor that is
capable of reading feedback information and generating
the appropriate output signal for the valve actuator.
– A block diagram for the closed-loop control system isshown below.
– If it is assumed that the force control of the system is a
regulation control problem and that the desired load force
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Dynamic System Investigation K. Craig 73
g p
Fd is constant, the dynamic equation for the closed-loopsystem is:
– This is first-order dynamic system. – The design objective for the force-control problem is to
shape the first-order response by designing the
appropriate time constant for the system.
– NOTE: This response is based on a slowly-moving
device. If the settling time becomes too short, higher-
order dynamics will become significant.
d
e
i af A B p
dFF F
dt
1 1K
K A A K
Case Study
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Dynamic System Investigation K. Craig 74
y
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Pump-Controlled Hydraulic Systems
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Dynamic System Investigation K. Craig 76
• A pump-controlled hydraulic system uses a pump as
opposed to a control valve for directing hydraulic power
to and from an actuator that is used to generate useful
output.
• Pump-controlled hydraulic systems exhibit an efficiency
advantage over valve-controlled systems due to the fact
that the control valve introduces a pressure drop thatresults in significant heat dissipation.
• The pump-controlled system does not use this valve; the
immediate power needs of the output are met directly by
the power source and that increases the overall
operating efficiency of the system.
• However, there are disadvantages of a pump-
controlled system:
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Dynamic System Investigation K. Craig 77
y
– The response characteristics of pump-controlled
systems can be slower due to the longer
transmission lines that usually are used for
reaching the output actuator and theaccompanying fluid compressibility effects.
– To eliminate long transmission lines, often times
the pump, actuator, and power source are too
bulky to be collocated.
– Pump-controlled systems consist of a single pump
that operates a single actuator . Multiple actuators
cannot share the power that is generated fromone pump, so the pump cost must be included
with the overall cost of a single actuator.
• In pump-control of a linear actuator, since the pump
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operates symmetrically as it sends flow to andreceives flow from the output actuator, only double-
rod linear actuators are suitable.
• Typical applications for these systems include
industrial robots and flight-surface controls in the
aerospace industry.
• Pump-controlled rotary actuators, used to drive a
rotating shaft, are often called hydrostatictransmissions. They are frequently used for lawn
tractors, off-highway earth-moving equipment, and as
a constant-speed drive for various aerospace flight
applications.
Fixed-Displacement Pump Control of a
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Dynamic System InvestigationK. Craig 79
Linear Actuator
speed controlleddouble-acting
F = load disturbance force
Vp = volumetric displacement per unit of rotation
• Comments: – Speed-controlled (driven by an input shaft rotating at a
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Dynamic System InvestigationK. Craig 80
variable angular velocity ω) fixed-displacement pumpwith volumetric displacement per unit of rotation Vd
– Double-rod linear actuator to facilitate symmetric action
of the actuator – Pressurized areas are the same on both sides of the
actuator
– Load is a single mass-spring-damper system with a load-
disturbance force
– Rod connects the load to the actuator piston
– Volumetric flow of hydraulic fluid into the actuator is
controlled by the output flow of the pump – For positive ω, pump flow is to side A of the actuator; the
load moves down and flow exits the actuator from side B
– For negative ω, pump flow is to side B of the
actuator; the load moves up and flow exits the
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Dynamic System InvestigationK. Craig 81
actuator from side A – Q A and QB are the volumetric flow rates into and
out of the actuator
– Shuttle Valve• Connects the low-pressure side of the hydraulic
control system to the reservoir
• It keeps the low-pressure side of the circuit at a
constant reservoir pressure, i.e., zero gage
pressure
• It keeps the fixed-displacement pump from
drawing a vacuum and causing fluid cavitation• It allows for the return flow to be cooled by a
low-pressure radiator (not shown)
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• Analysis
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Dynamic System InvestigationK. Craig 83
– Load Analysis
• ηaf is the force efficiency of the actuator
– Pressure Analysis
• Assume that the pressure transients that resultfrom fluid compressibility in the transmission
lines are negligible. This assumption is
especially valid for a system design in which
the transmission lines between the valve andactuator are very short, i.e., small volumes of
fluid exist on either side of the actuator.
af A Bmy cy ky A(P P ) F
af
A A
F
P A
• The omission of pressure transient effects is
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Dynamic System InvestigationK. Craig 84
also valid for systems in which the loaddynamics are much slower (seconds) than the
pressure dynamics (milliseconds) themselves.
• If there are long transmission lines between the
valve and actuator, or if the bulk modulus isreduced because of entrained air in the fluid, or
if the actuator dynamics are very fast, a
transient analysis of the pressure conditions on
both sides of the actuator may be necessary.
• Here, we assume that pressure transients may
be safely neglected.
• Therefore Aav
Ay
Q
ηav = actuator volumetric efficiency
• From the diagram we see that for an
incompressible fluid:
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Dynamic System InvestigationK. Craig 85
• The supply flow is given by:
• Combining equations results in:
• The shuttle valve is used to connect the low-
pressure side of the hydraulic circuit to thereservoir. The pressure on side B of the
actuator is therefore 0 gage pressure.
A s AQ Q KP
s pv pQ V
ηpv = pump volumetric efficiency
pv p A
av
V AP yK K
pv p
A
V AP y
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Dynamic System InvestigationK. Craig 86
• This equation shows a pump velocity and an
actuator velocity dependence for the fluid
pressure in side A. An adjustment of the pumpvelocity term will provide a control input to the
dynamic load equation. The linear velocity
term will be useful in providing favorable
damping characteristics.
• Analysis Summary
2
af pv paf
av
V A Amy c y ky FK K
avK K
– We see that the mechanical design of the linear
t t d th l t i di l t f th
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Dynamic System InvestigationK. Craig 87
actuator and the volumetric displacement of thepump have a decisive impact on the overall
dynamics of the hydraulic control system. The
design parameters help to shape the effective
damping of the system and provide an adequategain relationship between the input velocity of the
pump and the output motion of the load.
Design
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Dynamic System InvestigationK. Craig 88
• Actuator Design – We start the mechanical design of the hydraulic
control system by sizing the linear actuator inaccordance with the expected load requirements.
– We usually know the maximum working load and
maximum working pressure. Use these two
quantities along with the steady-state form of theload equation:
af A B
af A B
my cy ky A(P P ) F
reduces to: F A(P P )
– Since there is no pressure drop between the pump
d t t th fl id id A f th
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Dynamic System InvestigationK. Craig 89
and actuator, the fluid pressure on side A of thelinear actuator is the full supply pressure.
– Here Fw is the working force of the hydraulic
control system, and Ps is the supply pressure at
the working condition.
– Therefore, after substitution, the equation forsizing the pressurized area of the linear actuator
is:
– Other design considerations include the stroke of
the actuator or distance of piston travel.
w a s BF F P P P 0
w
af s
F A
P
• Pump Design
The p mp is a fi ed displacement p mp that
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Dynamic System InvestigationK. Craig 90
– The pump is a fixed-displacement pump thatproduces a volumetric flow rate that is proportional
to the angular input speed of the pump shaft ω.
– In order to size the required volumetric
displacement of the pump, it is common to specifya no-load velocity v0 requirement for the linear
actuator and to size the pump in such a way as to
achieve this velocity requirement.
– Set P A equal to zero for the no-load case,
efficiencies equal to unity, and ω equal to the
maximum pump speed. Then the equation below
becomes:pv p 0
A p
av
V Av AP y V
K K
– The pump may be any positive-displacement
pump that satisfies the volumetric displacement
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Dynamic System InvestigationK. Craig 91
pump that satisfies the volumetric displacementrequirement.
– Usually the pump construction type is selected
based on the required supply pressure of the
system and the desired operating efficiency of thepump.
• Input Power Design
The input power that is required to operate the
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Dynamic System InvestigationK. Craig 92
– The input power that is required to operate thehydraulic control system is given by the standard
power equation:
– ηp is the overall efficiency of the pump given by
the product of the volumetric and torque efficiencyvalues: ηp = ηptηpv. Ps and ω are the
instantaneous pressure and operating speed of
the pump. Use the maximum combination of
operating speed and pressure encountered in the
application.
p s
p
V PPower
Control
P l
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Dynamic System InvestigationK. Craig 93
• Position Control – The control objective is to position the load
accurately at a prescribed location within the
trajectory range of the actuator.
– Typically, this control function is carried out under
slowly moving conditions of the actuator.
– The plant description may safely neglect anytransient contributions that normally would be
significant during high-speed operations.
2af pv paf
av
V A Amy c y ky FK K
Neglect
– Use a standard PI controller:
e d i dK y y K y y dt
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Dynamic System Investigation
K. Craig 94
– The closed-loop system equation of motion is:
e d i dK y y K y y dt
d e
i af pv p
1 kKy y y K
K AV
• Velocity Control – Another possible control objective for the system
is velocity control The objective seeks to
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Dynamic System Investigation
K. Craig 95
is velocity control. The objective seeks to
establish a specific output velocity for the load
based on a desired velocity. This objective is
carried out for load systems that do not include aload spring. Therefore set k = 0.
2af pv paf
av
V A A
my c y ky FK K
Neglect
2
af pv paf
av
V A Amv c v FK K
– Use a standard PID controller:
dK K dt K
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Dynamic System Investigation
K. Craig 96
e d i d d ddK v v K v v dt K v vdt
– The closed-loop system equation of motion is:
2 2v 2 v v y
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Dynamic System Investigation
K. Craig 97
2 2
n n n d
2af av paf
e
avin
af av pd d n
af pv p
v 2 v v y
AV Ac K
K KKmK AV
K 2 m K AV K
• Force Control – Another control objective that is occasionally used
for the hydraulic control system is that of force
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K. Craig 98
for the hydraulic control system is that of force
control.
– Force-controlled systems are usually configured
very much like a velocity-controlled system withouta load spring, and when they are used, they often
switch between velocity and force control
depending on the immediate needs of the
application.
– Furthermore, force-controlled systems typically
are operated in slow motion so as to apply the
load force gradually to the object that theapplication is trying to resist.
– Under these slow-motion conditions, the load
inertia and viscous damping effects may be
ignored safely without sacrificing accuracy in the
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Dynamic System Investigation
K. Craig 99
ignored safely without sacrificing accuracy in the
modeling process.
– Us a standard PI controller.
2af pv paf
av
V A Amy c y ky F
K K
Neglect
e d i dK y y K y y dt
– In most applications, the instantaneous load force
F is measured by sensing the fluid pressures insides A and B of the linear actuator and
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Dynamic System Investigation
K. Craig 100
s easu ed by se s g e u d p essu essides A and B of the linear actuator and
calculating the load force.
af A B
F A(P P )
– The closed-loop system equations of motion is:
1 KF F F K
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Dynamic System Investigation
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• Summary – We have considered 3 control objectives: position,
velocity, and force control.
– For position and force control, an important slow-
speed assumption has been employed. Under
high-speed conditions, higher-order dynamics maymanifest and may produce an undesirable result.
d e
i af pv p
1 KF F F K
K AV