fluid power system investigation case study

8/19/2019 Fluid Power System Investigation Case Study

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Fluid PowerDynamic System Investigation

K. Craig 1


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K. Craig 2


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K. Craig 3

Dynamic SystemInvestigation

Process


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K. Craig 5Typical Fluid Power System


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K. Craig 6

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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K. Craig 7

– 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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K. Craig 8

• 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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K. Craig 9

• 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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K. Craig 10

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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K. Craig 11

– 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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K. Craig 12

– 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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K. Craig 13

– 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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Fluid PowerDynamic System Investigation K. Craig 14

– 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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Fluid Power

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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Fluid Power


Case Study


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Fluid Power



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Fluid Power


Control


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Fluid Power


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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Fluid Power


– 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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Fluid Power


Solenoid-Actuated Two-Stage Electrohydraulic Valve


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Fluid Power


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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Fluid Power


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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Fluid Power


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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Fluid Power


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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Fluid Power


Velocity Control of theFour-Way Valve-Controlled Linear Actuator

– If it is assumed that the velocity control of the


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Fluid Power


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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Fluid Power


– 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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Fluid Power


– 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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Fluid Power


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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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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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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y


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Pump-Controlled Hydraulic Systems


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• 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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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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Dynamic System InvestigationK. Craig 78

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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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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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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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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– 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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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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• 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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• 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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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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• 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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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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– 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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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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– 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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• 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:


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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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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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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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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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


– 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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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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K. Craig 101

• 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

fluid power system investigation case study

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