Hydrauliske Servomekanismer

Aalborg University

luid Power

d Analysis

FSystems

Modelling an

Pilot Stage Pilot Stage

Torben Ole Andersen Institute of Energy Technology

March 2003 2nd Edition

Preface

Preface The engineering effort in hydraulics is a problem of both educational and technical nature, but of a restrictive influence on the extension and successful use of hydraulics. The lack of well - established, systematic engineering methods to form the basic set-off in analysis and design of hydraulic control systems has been obvious. The employers within the branch - from which the majority started up without any special hydraulic education - are so busy in designing that no time is left to put theory behind the mainly empirical but often voluminous know-how gathered by many hydraulic engineers. No engineering discipline at university level, aimed at the professional engineering of hydraulic control systems, has been established on a par with classical disciplines. Furthermore, the theoretical tools needed for analysis of hydraulic problems are complicated through the following reasons: Many problems in hydraulic systems are of extremely dynamic nature and can only be described by differential equations. Hydraulic systems and components are inherently non-linear, making the describing differential equations non-linear and thus limits the applicability of linear control theory. Many hydraulic systems are in fact distributed-parameter systems. It takes some skill to make sound engineering judgements on when and how lumped-parameter approaches can be made. By the estimation of apparently simple hydraulic system-parameters a large number of details must be taken into consideration, such as laminar or turbulent flow, viscosity-temperature relations, friction factors, discharge coefficients, passage geometry, cavitation, air content, and so forth. Finally the development of valuable engineering theories very often must be based on experimental verification of hypothesis. It seems from above statements, that strong efforts in the elevation of the engineering methods ought to be emphasised as one means to ensure continuous ability to compete. Therefore it is rational efforts to seek the most intelligent ways of using hydraulic control. This demands a continuous effort in research and development, and in education of hydraulic engineers so their professional standing is adequate for dealing with actual hydraulic control problems. Accordingly the necessary, engineering tools have to be established and brought up to an operational level. In Figure 1 is shown hydraulic control engineering as a special subject and the most important contributing disciplines. “The-state-of the-art” is that 1, 2, 5, 6, and to some extent 7, have been close allies of hydraulics.

2nd Edition Page 1 of 3

Preface

10TECHNICAL

SALES &CONSULTING

9DIGITAL

SIGNAL CIRCUITDESIGN

MECHANICS(FEM)

8CONTINUUM

MASCHINE &MASCHINEPARTS

DESIGN

& INSTRUMENTATIONLAB. METHODS

(ELECTRONICS)

SIMULATIONAND GENERAL

COMPUTINGTECHNIQUES

11PRODUCTIONTECHNOLOGY

CONTROLENGINEERING

7FLUID

MECHANICS& DYNAMICS

(CFD)

6

HYDRAULIC

CONTROLENGINEERING

1

(EX. BACKHOE -LOADER)

APPLICATIONS5

DYNAMICS

GENERALSYSTEM

2

DIGITAL

4

3

The most essential special subjects contributing to hydraulic control engineering

Today it seems that hydraulic control engineering is to be considered as a specialised branch of control or systems engineering, whose well-established methods forms very powerful techniques both in analysis and in synthesis of hydraulic systems. Before control-engineering techniques can be applied it is required that hydraulic components and problems can be described in terms compatible to the “language” used in control engineering. This means, that the behaviour of components and systems must be expressed as static and dynamic characteristics, transfer functions etc. Thus the disciplines 2, 3, and 4 are needed for the physical-mathematical dynamic modelling (2), efficient handling and solution of the complex of describing equations (3), and experimental evaluation of models (4). Finally, all these activities loose most of their sense, if they are not based on a broad knowledge and experience in the intended application (5) of the hydraulic system. So the disciplines 1 to 5 form the important foundation of hydraulic control which - supplemented with basic engineering knowledge within the other mentioned disciplines - makes it possible to reach a high degree of competence in the engineering design of hydraulic control systems. The control engineering approach has here been legitimated through the above paragraph, and maybe the most characteristics feature is the use of dynamic models and simulation, which can be worked out to more or less sophistication depending on the level of analysis relevant to the actual problem. Hence, it is important to concentrate engineering research and development efforts in this fields, to establish some of the necessary tools. Generally, savings in work or cost are the essential reason why experiments are carried out with models in stead of real systems. However, some situations occur in which no alternative to model-experiments exist. This is particularly the case when dealing with systems in which eventual instability may initiate irreversably accelerating processes with hazardous and incalculable consequences. Other good reasons for using simulation in both analysis and synthesis are well known and can be stated as:


Preface

° The physical system is not available. Often, simulations are used to determine whether a projected system should ever be built.

° The cost of experimentation is too high. Often, simulations are used where real experiments are too expensive. The necessary measurement tools may not be available or are to expensive. The system might be used all the time and taking it “off-line” would involve unacceptable cost.

° The time constants of the system are not compatible with those of the experimenter. Often simulations are performed because the real experiment executes so quickly that it can hardly be observed (for example the reaction time for a spool valve towards neutral is important for safety reasons, but reaction times around 200 ms can not be noticed by an operator).

° Control variables (disturbances), state variables, and/or system parameters may be inaccessible. Often, simulations are performed because they allow us to access all inputs ad all state variables, whereas in the real system, some inputs may not be accessible to manipulation and some state variables may nor be accessible to measurement. Simulation allows us to manipulate the model outside the feasible range of the physical system.

° Suppression of disturbances. Often, simulations are performed because they allow us to suppress disturbances that are unavoidable in the real system. This allows us to isolate particular effects, and may lead to better insight into the system behaviour.

° The number of parameters to adjust in the experiment are too large, giving a huge number of possibilities and at best, result in a sub-optimal solution.

The most important strengths of simulation, but also its most serious drawbacks, are the generality and ease of its applicability. It does not require much of a genius to be able to utilize a simulation program. However, in order to use simulation intelligently, (having the handle to make the real world behave the way we want it to), we must understand what we are doing. Danger lies in forgetting, that the simulation model is not the real world, but that it represent the world under a very limited set of experimental conditions. Simulations are rarely enlightening. In fact, running simulations is very similar to performing experiments in the lab. We usually need many experiments, before we can draw legitimate conclusions. Correspondingly, we need many simulations before we understand how our model behaves. While analytical techniques (where they are applicable) often provide an understanding as how a model behaves under arbitrary experimental conditions, one simulation run tells us only how the model behaves under the one set of experimental conditions applied during the simulation run. Therefore, while analytical techniques are generally more restricted (they have a much smaller domain of applicability), they are more powerful where they apply. So, whenever we have a valid alternative to simulation, we should, by all means, make use of it. (for example, single-stage pressure control valves is applicable for analytical analysis while two-stage valves are not). However, as the level of analysis is elevated, as it for instance may happen in connection to development of new components or new systems, a barrier will appear at a certain level. The complexity which underlines the development of mechatronic systems should be recognised. Very few people are capable in making models of real mechatronic systems, and even though simulation programs are appearing on the market that claims they can simulate pretty much anything, there are still the number of parameters to adjust and the programs can not design the system for you.


Contents

Contents Chapter 1: Mathematical modelling 1.1 Introduction 1.2 Simple Models contra Complex Models 1.3 Systems Engineering and Component Design 1.4 Establishment of Mathematical Models Chapter 2: Steady state characteristics 2.1 Hydraulic Fluids 2.2 Flow Characteristics of Spool Valves 2.3 Valve Coefficients 2.4 Flow forces on Spool Valves Chapter 3: Dynamic Modelling 3.1 Introduction 3.2 Lumped Fluid Theory 3.3 Cylinders 3.4 Motors 3.5 Linear Characteristics of a Cylinder Chapter 4: Dynamic System Analysis 4.1 Introduction 4.2 Flow Control Valves 4.3 Power Elements Chapter 5: Advanced System and Component Studies 5.1 Introduction 5.2 Hydraulic Load Holding Circuit 5.3 Load Sensing Directional Valve 5.4 Two Stage Relief Valve Appendix A: Flow Force Compensation


Chapter 1

1 Mathematical Modelling

Fluid Power Systems (Hydraulisk Komponentanalyse) AaU ~ Forår 2003

1.1 1.2 1.3 1.4

Introduction................................………........................... Simple Models Contra Complex Models......................... Systems Engineering and Component Design.........……. Establishment of Mathematical Models............…………

1

2

2

3

1.1 Introduction The concept of models is probably one of the most fundamental concepts in science and in the scientific approach to the solution of practical problems. In a broad sense, a theory is an abstract model of, how things behave. To be of value, a model must come close to reality, and to the engineer nothing is more practical than a good theory, because it forms the basis on which realistic decisions are taken. In engineering, mathematical models are widely used because engineering is based on physical laws, and the language of physics is mathematics. Such mathematical models are a complex of equations describing the characteristic physical laws and effects of the system under consideration. A huge number of well-established methods are available for the handling of mathematical models. Computer technology has contributed momentous to this, particular regarding the solution of mathematical problems associated with engineering. In engineering, a model is a meaningful and operational description of the characteristics of a virtual technical system or process. The model is meaningful if it is an obvious fashion displays the most significant, useful causalities by which the engineer could make his system meet certain demands, and if it also clearly demonstrates significant, unwanted causal relations, which must be paid due attention and eventually acted upon to ensure, that the system resulting from the engineering will operate satisfactory under all predictable operation conditions. - The model is operational if it is fairly simple to carry out such investigations with the model from which conclusions about the virtual system can be drawn. Claming a model to be both meaningful and operational is much the same as to require, at the same time, accuracy and simplicity. As these properties usually are conflicting, one has to compromise, to chose an acceptable degree of accuracy, a certain “level of analysis”. The question generally turns out to be : Is it possible within acceptable time and cost to set up a model of the specific system which is sufficiently simple to be


Chapter 1

handled and still retains its meaning. If not, the engineering decision will, more or less, be based on guess-work. Generally, savings in work or cost are the essential reason why experiments are carried out with models in stead of real systems. However, some situations occur in which no alternative to model-experiments exist. This is particularly the case when dealing with systems in which eventual instability may initiate irreversably accelerating processes with hazardous and incalculable consequences. Finally, the model-concept and the associated simulation-techniques are very valuable pedagogical tools which are brought into use whenever technical knowledge, on any level, shall be transferred from one person to another in an efficient manner. 1.2 Simple Models Contra Complex Models Efforts shall always be used to investigate possibilities to reduce models. Reduce, because first step models should be established on a somewhat higher level than what is aimed at as the final level, simply to get a conscious and , if possible, quantitative based perception of what has been truncated. In case of oversimplification it is easier to understand discrepancies between predicted and found performance and, if wanted to try to correct the model by partwise returning to the higher level model. So, the modelling procedure will always be an iterative process. In much engineering work there is a natural tendency to overemphasise the claims for a simple and highly operational models and to start on a reduced level, probably because the time/economy break-even point in modelling-and-decision-making, averaged over a number of decisions, determines that slightly erroneous decisions may now and then be let behind to trial-and-error corrections, that is iterative modifications on the real systems. However, simple models are preferable with respect to gain a basic qualitative understanding of systems or components-behaviour. In fact, the human brain is not very capable to comprehend dynamic systems with more than a few variables, interrelated even in a simple structure. but using a computer, one can let the computer take care of all programmable interactions in even highly complex systems and concentrate the intellectual efforts on the understanding of the nature of the system through the study of simulated model-response on given inputs of disturbances. Computer-simulation is a very powerful way to “make experiments” with complex dynamic systems, and the accessibility of computer-facilities is highly determining the value of complex models. This means, that complex models are much more relevant to-day than they were just a few years ago, because it is possible to almost everyone, involved in mathematical modelling to get access to a computer and to be operative in a fairly short time by means of existing, high-level simulation languages. 1.3 Systems Engineering and Component Design Like all design activities the modelling process has a recurrent nature, and obviously many matters influence the level of analysis needed to make up a useful mathematical model of a component. However, two extremes exist, determined by two characteristic users of component-models.


Chapter 1

The systems-engineer normally wants to apply a “black-box” concept of the individual component. He wants to know the relevant input-output relations without being forced to care about internal phenomena. He wants to use the simplest, meaningful model representation of each component to keep the total system-model within acceptable limits. But his judgement of component representation may be an iterative process, because the needed level of analysis for each component depends on the properties and the interaction of the other components. The component-designer needs a detailed model, describing all internal effects with significant influence on the total performance. This type of model is useful during preliminary design of new components or when attempting to modify the performance of a given design by parameter variation. But it is necessary to be careful with complicated models, because they give the impression of a high level of analysis, while, in reality, they might be a compilation of conventionally used mathematical equations, poorly describing the virtual physical phenomena. The indication is, that it is often difficult to verify the type of complex, mathematical models established on a conventional empirical/theoretical analysis. The reason of this discrepancy is undoubtedly, that it is difficult to establish generally applicable adequate analytical expressions of flow-pressure-area relations, unless geometry is very simple and flow conditions are very well-defined. Flow condition, separation- and cavitation-effects must be taken into account as well as distributed parameters and several non-linear effects are characteristics of hydraulic components. Care must be taken, particular with flow-equations. 1.4 Establishment of Mathematical Models Basically two approaches exist: The analytical approach and the experimental approach. In principle they are very different and have individual purposes. In fact they are in many ways related and are complementary to one another: It is hardly possible to plan rational experiments without having formed any analytical ideas of a model (Hypothesis). Vice Versa, a complex analytical model without any experimental verification will often be met with some doubtfulness. Finally conventionally used flow-pressure-area equations, examplewise, usually contain empirical ingredients as the conceptual discharge coefficient, etc. Hence it is not relevant to talk about a pure analytical approach when modelling hydraulic systems. What often is designated ‘the analytical approach’ is in fact a pseudo-analytical approach with numerous empirical ingredients. Agreement between simulated and measured over-all performance of model and component, respectively, yields confidence in the validity of the model. But - depending on the type of model wanted, which one of the two approaches should preferably be chosen? In systems engineering, the most meaningful approach to the establishment of an over-all component-model is to approximate measured component characteristics with suitable and comparatively simple mathematical input-output relations.


Chapter 1

In component design, or whenever a detailed model should be established, it is natural to begin with the ‘analytical’ approach, even though experience has shown, that existing analytical tools are seldom adequate for the establishment of accurate models. Roughly, the ‘analytical’ approach can be used to establish the structure of the model, while a number of internal parameters must be measured. Also the structure of some flow-relations must often be laid down by experiments. This means, that normally we cannot speak of a proper analytical approach to the establishment of realistic models of hydraulic components even on a medium level of analysis. Instead an analytical/experimental approach must be applied.

----- oo0oo -----


Chapter 2

2 Steady State

Characteristics Fluid Power Systems (Hydraulisk Komponentanalyse) AaU ~ Forår 2003

2.1 2.2 2.3 2.4

Hydraulic Fluids................................................................Introduction • Fluid density • Viscosity • Dissolvability • Stiffness Flow Characteristics of Spool Valves......………….……Introduction • Flow through orifices • general valve analysis Valve Coefficients.............………………………………Introduction • A critically lapped valve with linear ports Flow forces on spool valves..........................……………

1

11

17

19

2.1 Hydraulic Fluids 2.1.1 Introduction The main purpose of the hydraulic fluid is to transport energy from the pump to the actuators. Secondary purposes involve the lubrication of the moving mechanical parts to reduce wear, noise and frictional losses, protecting the hydraulic components against corrosion and transporting heat away from its sources. The preferred working fluid in most applications is mineral oil, although in certain applications there is a requirement for water-based fluids. Water-based fluids and high water-based fluids provide fire resistance at a lower cost and have the advantage of relative ease of oil storage and fluid disposal. The recommended classification system is as follows: HFA – dilute emulsions, i.e. oil-in-water emulsions, typically with 95% water content. HFB – Invert emulsions, i.e. water-in-oil emulsions, typically with 40% water content. HFC – Aqueous glycols, i.e. solutions of glycol and polyglycol in water, typically with 40% water content. HFD – Synthetic fluids containing no water, such as silicone and silicote esters. The selection of the appropriate fluid will require specialist advice from both the component manufacture and the fluid manufacture.


Chapter 2

As the most commonly used hydraulic fluid is mineral oil and in the following sections it is the physical properties of commercial mineral oils that is discussed. The purpose of this chapter is to define certain physical properties which will prove useful and to discuss properties related to the nature of fluids. Because the fluid is the medium of transmission of power in a hydraulic system, knowledge of its characteristics is essential 2.1.2 Fluid density The mass density, ρ , of a hydraulic fluid is defined as a given mass divided by its volume, see Equation (2.1).

Vm=ρ (2.1)

where ρ mass density [ ]m/kg 3

m mass of the fluid [ ]kgV volume of the fluid [ ]m3

The mass density is both temperature and pressure dependent. It decreases with increasing temperature but increases with increasing pressure. A generally accepted empirical expression, the Dow and Fink equation, describes this:

( ) ( ) ( )( 20 ptBptA0.1)t(p,t ⋅−⋅+⋅ρ=ρ ββ ) (2.2)

where ρ mass density [ ]m/kg 3

0ρ mass density at atmospheric pressure [ ]m/kg 3

βA temperature dependant coefficient [ ]bar 1−

βB p

temperature dependant coefficient [ ]bar 2−

pressure [ ]bar

The density for a hydraulic fluid is normally (DIN 51757) given by the fluid manufacturer as the density at 15 and atmospheric pressure. This reference density lies between 0.85 and 0.91 g (850-910 kg ) for commercial hydraulic fluids.

Co

3cm/ 3m/The reference mass density in Equation (2.2) may be determined by:

( )( )15t1)t(

t

150 −⋅α+

ρ=ρ (2.3)

where 0ρ mass density at atmospheric pressure [ ]m/kg 3

15ρ mass density at atmospheric pressure and 15 o [ C ]m/kg 3

tα thermal expansion coefficient [deg ]1−

t temperature [ ]Co

The thermal expansion coefficient is normally regarded as independent of temperature and pressure and lies within the range of 0.0065 to 0.007 deg-1.


Chapter 2

The two coefficients in Equation (2.2) are normally referred to as the Dow and Fink coefficients. They have experimentally been found to:

( ) 624 1002.36T53.0T1072.6A −−β ⋅−⋅+⋅⋅−= (2.4)

( ) 924 1017.57T24.0T1084.2B −−β ⋅+⋅−⋅⋅= (2.5)

where βA temperature dependant coefficient [ ]bar 1−

βB temperature dependant coefficient [ ]bar 2−

T absolute temperature [ ]K The variation of the Dow and Fink coefficients with temperature is displayed graphically in Figure 2.1

15 bar10A −⋅β 29 bar10B −⋅β

50

52

54

56

58

60

62

64

66

68

-20 0 20 40 60 80 100 1208

9

10

11

12

13

14

15

16

17βA

βB

[ ]Ct 0

Fig. 2.1 The variation of the Dow and Fink coefficients with temperature

Inserting Equations (2.3)..(2.5) in Equation (2.2) means that the density can be determined by calculations only (no measurements), for any pressure and temperature combination, as long as the reference mass density, ρ15, is known. The variation of the mass density with temperature and pressure is displayed graphically in Figure 2.2. In Figure 2.2 the mass density is displayed relative to the reference mass density. 2.1.3 Viscosity The most important of the physical properties of hydraulic fluids is the viscosity. It is a measure of the resistance of the fluid towards laminar (shearing) motion, and is normally specified to lie within a certain interval for hydraulic components in order to obtain the expected performance and lifetime. The definition of viscosities is related to the shearing stress that appear between adjacent layers, when forced to move relative (laminarly) to each other. For a newtonian fluid this shearing stress is defined as:

dyxd

xy&

µ=τ (2.6)


Chapter 2

where τxy shearing stress in the fluid, [N/m2] µ dynamic viscosity, [Ns/m2] x& velocity of the fluid, [m/s] y coordinate perpendicular to the fluid velocity, [m]

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 100 200 300 400 500 600 700 800

15ρρ

p [bar]

0 °C

20 °C40 °C60 °C80 °C

100 °C

Fig. 2.2 The variation of the mass density with temperature and pressure In Figure 2.3 the variables associated with the definition of the dynamic viscosity are shown.

x(y)

fluid

y

=x

dy

x

dxτ xy

τxy

Fig 2.3 Deformation of a fluid element The usual units for the dynamic viscosity is P for Poise or cP for centipoise. Their relation to the SI-units are as follows: 1 . 2m/Ns1.0cP100P ==For practical purposes, however, the dynamic viscosity is seldom used, as compared to the kinematic viscosity that is defined as follows:

ρµ=ν (2.7)


Chapter 2

where ν kinematic viscosity, [m2/s] µ dynamic viscosity, [Ns/m2] ρ density, [kg/m3]

The usual unit used for ν is centistoke, cSt, and it relates to the SI units as follows:

smm1

sm10cSt1

226 == − .

A low viscosity corresponds to a "thin" fluid and a high viscosity corresponds to a "thick" fluid. The viscosity depends strongly on temperature and also on pressure. The temperature dependency is complex and is normally, DIN51562 and DIN51563 described by the empirical Uddebuhle-Walther equation:

( ) a101010 tlogmC8.0loglog ⋅−=+ν νν (2.8)where

ν kinematic viscosity, [cSt] Cν,mν constants for the specific fluid ta absolute temperature, [K]

This dependency is normally shown in specially designed charts, where the kinematic viscosity shown as function of the temperature becomes a straight line, see Figure 2.4.

1000

500300

2000

5000

2001501008060504030252016

1210

8765

43.5

32.7

ν [cSt]

-50 -40 -30 -20 0 120100908070605040302010-10t [°C]

ISO VG 10

ISO VG 22

ISO VG 32

ISO VG 46ISO VG 68

ISO VG 100

Fig. 2.4 Uddebuhle-chart: The temperature dependency for some of the most commonly used

mineral oils. The ISO VG standard refers ν at 40°C


Chapter 2

The vertical axis of an Uddebuhle chart is a mapping of log log(ν+0.8), i.e., approximately a double logarithmic axis (especially at higher values of ν). The horizontal axis is a mapping of logT, i.e., a logarithmic axis. A hydraulic fluid is, in general, referred to by its kinematic viscosity at 40°C. A different way of describing a hydraulic fluid is by means of the viscosity index, where the temperature dependency is related to a temperature sensitive fluid and a temperature insensitive fluid. The hydraulic fluid to be indexed and the 2 reference oils must have the same viscosity at a temperature of 210°F. If that is fulfilled, the viscosity index, V.I., may be determined as:

%100HLULVI ⋅

−−= (2.9)

where VI viscosity index L kinematic viscosity at 100°F for the temperature sensitive fluid U kinematic viscosity at 100°F for the fluid to be indexed H kinematic viscosity at 100°F for the temperature insensitive fluid

Different standards, e.g. DIN ISO 2909, offer a list of reference fluids with different kinematic viscosities at 210°F to pick from. The method dates back to 1929 and the improvement in mineral oil destillation and refining means that many hydraulic fluids come out with an index above 100. Beside the temperature dependency the viscosity also depends on pressure, especially at higher levels. The general accepted expression is as follows:

pB0 e η⋅µ=µ (2.10)

where µ dynamic viscosity, [Ns/m2] µ0 dynamic viscosity at atmospheric pressure [Ns/m2] Bη temperature dependant parameter, [bar-1] p pressure, [bar]

The parameter Bη may, within temperature ranges from 20°C to 100°C, be determined empirically as:

t100026.0B 5 ⋅−=η (2.11)where

Bη temperature dependant parameter, [bar-1] t temperature, [°C]

The pressure dependency may be rewritten to cover kinematic viscosities:

pB0 e η⋅ρµ

=ν (2.12)

where

ν kinematic viscosity, [m2/s]


Chapter 2

µ0 dynamic viscosity at atmospheric pressure [Ns/m2] ρ density, [kg/m3] Bη temperature dependant parameter, [bar-1] p the pressure, [bar]

In the above it should be remembered that the density increases with pressure, thereby making the kinematic viscosity less sensitive to pressure rise. 2.1.4 Dissolvability The capability of dissolving air (saturation point) varies strongly for hydraulic fluids with pressure. For pressure levels up to approximately 300 bar, the Henry-Dalton sentence applies:

atm

aFVa p

pVV ⋅⋅α= (2.13)

where Va volume of dissolved air in the oil, [m3] αV Bunsen coefficient, approximately constant at 0.09 VF volume of the fluid at atmospheric pressure, [m3] pa absolute pressure, [bar] patm atmospheric pressure ≈ 1 bar, [bar]

The capability of hydraulic fluids to absorb air is a problem, because the subsequent release of air at lower pressures leads to reduced fluid stiffness. 2.1.5 Stiffness When pressurized a hydraulic fluid is compressed causing an increase in density. This is described by means of the compressibility which is defined as

p1

F ∂ρ∂⋅

ρ=κ (2.14)

where κF compressibility of the fluid, [bar-1] ρ mass density, [kg/m3] p pressure, [bar]

The reciprocal of κF is defined as the stiffness or bulk modulus of the fluid:

FF

1κ

=β (2.15)

where κF compressibility, [bar-1] βF bulk modulus, [bar]


Chapter 2

Based on the above definition it can be shown that for fixed temperature the stiffness is proportional to the pressure rise caused by a compression of the fluid:

0

F

VdVdp ⋅β= (2.16)

where dp increase in pressure, [bar] βF bulk modulus of the fluid, [bar] dV the compression, i.e., decrease in volume, [m3] V0 the volume corresponding to the initial pressure, [m3]

Just like density the bulk modulus and the compressibility are functions of temperature and pressure. Inserting Equation (2.2) in Equation (2.14) and Equation (2.15) leads to:

( ) ( ) ( )( ) ( ) ptB2tA

ptBptA0.1p,t

2

F ⋅⋅−⋅−⋅+

=βββ

ββ (2.17)

where βF stiffness of the fluid, [bar] Aβ temperature dependant coefficient, [bar-1] p pressure, [bar] Bβ a temperature dependant coefficient, [bar-2]

Where the temperature dependant coefficients can be determined from and. It should be noted that Equation (2.17) implies that the fluid stiffness may be calculated for any temperature and pressure combination regardless of the specific type of mineral oil. The variation of the fluid stiffness with temperature and pressure is displayed graphically in Figure 2.5.

12000

14000

16000

18000

20000

22000

24000

26000

28000

30000

32000

0 100 200 300 400 500 600 700 800

]bar[Fβ

p [bar]

0 °C

20 °C

40 °C60 °C80 °C

100 °C

Fig. 2.5 The variation of the fluid stiffness with temperature and pressure


Chapter 2

In real systems air will be present in the fluid. The volume percentage at atmospheric pressure will go as high as 20 %. As air is much more compressible than the pure fluid it has, potentially, a strong influence on the effective stiffness of the air containing fluid. If the air, however, is dissolved in the fluid there is no significant effect on the compressibility. Hence, it is the amount of free or entrapped air in the fluid that markedly reduces the effective stiffness. Taking the presence of air into account the effective stiffness of the fluid becomes:

( )

A

A

FFAA

F

Aeff 11

1111,p,t

βε

+β

≈

β

−β

ε+β

=εβ (2.18)

where βeff effective stiffness of the fluid-air mixture, [bar] εA the volumetric ratio of free air in the fluid βF stiffness of the pure fluid according to, [bar] βA the air stiffness according to, [bar] p pressure, [bar]

The volumetric ratio is defined as:

AF

AA VV

V+

=ε (2.19)

where

εA volumetric ratio of free air in the fluid VA the volume of air, [m3] VF volume of the fluid, [m3]

Assuming adiabatic conditions the volume and stiffness of the air may be determined as:

adc1

a

atm0AA p

pVV

⋅= (2.20)

aadA pc ⋅=β (2.21) where

VA volume of air, [m3] VA0 volume of air at atmospheric pressure, [m3] patm atmospheric pressure ≈ 1 bar, [bar] pa absolute pressure, [bar] cad adiabatic constant for air, 1.4

The volume of the fluid is determined from:

( ) ( )( )p,t

tVp,tV 00

0FF ρρ

⋅= (2.22)


Chapter 2

where

VF volume of the fluid, [m3] VF0 volume of the fluid at atmospheric pressure and a reference

temperature, [m3] ρ0 mass density at atmospheric pressure according to Equation

(2.3), [kg/m3] ρ the mass density according to Equation (2.2), [kg/m3] t0 reference temperature, [°C] t temperature, [°C] p pressure, [bar]

From Equation (2.20) and Equation (2.22) it is clear, that the volumetric ratio varies with both temperature and pressure. A reference volumetric ratio at atmospheric pressure is defined:

0A0F

0A0A VV

V+

=ε (2.23)

where 0Aε the reference volumetric ratio of free air in the fluid at

atmospheric pressure VA0 volume of air at atmospheric pressure, [m3] VF0 volume of the fluid at atmospheric pressure and a reference

temperature, [m3] Knowing this reference, volumetric ratio together with the reference temperature, t0, may be rearranged to yield an expression for the volumetric ratio directly obtainable from temperature and pressure:

( )( )

( ) 0.1p

pp,tt0.1

0.1p,tadc1

a

atm00

0A

0A

A

+

⋅

ρρ

⋅

ε

ε−=ε −

(2.24)

In Figure 2.6 the variation of the effective stiffness according to Equation (2.18) is displayed. The variation of the stiffness is dramatic for small pressure levels. The curves in Fig. 2.6 do not take into account the effect of the Henry-Dalton sentence, Equation (2.13), according to which the free air should dissolve at a few bars pressure and subsequently have no effect on the effective stiffness. The Henry-Dalton sentence, however, is for static conditions and in a hydraulic system the pressure variations outside the tank reservoirs are typically so fast, that the hydraulic fluid does not have time to dissolve the free air. Naturally, some air is dissolved, meaning that the curves shown in Fig. 2.6 represents worst case, i.e., instantaneously pressure build up. Stiffness plays a central role w.r.t. to dynamic performance of hydraulic systems and should be determined/predicted as precisely as possible. This is, however, not an easy task. The sum VF0+VA0 in Equation (2.23) is relatively easily determined, whereas VF0 or VA0 are more elusive.


Chapter 2

As a rule of thumb, the stiffness under working conditions used for modelling a system should not be set above 10000 bar, unless verified by means of testing.

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

0 20 40 60 80

p [bar]

]bar[effβ

100

00A =ε

005.00A =ε01.00A =ε02.00A =ε05.00A =ε1.00A =ε

Fig. 2.6 Variation of effective stiffness of fluid-air mixture with respect to pressure and volume ratio of free air at atmospheric pressure. The temperature of the fluid is 40 °C and the

compression of the free air is assumed adiabatic 2.2 Flow Characteristics of Spool Valves 2.2.1 Introduction Hydraulic control valves are devices that use mechanical motion to control a source of fluid power. They vary in arrangement and complexity, depending on their function. Because control valves are the mechanical to fluid interface in hydraulic systems, their performance characteristics are essential. Although emphasis is placed on a principal type of spool valve, the principles apply equal well to other valves, such as different kinds of pressure valves and flow control valves. The most common used control valve is the spool valve. Two typical spool valve configurations are shown in Figure 2.7. One in the pilot stage and one in the main stage. Spool valves can broadly be classified by the number of “ways” the flow can enter and leave the valve, and the type of centre when the valve spool is in neutral position.


Chapter 2

P

Pload supplyP

tankP Ppilot

tankP

Pilot stage

Main stage

tank

Fig. 2.7 Typical spool valve configurations Because all valves require a supply, a return, and at least one line to the load, valves are either three-way or four-way (see Figure 2.7). Two-port valves are also available. However, two-way valves cannot provide a reversal in the direction of flow. If the width of the land is smaller than the port in the valve sleeve, the valve is said to have an open centre or to be underlapped. A critical centre or zero lapped valve has a land width identical to the port width and is a condition approached by practical machining. Closed centre or overlapped valves have a land width greater than the port width when the spool is at neutral. The above examples serve the purpose of illustrating how flow paths may be created using a variety of restrictions. The actual displacement of the spool which cause the flow restriction is usually of such small value relative to port diameter that the pressure/flow equations obey the Bernoulli equation. 2.2.2 Flow through orifices The flow restrictions or orifices are a basic means for the control of fluid power. An orifice is a sudden restriction of short length in a flow passage and may have a fixed or variable area. In fluid power is it only inertia and viscous forces that matters. Experience has shown that it is either the inertia forces or the viscous forces that dominate, giving two types of flow regimes. Therefore, it is useful to define a quantity which describes the relative significance of these two forces in a given flow situation. The dimensionless ratio of inertia forces to viscous force is called Reynolds number and defined by

µρ

= hduRe (2.25)

where ρ is fluid mass density, µ is absolute viscosity, u is the average velocity of flow, and is a characteristic length of the flow path. hdIn our case is taken to be the hydraulic diameter which is defined as: hd

perimeterflowareaflow4d h

×= (2.26)


Chapter 2

Flow dominated by viscosity forces is referred to as laminar or viscous flow. Laminar flow is characterised by an orderly, smooth, parallel line motion of the fluid. Inertia dominated flow is generally turbulent and characterised by irregular, eddylike paths of the fluid. In some cases viscosity is only important in the boundary layer, while the main flow outside the boundary layer is dominated by inertia and behaves like laminar flow. If the boundary is neglected, the resulting flow is called potential flow. Potential flow has no losses while it is frictionless, so Reynolds number is infinite. For potential flow the Navier-Stokes equations reduce to

2up 2

+ρ

= constant (2.27)

Equation (2.27) is Bernoulli’s equation with negligible gravity forces. As an important case where Equation (2.27) is used consider flow through an orifice (see Figure 2.8).

1

0A

2

A 2

3

Fig. 2.8 Flow through an orifice; turbulent flow Since most orifice flow occur at high Reynolds numbers, this region is of great importance. Experience has justified the use of Bernoulli’s equation in the region Between point 1 and 2. The point along the jet where the area becomes a minimum is called the vena contracta. The ratio between the area at vena contracta and the orifice area defines the so called contraction coefficient

2A

0A cC .

02c A/AC = (2.28) After the fluid has passed the vena contracta there is turbulence and mixing of the jet with the fluid in the downstream region. The kinetic energy is converted into heat. Since the internal energy is not recovered the pressures p and are approximately equal. 2 3pNow it is possible to use Bernoulli’s equation (2.27) to calculate the relation between the upstream velocity to the velocity u in vena contracta. Therefore 1u 0

)pp(2uu 2121

22 −

ρ=− (2.29)

Applying the continuity equation for incompressible flow yields


Chapter 2

332211 uAuAuA == (2.30) Combining Equation (2.29) and Equation (2.30) and solving for gives 2u

)pp(2AA1u 21

2/12

1

22 −

ρ⋅

−=

−

(2.31)

In the real world there will always be some viscous friction (and deviation from ideal potential flow), and therefore an empirical factor is introduced to account for this discrepancy. is typically around 0.98. Since the flow rate at vena contracta becomes, by using Equation (2.31).

vCQvC 22uA=

)pp(2)A/A(1

ACQ 212

12

2v −ρ−

= (2.32)

Defining the discharge coefficient in Equation (2.32) it is possible to express the orifice flow by the orifice area.

dC

210

2c

cvd

)A/A(C1

CCC

−= (2.33)

Now, combining Equation (2.28), (2.32), and (2.33) the orifice equation (in Danish blændeformlen) can be written

)pp(2ACQ 210d −ρ

= (2.34)

Normally is much smaller than A and since C , the discharge coefficient is approximately equal to the contraction coefficient. Different theoretical and experimental investigations has shown that a discharge coefficient of is often assumed for all spool orifices.

0A 1 1v ≈

6.0Cd ≈

200

0

0.2

10

0.4

Cd

0.8

0.6

Re30 40 50

Fig. 2.9 Plot of a discharge coefficient versus Reynolds number for an orifice


Chapter 2

At low temperatures, low orifice pressure drop, and/or small orifice openings, the Reynolds number may become sufficiently low to permit laminar flow. Although the analysis leading to Equation (2.34) is not valid at low Reynolds numbers, it is often used anyway by letting the discharge coefficient be a function of Reynolds number. For Re < 10 experimental results shows that the discharge coefficient is directly proportional to the square root of Reynolds number; that is Red δ=C . A typical plot of such a result is shown in Figure 2.9. 2.2.3 General valve analysis In this section we define some general performance characteristics, such as pressure-flow curves and valve coefficients, which are applicable to all types of valves. Although the analysis is illustrated with a spool type valve, the principles involved are quite general. Consider the four-way valve shown in Figure 2.10. It is assumed that the valve is connected to a symmetric load, i.e. a rotating motor or a equal area cylinder. The valve geometry is assumed ideal, implying that the orifice edges are perfectly square with no rounding and that there is no radial clearance between the spool and sleeve. It is also assumed that the discharge coefficients for the orifices are equal. The return line pressure is neglected because it is usually much smaller than the other pressures involved.

Rp

LOAD

2X

2P QL

3

Return

Supply

P QSR

QPS S

LQ

1

P1

4

Fig. 2.10 Four-way spool valve Let the spool be given a positive displacement from the null or neutral position, that is the position x = 0, which is chosen to be the symmetrical position of the spool in its sleeve. This allows the supply flow to travel to the load as , the difference being only leakage flow present, , across the other land. The flow from the load returns as Q , which with the possible addition of the leakage flow , then forms the return flow. Because we are only interested in the steady-state characteristics, the compressibility flows are zero and the flow continuity equations for the valve chambers are

SQ 1Q

2

4Q 3

Q


Chapter 2

23L

41L

QQQQQQ

−=−=

(2.35)

where is the flow through the load. The load pressure differential is defined as the pressure drop across the load.

LQ

21L ppp −= (2.36)

Flows through the orifices are described by the orifice Equation (2.34). Therefore

)pp(2ACQ 1S1d1 −ρ

=

23d3 p2ACQρ

=

)pp(2ACQ 2S2d2 −ρ

=

14d4 p2ACQρ

= (2.37)

In the vast majority of cases the metering orifices are made so that they are matched and symmetrical. Matched orifices require that

31 AA = ; 42 AA = (2.38) And symmetrical orifices require that (x is the spool position)

)x(A)x(A 21 −= ; )x(A)x(A 43 −= (2.39) This means, that in neutral position (x = 0), all four orifice areas are equal ( ≈ ). If further the orifice areas varies linear with the stroke, as is usually the case, the areas can be described by only one parameter w, defining the width of the slot in the valve sleeve. w is the area gradient.

0A

wxAAA;wxAAA 042031 ⋅−==⋅+== (2.40)

The condition that the orifices are matched and symmetrical, gives that

31 QQ = ; 42 QQ = (2.41) Substituting Equation (2.37) ( Q and ), and Equation (2.38) into Equation (2.41) yields

1 3Q

21S ppp += (2.42) Now, Equation (2.36) and Equation (2.42) can be solved simultaneously to obtain

2ppp LS

1+

= ; 2

pp LS2

−=p (2.43)


Chapter 2

With the relations in Equation (2.41), and Equation (2.43) together with the equations in Equation (2.35) it is possible to find an expression for the load flow as a function of the load pressure.

)pp(1AC)pp(1ACQ LS2dLS1dL +ρ

−−ρ

= (2.44)

Equation (2.44) represent the general steady-state valve equation for a symmetric matched four-way spool valve applied to a symmetric load. 2.3 Valve Coefficients 2.3.1 Introduction It will be found necessary in a dynamic analysis that the non-linear algebraic equation which describe the pressure-flow curves to be linearised. From Equation (2.44) the load flow can be written as a function of the spool position and the load flow

. If and p only changes by a small amount about a operating point ( Q ) the general expression for the load flow can be expressed by a Taylor’s series. We only consider the first order terms, assuming that the higher order infinitesimals are negligible small. Hence,

)p,x(QQ LLL =

00L0L x,p,x L

...ppQ

xx

QQQ L

0L

L

0

L0LL +∆

∂∂

+∆∂

∂+= (2.45)

The partials in Equation (2.45) defines the two most important parameters for a valve. The flow gain is defined by

xQ

k Lq ∂

∂≡ (2.46)

The flow-pressure coefficient is defined as

L

Lqp p

Qk∂∂

≡ (2.47)

Another useful quantity is the pressure sensitivity defined by

xpk L

p ∂∂

≡ (2.48)

There is the following relation between the quantities

LL

LL

p/Qx/Q

xp

∂∂∂

=∂∂ or

qp

qp k

k=k (2.49)


Chapter 2

The three quantities are called valve coefficients and are extremely important in the dynamic analysis of valves in combination with actuators. express the ability to of a valve-actuator combination to breakaway large friction loads. k has a direct influence on the damping in the valve-actuator combination. k directly affects the open loop gain in a system and therefore has influence on system stability. The valve coefficients evaluated in the neutral position of the valve are called the null valve coefficients. This operating point is the most critical point from a stability viewpoint, while the flow gain is largest, giving high system gain, and the flow-pressure coefficient is smallest, giving a low damping.

qk , qpk , pk

pk

) =

qp

0,0(

q

p, 0L )0,x,Q( 00L

2.3.2 A critically lapped valve with linear ports Many valves are manufactured with a relative linear flow gain near null position, meaning that in Equation (2.40). Assuming the valve to have ideal geometry the leakage flows are zero. For such a valve the load flow can be expressed by

0A0 =

ρ−

⋅= LSdL

ppxwCQ ; 0x > (2.50)

while and in Equation (2.44). xwA1 ⋅= 0A2 =

ρ+

⋅−= LSdL

ppxwCQ ; 0x < (2.51)

with and in Equation (2.44). xwA 2 ⋅−= 0A1 =Equation (2.50) and (2.51) can be combined into a single equation:

)pxxp(1xwCQ LSdL −

ρ⋅= (2.52)

This is the general equation for the pressure-flow curves of an ideal critical centre valve with matched and symmetrical orifices. The Equation (2.52) is plottet in Figure 2.11.

-PS

QL

PL

PS

x

Fig. 2.11 Pressure-flow curves of critical centre four-way valve


Chapter 2

The valve coefficients for the important case of an ideal critical centre valve can be obtained by differentiation of Equation (2.44), and are given below in Figure 2.12.

General valve coefficients Null valve coefficients

qk

)pp(1wC 0LSd −ρ

ρ

Sd

pwC

qpk

)pp(2xwC

0LS

0d

−ρ⋅

0

pk

0

0LS

x)pp(2 − ∞

Fig. 2.12 Valve coefficients for a critical centre four-way valve

2.4 Flow Forces on Spool Valves Consider the steady-state flow through a spool valve as shown in Figure 2.13. When the fluid is flowing through the valve there will be induced some forces acting on the valve.

FR

Face a

X 1V

L

Face b

1p

elementFluid

p

Θ

2V2

Fig. 2.13 Flow forces on a spool valve due to flow leaving a valve chamber

These forces are normally calculated using a mathematical formulation of Newton’s second law suitable for application to a control volume.

∫∫ ⋅ρ+ρ∂∂=+=

CSCVBS AdVVVdVt

FFFrrrrrrr

(2.53)

This equation states that the sum of all forces (surface and body forces) acting on a non-accelerating control volume is equal to the sum of the rate of change of momentum inside the control volume (CV) and the net rate of efflux of momentum through the control surface (CS).


Chapter 2

Since we are looking for the horizontal force the body force is zero, and the only surface force in horizontal direction is the force F , which is the force of the spool on the control volume. Change of momentum inside the control volume occur when the spool position is suddenly changed, say to the right, as shown in Figure 2.13. If the fluid element is being accelerated, the pressure on the left side of the element must be greater than the pressure on the right side. Therefore, the pressure on face a must be greater than the pressure on face b. Thus the transient flow force is due to acceleration of the fluid in the annular valve chamber. The direction of this force for the case shown in Figure 2.13 is such that it tends to close the valve – however this is not the general rule. A movement of the spool can also cause the fluid to be decelerated. Applying the momentum equation in the horizontal direction gives

R

)cos(QVdt

)A/Q(dLAFF 2n

nR θρ+ρ== (2.54)

where is the volumetric flow rate and A is the annular area of the spool. The last term in Equation (2.54) can be rewritten as

Q n

0c

2

2

2

2 ACQ

AQ)cos(QV ρ=ρ=θρ (2.55)

Where is the orifice area. The flow can be described by the orifice equation as 0A Q

)pp(2ACC)pp(2ACQ 210vc210d −ρ

=−ρ

= (2.56)

Obtaining dQ from Equation (2.56), the transient flow force, , becomes dt/ tF

dt)pp(d

)pp)(/2(wxLC

dtdx)pp(2wLCF 21

21

d21dt

−−ρ

+−ρ= (2.57)

The last term in Equation (2.57) is normally neglected. The velocity term is more significant because it represents a damping force. The quantity L is the axial length between incoming and outgoing flow and is called the damping length. Inserting Equation (2.56) into Equation (2.55), the steady-state axial flow force acting on the valve spool can be obtained as

)cos()pp(ACC2F 210vds θ−= (2.58) For a spool with no radial clearance it is well known, and usually assumed, that the jet leaves the control port at an angle of . The sum of Equation (2.57) and Equation (2.58) give the total flow force, steady-state and transient, opposing the spool motion, while the force from the fluid on the spool is opposite F , which is the external force acting on the control volume.

o69=θ

R

----- oo0oo -----


Chapter 3

3 Dynamic Modelling

Fluid Power Systems (Hydraulisk Komponentanalyse) AaU ~ Forår 2003

3.1 3.2 3.3 3.4 3.5 3.6

Introduction....................................................................... Lumped Fluid Theory..............................………….…… Flow continuity equation • Momentum equation • Application of fluid theory to a line Cylinders...........................……………………………… Motors...........................................................…………… Linear Characteristics of a Cylinder................................. Over Centre Valve............................................................ Introduction • Functional description • Mathematical model

1

3

5

6

8

11

3.1 Introduction Hydraulic control systems are used to control the position or speed of resisting loads. The final drive is usually either a linear motion hydraulic cylinder or a rotary-motion hydraulic motor. Figure 3.2 illustrates two simple hydraulic control systems. The actuator develops its force or torque by receiving liquid from a positive displacement pump at a relatively high pressure, and the actuator develops its motion by receiving flow rate of liquid from the pump. It is not sufficient for the designer of a hydraulic system to know that the system will move. He also need to know how it will move. He needs to appreciate not only the initial and final states of the responses, but also the time-domain path between these states. He should know if the system response is stable, if it is fast enough or perhaps too fast, if it is oscillatory, etc. A meaningful dynamic analysis cannot be made without including the dynamics associated with the linear cylinder or the rotary motor. One of the elements the designer must take into account is the natural frequency of the system. From the laws of physics we know that the formula for undamped, natural frequency is

m/cn =ω

where c the spring constant m the moving mass.

Fig. 3.1 Spring-mass system


Chapter 3

4-Way Control Valve

Filter

p

p

T

n

pumpHydraulic

Reservoir

Pressure Relief Valve

150 bar

Hydraulic Cylinder

Vc

Fc

Flow ControlValve

TmnHydraulic Motor

m

Fig. 3.2 A linear and rotary drive hydraulic control system

The spring constant for a hydraulic system can be related directly to the oil volume trapped between the controlling element, typical a valve, and the actuator. The natural frequency determine how fast a load can be accelerated and decelerated without causing instability and subsequent damage to the system. This frequency can be calculated mathematically to tell us how fast this weight can be moved back and forth without having the weight of the object directly oppose the input to the spring. For example, the input to the spring-mass system (see Figure 3.1) could be someone’s hand moving the spring-mass system up and down a certain distance. As long the spring is moved more slowly than the natural frequency of the total spring-mass system, the weight will follow the movement of the spring. There will be very little difference between the movement of the spring and the weight. The faster the input, or hand movement to the spring, the more the weight will lag. If the input to the spring is at the same frequency as that of the total spring-mass system, as one’s hand moves down, the weight moves up. Likewise, as the hand moves up, the weight moves down. The weight would then be in direct opposition to the movement of the spring. This would result in a system performing a function opposite of that required. This is called instability, or resonance. To put this concept in perspective with respect to a hydraulic system, the natural frequency can be calculated and trying to accelerate or decelerate high inertia loads too quickly is likely to cause a cylinder or motor to become unstable. In the following sections the nonlinear equations of a hydraulic cylinder and rotary motor will be derived, and the natural frequency and damping of a symmetrical cylinder. But first some necessary fluid theory will be introduced.


Chapter 3

3.2 Lumped Fluid Theory 3.2.1 Flow continuity equation The physical principle used to derive the continuity equation is conservation of mass. It is intuitive that mass neither can be created nor destroyed; if the flow rate of mass into a control volume exceeds the rate of flow out, mass will accumulate within the control volume. Conservation of mass requires that the sum of the rate of change of mass within the control volume (CV) and the net rate of mass outflow through the control surface (CS) be zero.

∫∫ ⋅ρ+ρ∂∂=

CSCVAdVVd

t0

rr (3.1)

In Equation (3.1) the first term represents the rate of change of mass within the control volume; the second term represents the net rate of mass efflux through the control surface. Now consider the control volume given in Figure 3.3 and the Equations in (3.2).

Control V

olume

Qρ

Qp

in

min

in

in

V Q

pρ out

mρout

outp

Fig. 3.3 Fluid volume

outinCVmmVd

t0 && −=ρ

∂∂= ∫

⇓

)V(t

QQ outoutinin ρ∂∂=ρ−ρ

(3.2)

If a mean density, ρ , is assumed throughout then expanding Equation (3.2) gives

dtdV

dtdVQQ outin

ρρ

+=− (3.3)

Equation (3.3) needs to be transformed to a more usable form. This can be achieved by using Equation (2.14) and Equation (2.15). The latter defining the bulk modulus of the fluid.

F

dpdβ

=ρρ (3.4)

Combining Equation (3.3) and Equation (3.4) gives

dtdpV

dtdVQQ

Foutin β

+=− (3.5)

The first term on the right side is the flow consumed by expansion of the control volume; if the volume is fixed this term is zero. The second term is the compressibility flow and describes the flow resulting from pressure changes.


Chapter 3

Normally expansion of the control volume due to wall deformation in the different components is assumed to be contained in the definition of the effective bulk modulus

Fβ . Equation (3.5) will be referred to as the flow continuity equation. 3.2.2 Momentum equation Newton’s 2nd law of motion states:

applied force - resisting forces = mass × acceleration ∑ (3.6) Applying the equation to motion of a fluid volume gives:

∑ =−−dtdvMF)ApAp( ioutoutinin (3.7)

where is the input net cross-sectional area and is the output net cross-sectional area. M is the mass of the fluid being accelerated. To change the velocity in for example a pipe or a cylinder the fluid needs to be accelerated. The necessary force is generated by a pressure difference.

inA outA

The application of Equation (3.6) depends upon the component being considered. In the following it is applied to a fluid line. 3.2.3 Application of fluid theory to a line In this section we look at a circular pipe of uniform cross-sectional area as given in Figure 3.4.

ρm

A

Fig. 3.4 Circular pipe

From the flow continuity equation (3.5) we can write an expression for the pressure in the pipe

dtdpVQQ

Foutin β

=− ; 4dA,LA

2π==V (3.8)

Where d is the internal diameter of the pipe and L its length. The momentum equation, from Equation (3.7), becomes

∑ ρ=−−dtdQLFA)pp( ioutin (3.9)

The resisting force is assumed to be entirely due to fluid viscosity effects. Other effects are due to flow in a piping system may be required to pass through a variety of

∑ iF


Chapter 3

fittings, bends, or abrupt changes in area. The loss can be expressed as a pressure loss, meaning

ApFi∑ ∆= ; where 2

vdlp

2

ρλ=∆ (3.10)

Where v is the fluid velocity. The friction factor is determined experimentally, and can be expressed as a function of Reynold’s number :

λµρ= /dvRe

For laminar flow (Re < 2300): Re64=λ

For turbulent flow (Re > 2300): 25.0Re3164.0=λ

(3.11)

where µ is the dynamic (or absolute) viscosity. 3.3 Cylinders Consider the asymmetric cylinder shown in Figure 3.6. We want to write up the equations describing the dynamics of the cylinder and load.

Q1

Q

1p

2p

V1 V2

2AP AR

M L

FL

Fig. 3.5 Asymmetric linear actuator

Applying the momentum equation (3.6) to the cylinder motion gives

LR2P1L FApApdtdvM −−= (3.12)

Consider next the flow continuity equation (3.5) applied to the actuator. Extending

dtdpV

dtdV0Q 1

F

111 β

+=−

dtdpV

dtdVQ0 2

F

222 β

+=−

Retracting

dtdpV

dtdV0Q 2

F

222 β

+=−

dtdpV

dtdVQ0 1

F

111 β

+=−

(3.13)


Chapter 3

Simplifying Equation (3.13) gives Extending

dtdpVvAQ 1

F

1P1 β

+=

dtdpV

vAQ 2

F

2R2 β

−=

Retracting

dtdpV

vAQ 2

F

2R2 β

+=

dtdpVvAQ 1

F

1P1 β

−=

(3.14)

If we are consistent about the notation, it is possible to reduce Equation (3.14). If the flow directions shown in Figure 3.6 are termed positive and the extending velocity is termed positive, then Equation (3.14) reduce to

)vAQ(Vdt

dpP1

1

F1 −β

=

)QvA(Vdt

dp2R

2

F2 −β

= (3.15)

A cylinder is a full stroke component and dynamically both volumes and V vary with piston motion. In linearised analysis we are often interested only in the transient behaviour around some operating point. In this situation it is ok to neglect the variation of the volumes. In a nonlinear simulation model, however, the variation must be taken into account.

1V 2

Also, other phenomena like viscous friction and spring terms can be added to the momentum equation. 3.4 Motors The rotary hydraulic motor (see Figure 3.6) is an important element in hydrostatic transmissions. The motor transforms, as earlier described, hydraulic power to mechanical power. Newton’s 2nd law of motion for a rotating device states:

generated torque - resisting troques = inertia × angular acceleration ∑ (3.16) Now, let us look at the elements in Equation (3.16). The ideal generated torque is

)pp(DT 21mg −= (3.17) where , by definition is the flow through the motor ( ) divided by the shaft speed of the motor (

mD LQω);

ω≡ /QD Lm (3.18)

and is called the volumetric displacement (or simply displacement) of the motor, [ cm ]. In Equation (3.17) the term is the pressure difference across the motor.

rad/321 pp −


Chapter 3

Q2p

2

Q p1 1

JL

ω

TL

Fig. 3.6 Rotary motor with inertia load and load torque However, there are at least three sources of resisting torque losses which detract from the generated torque. The first one is viscous damping due to shearing in the fluid in the tight clearances between the mechanical elements in relative motion. With as a viscous damping coefficient, this damping torque can be written as

mB

ω= md BT (3.19)

Investigation into the movement and forces of each piston gives a friction force opposing motion that is proportional to the pressure acting on the piston.

)p,p,(fT 21f ω= (3.20) This nonlinear term in Equation (3.20) has to be experimentally determined. Now, the torque equation (3.16) can be written as

Lfd21mL TTT)pp(DdtdJ −−−−=ω (3.21)

In the torque equation the load inertia and load torque T are included, see Figure 3.6. Looking at the fixed displacement axial piston motor schematical represented in Figure 3.7, we can write down the continuity equations for the two motor chambers.

LJ L

Q

lines

Stationaryvalving plate

P2 Q Q2 il

Drain line

Rotating cylinder barreland drive shaft

Q

Fluid

P1 1

Housing

Q

+el1 el2Q

el2

slide on platePiston shoes

Qel1

Stationary cam plateat fixed angle

Fig. 3.7 Fixed displacement axial piston motor


Chapter 3

Only two chambers is shown in Figure 3.7, but the leakage flows from all the pistons are lumped at these two pistons. From the figure we can see that there are at least two types of leakage flows; internal or cross-port leakage between the pressure side and the return side, and external leakage resulting from oil passing the pistons. Due to the very small clearances in hydraulic motors the leakage flows are laminar and therefore proportional to the pressure drop. Thus, the internal leakage can be written

)pp(CQ 21ilil −= (3.22) where is the internal og cross-port leakage coefficient, and the pressure difference across the motor.

ilC 21 pp −

The external leakage is though proportional to the particular chamber pressure and may be written

1el1el pCQ = ; 2el2el pCQ = (3.23) where is the external leakage coefficient. elCThe compressibility flow is normally neglected when writing the continuity equation (3.5) for the motor, since the small internal volumes on either side of the motor usually are added to the line volumes connecting the valve/pumpe and motor. The steady-state continuity equations for the motor chamber then are

1el21ilm1 pC)pp(CDQ +−+ω= (3.24)

2el21ilm2 pC)pp(CDQ −−+ω= (3.25) These two equations completely describe the flows in the motor. If leakage is zero, we have that ω== m21 DQQ (ideal motor). 3.5 Linear Characteristics of a Cylinder In this section we will derive the dynamic characteristics of a cylinder. In Figure 3.8 is schematically shown a double-acting cylinder and the related parameters that will be used in the section.

A

α

P2

2Q

F x

P1

Q1

m

Fig. 3.7 A double-acting piston actuator


Chapter 3

The dynamic characteristics will be expressed by the natural frequency and damping ratio. We consider a situation where the cylinder ports are blocked. Newton’s 2nd law of motion of the piston states:

)pp(AxFxm 21 −−α−= &&& (3.26) where α is a viscous damping coefficient. The flow continuity equations for the two cylinder chambers can be written as

0dt

dPVxAQ 1

F

11 =

β−= & (3.27)

0dt

dPVxAQ 2

F

22 =

β+= & (3.28)

where the chamber volumes and are functions of the piston displacement. The equations (3.27) and (3.28) are non-linear, and we need to linearise the equations around an operating point – the index o meaning evaluated in the operating point. To linearise the product term we utilise Taylor expansion for functions of more than one variabel (see also Equation (2.45)).

1V 2V

)yy(yf)xx(

xf)y,x(f)y,x(f o

oo

ooo −

∂∂+−

∂∂== (3.29)

Applying Equation (3.29) to Equation (3.27) and Equation (3.28) we get

xV

Adt

dp

10

F1 &β= (3.30)

xVA

dtdp

20

F2 &β−= (3.31)

Equations (3.26), (3.30) and (3.31) represent the linear equations describing the dynmics of the cylinder. After Laplace transforming the equations we have

))s(p)s(p(A)s(x)s(f)s(xsm 21 −−α−= && (3.32)

)s(xsV

A)s(p10

F1 &

β= (3.33)

)s(xsV

A)s(p20

F2 &

β−= (3.34)

Equations (3.32), (3.33) and (3.34) may be solved simultaneously to obtain


Chapter 3

)s(f)

V1

V1(

mAs

ms

sm1

)s(x

2010

F2

2 +β+α+=& (3.35)

Comparing the coefficients in the denumerator with a standard second order system we get the natural frequency and damping ratio as

)VVVV

(m

A)V1

V1(

mA

2010

2010F

2010

Fn

+β=+β=ω (3.36)

)VV(mVV

A2 2010F

2010

+βα=ς (3.37)

The total volume under compression is defined V . Inserting in equations 3-36 and 3-37 gives

2010T VV += TV

)VV(VV

mA

10T10

TFn −

β=ω (3.38)

TF

10T10

Vm)VV(V

A2 β−α=ς (3.39)

Now, it is easy to verify that the minimum natural frequency, is when the piston is in its mid position, meaning . In this position we obtain from Equation (3.38) and Equation (3.39).

2/VVV T2010 ==

T

Fminn mV

A2β

=ω and F

T2/V m

VA42 βα=ς (3.40)

EXAMPLE

Given the following values for a double-acting cylinder Area of the piston : 2cm45A =Piston travel : L mm640=Bulk modulus : bar7000F =β Mass of the load and piston : kg100m =Viscous friction coefficient : m/Ns6.175=α

Inserting these values in Equation (3.40) gives

Hz348secrad2190

minn ==ω ; 2.02/VT

=ς


Chapter 3

3.6 Over Centre Valve 3.6.1 Introduction Over centre valves are useful in a variety of mobile fluid power applications. For example they are often used in mobile cranes to ensure that a broken pipeline will not cause the load to drop, see Figure 3.8. Other applications are on vehicle and winch drives, to ensure that the load does not run ahead of the pump flow, creating cavitation inside the motor. When the load is reversing they allow normal operation using by-pass elements and has built in relief valve protection. The valve also provides velocity control on descending load.

Directional Control

LS

T P

Valve

Over CenterValve

Fig. 3.8 Hydraulic sub system for a single actuation

Today, almost all systems including over centre valves, are based on a similar configuration. The flow to the actuator is controlled by a directional control valve and the overcenter valve is placed at the outlet of the actuator, pilot-operated from the pressure in the inlet connection. Over centre valves, are closely related to pressure relief valves and check valves opened by a pilot pressure. An over centre valve is, in effect, a pilot opened pressure relief valve. It is well known that systems equipped with over centre valves are prone to oscillations in the load and can also become unstable. Another unwanted behaviour is the tendency to abruptly stop when the speed of the load is retarded. In many cases, these types of instability can lead to harzardous conditions. Since safety reasons and legislations makes them a necessary element in many systems, they are used despite the drawbacks of their dynamic behaviour, and the method used to get a satisfactory operation is often trial end error. In particular, it is difficult to combine this kind of valve with load sensing systems or systems with constant flow characteristics (see section 5.2). In this section the over center valve will be modelled.


Chapter 3

3.6.2 Functional description The over center valve considered in this section is shown in Figure 3.9. It consist of the valve body (3), control plunger (4), check valve (2), and pilot piston (1).

Valve PortPilot Port 4

12

3

Cylinder Port

Fig. 3.9 Typical over center valve When lifting the load, the fluid passes from the valve port V through the check valve and port C to the consumer. During sinking, the flow directions is opposite – from C to V . The check valve is closed and the flow to the output V is not possible until after the control plunger is lifted due to the load pressure and the pilot pressure. The pilot, or load lowering pressure acts on the pilot piston ( A ), see Figure 3.10, and pushes open the control plunger. Of course also the load induced pressure, which acts directly on the control plunger ( ), tends to push the plunger in the open position.

P

CA

Av1 Av

Ac

2

Ap

AR

Fig. 3.10 Areas for combined actions in opening direction The displacement of the control plunger is proportional to the cracking pressure corresponding with the load. While the load pressure alone can open the valve the preload of the spring determines the relief setting. An important parameter characterising over center valves is the Pilot ratio ( PR ) defined as

)A(PlungerControlofareaalDifferenti)A(PistonPilotofAreaPR

C

P= (3.41)


Chapter 3

A high pilot ratio permits to lower the load with little pilot pressure, allowing a quicker operation of the machine combined with energy saving. It is best suited for applications where the kinematic motion of the structure maintains the load induced pressure relatively constant. A low pilot ratio requires a high pilot pressure in order to lower the load, but it permits more precise and smooth control of motion. It is recommended for applications where the load induced pessure varies during motion and can induce instability on the machine. 3.6.3 Mathematical model In deriving the mathematical model we only consider the case where the load needs to be counter balanced. In the other direction we assume that there is no loss when oil is flowing through the check valve. It normally has a very soft spring and a large opening area creating a negligable pressure drop. Defining the load pressure as , the pilot pressure as , and the back pressure we are able to describe the coverning equations for the over center valve.

Cp Pp Vp ,

The flow trough the valve from the load to downstream the over center valve can be desribed by the orifice equation

)pp(2)x(ACQ VCoodo −ρ

= (3.42)

The opening area can be calculated as shown in Figure 3.11.

xα

do__

o

2

)cossinxsinxd()x(A 22ooooo αα−απ=

Fig. 3.11 Discharge area for a seat valve Normally the opening of such valves are relatively small, meaning that is small. This means that Equation (3.42) can be rewritten as

oo d/x

)pp(2sinxdCQ VCoodo −ρ

απ= (3.43)

Newton’s second law applied to the plunger yields;

CoooSRFPLPVVCCPPoo FxBxKFF)AA(pApApxm −−−−−−++= &&& (3.44) where Figure 3.10 defines the areas where pressure is acting, and

R2V1VV AAAA +−= (3.45)


Chapter 3

The other terms in Equation (3.44) are mass of plunger om preloaded spring force on the plunger PLF

FF flow force

SRK spring rate

oB viscous damping coefficient

CF coulomb friction force If is zero, we say that the over center valve is compensated to back pressure with respect to the relief function. The back pressure still opposes the pilot piston, thus increasing the pilot pressure needed to open the valve and to lower the load.

VA

The preloaded spring force on the plunger is equal to

SRini_oPL KxF = (3.46) where is the initial compression of the spring. ini_oxThe flow force can be found applying the theory from section 2.4. In Figure 3.12 is shown a control volume for the seat valve

V2

Control volume

FF

Ao

V1

FF

Fig. 3.12 Control volume for calculating the flow force Applying the momentum equation (2.53) in the horizontal direction gives

)VV(cosQF 21oF −αρ−= (3.47) Using we have that the force opposing the spool motion can be written 21 VV >>

α−= cos)pp()x(AC2F VCoodF (3.48) Normally static friction from O-rings is modelled as Coulomb friction.

F)x(signF oC &= (3.49) where F is the value of the friction force, but this type of friction is of such complexity that it mostly requires experimental investigation to find the friction characteristic.

----- oo0oo -----


Chapter 4

4 Dynamic System

Analysis Fluid Power Systems (Hydraulisk Komponentanalyse) AaU ~ Forår 2003

4.1 4.2 4.3

Introduction....................................................................... Flow Control Valves................................………….…… Power Elements.................………………………………Introduction • The electro-hydraulic valve • Valve-motor combination • Valve-cylinder combination

1

1

4

4.1 Introduction One of the major concerns of this note is that it attempts to reconcile linearised analysis with dynamic simulation and measurements on a real system. Normally, linearised analysis is extremely useful in understanding the behaviour of systems like those mentioned above, because it shows how parameter variations will affect the system. Dynamic simulation will then make it possible to examine the actual response of the system including non-linearities. It is then possible to predict the behaviour of a system after a design change, or in some other operating points. Finally, actual measurements on a system similar to that studied by dynamic simulation will help improve the simulation model, and make sure that it is realistic. 4.2 Flow Control Valves In Figure 4.1 below is given a restrictor type (two-way) pressure compensated flow control valve.

33p Q

x

2

p2Q

p1 Q1

Fig. 4.1 Pressure compensated flow control valve


Chapter 4

Flow control valves which are constructed in such a way that pressure difference across control orifice is maintained constant, independently of pressure changes at the entry and the outlet port of the valve. Because flow characteristics of these valves are relatively insensitive to changes in supply or load pressure they are referred to as ”pressure compensated” flow control valves. The valves are used in systems which require constant flow delivery to actuators, thus maintaining their constant speed independently of load variations. The dynamic equations that describe the dynamics of the valve can be stated as:

)pp(2xCQ 21d1 −ρ

ω=

)pp(2ACQ 32dd2 −ρ

=

Flow equations (4.1)

(4.2)

dtdpV

dtdxAQQ 22

s12 β−+=

dtdpV

dtdxAQQ 33

s23 β−−=

Continuity equations (4.3)

(4.4)

dtdxB)pp(A)xx(KK

dtxdM 32soo2

2

−−−−+=

Force equation

(4.5)

Where and are the areas of the spool end and the fixed orifice, respectively. is the initial spring force.

sA dA oK K is the spring constant, and is a viscous damping

coefficient. B

ω is the area gradient. The maximum spring displacement is , and is the mass of the compensator spool. and are the volumes on both sides of the restriction.

ox M

1V 3V

The valve in Figure 4.1 is a two-way type. When p the flow through the valve is zero. The initial spring force will move the spool to the right ( . If

is increased or is decreased, there will be a flow through the valve, and . When the spool will move to the left and decrease the metering area of the spool. If the pressure difference p is increased further, then it can be seen from the force equation that the pressure difference p will be proportional to the spring force (in steady state)

)p(p 231 ==

32 p−

)K( o

1

)xx o=

2p >1p 3p

) ≥3p

o32s Kpp(A −

3p−

)xx(KK)pp(A oo32s −+=− (4.6)

From the above equation it can be seen that by choosing the spring constant properly, the flow will approximately be constant despite variations in the pressure difference

. 31 pp −


Chapter 4

In linearising the set of equations it is assumed that p and are constant, - and compressibility of oil is neglected. It is only the two flow equations that are non-linear. Linearising these two equations around some operating point i, gives

1 dA

2211 pKxKq += (4.7) 34232 pKpKq += (4.8)

xAqq s12 &+= (4.9) xAqq s23 &−= (4.10)xB)pp(AKxxM 32s &&& −−−−= (4.11)

where the coefficients K are . {Notice that ( )} 41 K− 43 KK −=

i

21di

11 )pp(2C

xQ

K −ρ

ω=∂

∂= ;

i21

d

i2

12 )pp(2

xCpQK

−ρω−

=∂∂=

i32

dd

i2

23 )pp(2

ACpQK

−ρ=

∂∂= ;

i32

dd

i3

14 )pp(2

ACpQK

−ρ−

=∂∂=

Varying loads on the actuator will cause changes in the pressure drop across the valve which in turn will alter the flow and, therefore, the speed of the actuator. To analyse this effect we which to find the transfer function . )s(p/)s(q)s(H 33= Using Laplace transformation on the above linearised equations gives

)s(pK)s(xK)s(q 2211 +=

(4.12) ))s(p)s(p(K)s(q 3232 −= (4.13)

)s(sxA)s(q)s(q s12 +=

(4.14) )s(sxA)s(q)s(q s23 −= (4.15)

)s(Bsx)s(p)s(p(A)s(Kx)s(xMs 32s2 −−−−= (4.16)

There are six variables and q and the task is now to eliminate the variables and . Using in the following a notation without , one have from Equation (4.16)

2132 q,q,p,p,x

2q3

12 q,p,x )(s

KBsMs)pp(A

x 232s

++−−

= (4.17)

Equation (4.14) and Equation (4.15) gives that q . Using this and inserting Equation (4.13) into Equation (4.14) gives

31 q=

333s23 pKqsxApK +=− (4.18)

And from Equation (4.12), again using 31 qq =

3122 qxKpK =− (4.19)


Chapter 4

Solving the Equations (4.18) and Equation (4.19) for and gives x 2p

sKAKKsqAK)pKq(p

2s13

3s13332 +

++= ;

sKAKK)pKq(KqKx

2s13

333233

++−

=

The expressions for and p can now be put into Equation (4.17) and the transfer function asked for can be found to

x 2

)KK(KKBs)A)KK(B(s)KK(MKKKs)KKBKA(sKKM

)s(p)s(q)s(H

2312s23

223

323222s

232

3

3

−+++−+−+++

== (4.20)

Using Equation (4.20) it is now possible to utilise transient-response analysis and frequency-response methods to obtain the dynamic characteristics of the flow control valve as a function of the parameters defining the valve. 4.3 Power Elements 4.3.1 Introduction The interconnection between a directional/servo valve is well described in the literature. The actuator may be either linear or rotary, and these two actuators may be driven by either a pump or a valve. In this section we will only consider the valve-actuator combination see Figure 4.2. It is impossible to present a generalised all-parameter set of results due to the variety of valve constructions and thereby different dynamics, and also the actuator may be applied to different dynamics.

x1Q

V1

2Q

V2

ps

v

p1 2p

ω

TL

JL

Q

1Q

V1

ps

vx2

V

p

p1

x

2

2

MFL

L

Fig. 4.2 Valve-actuator combinations

The purpose of this section is to define parameters and transfer functions of the valve-actuator combinations. These analyses yield a description of the dynamic performance of the power elements.


Chapter 4

4.3.2 The electro-hydraulic valve As an example is shown a single stage electro-hydraulic valve in Figure 4.3.

Supply

Return

LOAD

Fig. 4.3 Single stage electro-hydraulic valve

If we assume an ideal critical centre valve with matched and symmetrical orifices it was shown in section 2.3 that the linearised valve flow could be written as Equation (2.45)

LqpvqL pkxkQ ∆−∆=∆ (4.21) where is the load average load flow and the load pressure difference. The opening of the valve is described by the spool position . In Equation (4.21) the flow-pressure coefficient is defined by k , while is always negative, making the flow-pressure coefficient always a positive number. An electro-hydraulic actuated valve must also be treated as a dynamic system. Knowing the parameters of the electrical activation it is possible to build a analytical model of the valve, but often the dynamics is obtained by measurement using frequency-response methods. From the frequency response a linear approximate model can be estimated. If the frequency response is approximated with a second order system, then Equation (4.21) can be Laplace transformed and written as (the is omitted in the following)

LQ Lp

L /Qvx

L

∆

qp p∂−∂= LL p/Q ∂∂

)s(pk)s(u1s/2)/s(

k)s(Q Lqp

n2

n

qL −

+ως+ω= (4.22)

where u(s) is an electrical input to the electrical activation. 4.3.3 Valve-motor combination Having described the dynamics of the valve, we need the actuator. In Figure 4.2 is shown the two combinations considered in this text. Consider the left configuration. Applying the continuity equation (3.5) to each motor chamber yields

dtdpVpC)pp(CDQ 1

F

11el21ilm1 β

++−+ω= (4.23)


Chapter 4

dtdpVpC)pp(CDQ 2

F

22el21ilm2 β

−−−+ω= (4.24)

where the two volumes and includes the volumes in the valve, connecting line and/or manifold, and volume in the motor. The other parameters are defined in section 3.4.

1V 2V

Define the average contained volume of each motor chamber , and the total contained volume of both chambers V . Then we can write

0V

t

021t V2VVV =+= (4.25)

It is desirable to express the continuity equations (4.23) and (4.24) in terms of the load flow, as done in Equation (4.24). Using that we get from Equation (4.23) and Equation (4.24) 2/)QQ(Q 21L +=

( ) ( )dt

ppd2Vpp

2CCDQ 21

F

021

elilmL

−β

+−

++ω= (4.26)

Equation (4.26) is linear and can be Laplace transformed to give

)s(ps4V)s(pC)s(D)s(Q L

F

tLtlmL β

++ω= (4.27)

were is the total leakage of the motor. 2/CCC eliltl += The final equation is the torque equation (3.21).

)s(T)s(B)s(sJ)s(pD LmLLm +ω+ω= (4.28) Non-linear friction terms may also be present, but is neglected in the linear analysis. The equations (4.22), (4.27), and (4.28) define the valve-motor combination. Normally, the valve dynamics is much faster than the actuator and load dynamics, so it can be neglected. Combining the above describing equations then yields

++

β

++β

β+−

=ω

2m

qptm2mF

tm2m

Lqpt22mF

Lt

LqptF

t2m

qptv

m

q

DkB

1sD4VB

DJk

sD4JV

)s(Tsk4

V1Dk

)s(xDk

)s( (4.29)

Where is the total flow pressure coefficient. Equation (4.29) describe the dynamics of the valve-motor combination – the output being the motor shaft velocity, due to the two inputs, valve spool position and the load torque. The term

tlqpqpt Ckk +=


Chapter 4

2mqpt D/k is usually much smaller than , meaning that the term is small

compared to unity and can therefore be neglected. mB 2

mqptm D/kB

)s( =ω )s(Ts

L⋅

s(ω

L

t

J=

p1 −

p1 −

pA

1V

10V

Splitting up the transfer function in Equation (4.29) due to the two inputs we get

)s(x1s2s

Dk

v

n2n

2m

q

⋅+

ως+

ω

and 1s2s

k4V

1Dk

)

n2n

2qptF

t2m

qpt

+ω

ς+ω

β+−

= (4.30)

where the natural frequency nω and damping ratio ς is given by

Lt

2mF

n JVD4β

=ω and Fm

m

t

LF

m

qpt VD4B

VJ

Dk

β+βς (4.31)

The response due to the valve input is often used in analysis and design and knowledge of the involved parameters is essential. It is worth mentioning that the flow-gain directly affects the open loop gain of the system, and that the flow-pressure coefficient directly affects the damping. The response of the system is determined by the natural frequency, which is dominated by the inertia load and the trapped oil springs between valve and motor. 4.3.4 Valve-cylinder combination The approach describing the dynamics of the valve-cylinder combination shown to the left in Figure 4.2 is the same as for the valve-motor combination. Applying the continuity equation (3.5) to each of the piston chambers yields

dtdpV)p(CxAQ 1

F

12tlp1 β

++= & (4.32)

dtdpV)p(CxAQ 2

F

22tlp2 β

−+= & (4.33)

where is the piston area, and is the total leakage coefficient. The two volumes

and includes the volumes in the valve, connecting line and the cylinder chambers. Actually the volumes is not constant but varies with the piston position as

tlC

2V

xAVV p101 += and xAVV p202 −= (4.34)

where and are the initial volumes in the cylinder chambers. However, it will be assumed that the piston is centred in the analysis. This is ok from a stability point of view, while it was shown in section 3.5 that the minimum natural frequency is when the piston is centred. As in the previous section we define the total volume under compression

20V

tV

201021t VVVVV +=+= 4-35


Chapter 4

Now following the development of Equation (4.27) we get

)s(ps4V)s(pC)s(xA)s(Q L

F

tLtlpL β

++= & 4-36

Using Newton’s second law to the piston and load, the resulting force equation, Laplace transformed is

LpLLp F)s(xB)s(xsM)s(pA ++= && 4-37 where is a viscous damping coefficient of piston and load, and an arbitrary load force. Equations (4.21), (4.36), and (4.37) are the basic equations describing the dynamics of the valve-cylinder arrangement. Combining these equations we get

pB LF

++

β++

β

β+−

=

2p

qptp2pF

tp2p

Lqpt22pF

Lt

LqptF

t2p

qptv

p

q

AkB

1sA4VB

AMk

sA4

MV

)s(Fsk4

V1Ak

)s(xAk

)s(x& 4-38

Where is the total flow pressure coefficient. Equation (4.38) describe the dynamics of the valve-motor combination – and gives the response of the piston, due to the two inputs, valve spool position and the load force.

tlqpqpt Ckk +=

The term is usually much smaller than , meaning that the term is small compared to unity and can therefore be neglected.

2pqpt A/k pB 2

pqptm A/kB

Splitting up the transfer function in Equation (4.38) due to the two inputs we get

)s(x1s2s

Ak

)s( v

n2n

2p

q

⋅+

ως+

ω

=ω and )s(F1s2s

sk4

V1

Ak

)s( L

n2n

2qptF

t2p

qpt

⋅+

ως+

ω

β+−

=ω 4-39

where the natural frequency nω and damping ratio ς is given by

Lt

2pF

n MVA4β

=ω and LF

t

pt

LF

p

qpt

MV

A4Bp

VM

Ak

β+β=ς 4-40

Comparing the transfer functions with those found for the valve-motor combination they are very much alike – the displacement for the motor being replaced with the piston area, and the inertia load replaced by the mass of the load. The same conclusions can then be drawn, and care must be taken, especially considering the natural frequency and damping ratio that varies a lot with different operating points.

----- oo0oo -----


Chapter 5

5 Advanced System and

Component Analysis Fluid Power Systems (Hydraulisk Komponentanalyse) AaU ~ Forår 2003

5.1 5.2 5.3 5.4

Introduction....................................................................... Hydraulic Load Holding Circuit...............……….……... Introduction • Mathematical model for the circuit • Transfer function of the mathematical model • Stability analysis Load Sensing Directional Valve...……………………… Introduction • Function of the PVG 32 valve • Modelling of the PVG 32 valve • The pump side module (PVP) • The pressure compensator and main spool (PVB) Two Stage Relief Valve.................................................... Performance characteristics of a PCV • Modelling af the two-stage PCV • A detailed dynamic model (Model I) • Simplifications of Model I (Model II) • Linearisation of the Model (Model III)

1

1

12

19

5.1 Introduction The object in this chapter is to relate the material presented in the previous chapters to more in depth analysis of more complex components and systems. The following studies are presented: ° Stability analysis of hydraulic load holding circuits. From this, stability criteria can

be obtained which can give guidelines on how the over centre valve should be dimensioned to ensure stability.

° A mathematical model of a pressure compensated mobile proportional valve. The

model can for example be used to simulate the operation of a hydraulic system that includes proportional valves.

° Modelling and analyses of a two-stage pressure relief valve. Hereby getting a

complete understanding of how the valve works and how internal and external parameters affects the performance.

A variety of modelling and control techniques are therefore presented by these studies. 5.2 Hydraulic Load Holding Circuit


Chapter 5

5.2.1 Introduction The purpose of this chapter is twofold. First of all it is to define parameters and determine the transfer function of a basic hydraulic load holding circuit with varying load. This analysis yields a description of the dynamic performance, knowledge of which is absolutely essential in the rational design of load holding circuits. Second the purpose is to illustrate a method used in analysing such systems, giving an understanding of the problem. The most important asset of a hydraulic system is stability, and therefore stability should be based on hard quantities. Quantities, that can be easily identified, and determined with fair precision and whose values remain relatively constant. For a more in depth analysis performed, the performance index will more or less depend on soft quantities, making stability and performance judgement more difficult, and computer simulation becomes necessary. For example, the flow-pressure coefficients of orifices, and most damping ratios are soft quantities. The differential equations that describe hydraulic components and systems are non-linear and, in most cases, of high order. Preliminary dynamic analysis is therefore necessarily restricted to linear differential equations, because only they may be solved without great difficulty. Also because general performance indices have only been developed for linear systems, makes linear analysis perhaps the only tool. 5.2.2 Mathematical model for the circuit The hydraulic circuit studied in this section is given by the schematic design in Figure 5.1. Note that the system is quite general. Nothing is said about the components in the system, only how they are connected.

Spool position (Xo)

Piston position (Xp)

Volume (Vr)

Cylinder ratio (CR) = Ap/Ar

Pressure compensatedflow control valve

Pressure (Pr)Flow (Qr)

Pilot ratio (PR) = Ac/Ao

Ao Ac

Pressure (Pb)

Spring constant (Ko)

Volume (Vf)

Pressure (Pf )Flow (Qf )

T

LS

P

Ap Ar

Load (Mt)

Fig. 5.1 Schematic design of hydraulic circuit


Chapter 5

This kind of system is non-linear because of the flow-pressure relation for the orifice in the over centre valve and friction both in the over centre valve and in the cylinder. Though a linear analysis will not be especially helpful in designing a specific system, it can still give guidelines to the behaviour caused by modifications of different parameters of the system. Stability problems will obviously be present even in completely friction-free systems. The over centre valve illustrated in Figure 5.1 uses a spring-loaded poppet element to control the flow of fluid from the load line to the tank line. When the plunger is moved to the left by the application of the pressures and , the meter out and return port are connected. The actual displacement of the plunger element, which cause the flow restriction is usually of such small value relative to port diameter that the pressure/flow equations obey the Bernoulli form:

fP rP

)PP(2wxCQ br0dr −ρ

= (5.1)

where return flow rQ

0x spool displacement from neutral ρ density of oil

dC flow coefficient area gradient w other symbols used in the following are defined in Figure 5.1. The flow through the over centre valve is though given by Equation (5.1). In this analysis it is assumed that the back pressure is constant, and can therefore be omitted in the dynamic analysis. It will be found necessary in the analysis of dynamic control system equations to determine the system linear coefficients. Using Taylor series expansion theory, assuming first order dominance for differentials, the linear flow equation is

bP

rqp0qr PKxKQ += (5.2)

with rdP,x0

rq P2wC

xQ

Kr0

ρ=

∂∂

= ; r

0d

P,xr

rqp

P2

xwCPQ

Kr0

ρ=

∂∂

=

where qK valve flow gain

qpK valve flow-pressure coefficient

r0 P,x referring to operating conditions Looking at the cylinder chambers, it is assumed that the pressure in each chamber is everywhere the same and does not saturate or cavitate, line phenomena are absent, and temperature and density are constant. Also, internal and external leakage of the piston are neglected. Applying the continuity equation to each of the piston chambers yields


Chapter 5

dtdPV

dtdVQ f

F

fff β

+= (5.3)

dtdPV

dtdVQ r

F

rrr β

+=− (5.4)

where fV volume of forward chamber

(includes valve, connecting line and piston rod volume) volume of return chamber rV (includes valve, connecting line, and piston volume) Fβ effective bulk modulus of system

(includes oil, entrapped air, and mechanical compliance of chambers) The volumes of the piston chambers may be written

proff xAVV += (5.5)

pporr xAVV −= (5.6)where

rA area of piston at the rod side.

PA area of piston.

Px displacement of cylinder piston.

ofV initial volume of forward chamber.

orV initial volume of return chamber. Substituting Equation (5.5) and Equation (5.6) into Equation (5.3) and Equation (5.4), we have

dtdPxA

dtdPV

dtdx

AQ f

e

Prf

e

ofPrf β

+β

+= (5.7)

dtdPxA

dtdPV

dtdx

AQ r

e

PPr

e

orPPr β

−β

+−=− (5.8)

It is desirable and possible to express the continuity equations in a more useful form. The displacement flow of the piston is defined by . The last terms in Equation (5.7) and Equation (5.8) may be neglected by assuming that V >>

)dt/dxA(Q PrP ⋅=

of Pr xA , and >>orV PP xA

fC. By introducing the cylinder ratio CR , and the fluid

capacitances rP A/A=

Fof /V β= , Forr /VC β= , equations (5.7) and (5.8) can now be Laplace transformed to give

Pfff QQsPC −= (5.9)

rPrr QQCRsPC −= (5.10)


Chapter 5

The final equations arise by applying Newton's second law to the forces on the cylinder piston and spool in the over centre valve. In considering the cylinder piston and the over centre valve spool, friction forces and viscous damping is omitted. The resulting force equation for the piston, Laplace transformed, is

LrfrP2

t F)PCRP(AxsM +−= (5.11)where

tM total mass of piston and load referred to piston. arbitrary load force on piston. LF Considering the over centre valve, and assuming that only the forces produced by pressures, flow (the flow force is included in the spring constant), and spring are meaningful terms of the static and dynamic behaviour of the valve, it is possible to neglect the inertial forces of the spool and the friction force between the spool and housing. Thus, the force equation describing the over centre valve takes the following form

ooorcf xKAPAP =+ (5.12)where

cA area applied to pilot pressure. area applied to load pressure. oA spool displacement. ox spring gradient. oK Equations (5.2), (5.9), (5.1), (5.11), and (5.12) define the system given in Figure 5.1. 5.2.3 Transfer function of the mathematical model In this section the equations defined in the previous section 5.2.2. will be combined to obtain physically interpretable results and an over-all transfer function. We seek the cylinder response to a flow input. While the cylinder velocity is proportional to the displacement flow Q , the transfer function of interest is the one describing the relation between Q and . Combining Equation (5.2) and Equation (5.12) gives

P

Pf Q

fo

cqrqp

o

oqr P

KAK

PKK

AKQ +

+= (5.13)

Introducing , and the pilot ratio , means Equation (5.13) can be written as

ooqqo K/AKK = oc A/APR =

( ) fqorqpqor PKPRPKKQ ++= (5.14)

Usually force loads are omitted in system designs so that only the transfer function from input is of interest. Using Equation (5.11), written as , means that the displacement flow can be written as

)PCRP(A)sx(sM rfrPt −=⋅


Chapter 5

( )rft

2r

PrP PCRPsM

AsxAQ −== (5.15)

Introducing the three transfer functions

sC1)s(Hf

f = (5.16) sC1)s(H

rr = (5.17)

sMA

)s(Gt

2r

M = (5.18)

and inserting Equation (5.16) and Equation (5.17) into Equation (5.9) and Equation (5.10) the basic equations, describing the dynamics of the system in Figure 5.1, are

( )Pfff QQ)s(HP −= (5.19) ( )rPrr QQCR)s(HP −= (5.20)

( rfMP PCRP)s(GQ −= ) (5.21) ( ) fqorqpqor PKPRPKKQ ++= (5.22) For purpose of reducing the equations, they can conveniently be represented in a block diagram, as shown in Figure 5.2.

K

+

P

+

+Kqo

(s)+

QfHf

qoK PRPr

Qr +

HHHqp (s)r

CR

CR

f +(s)GM

PQ

Fig. 5.2 Block diagram representing equations (5.19) → (5.22) Closing the inner loop (and thereby defining ) gives the transfer function )s(Hrq

qpqor

rq

KK)s(H

11)s(H

++=

(5.23)

By using H , and closing the two remaining inner loops, the block diagram in Figure 5.2 can further be reduced to the one shown in Figure 5.3. Block 1 consist of an integrator and a gain, depending on the capacitance C . Block 2 contains the feed forward of P through the pilot line, and Block 3 includes the load, cylinder and over centre valve, and the volume between the two.

)s(rq

f

f


Chapter 5

Go(s)

(s)f+

QfH 1+ CRK PRqo

M

1G (s) +

H (s)rq

rqHCR (s)2

1 QP

Fig. 5.3 Reduced block diagram To allow for an analytical study of the system, we have to write out the terms in the three blocks, shown in Figure 5.3. The transfer function of the first block is simply

, and is given in Equation (5.16). Inserting Equation (5.17) into Equation (5.23), can be written as

)s(H)s(G f1o =)s(Hrq

qpqorrq KKsC

1)s(H++

= (5.24)

with given as in Equation (5.24) the transfer function of block 2 can be expressed as

)s(Hrq

qpqor

qpqor

qpqorqo2o KKsC

K)CRPR1(KsCKKsC

1CRPRK1)s(G++

+++=

+++= (5.25)

which can be written as

n2

t22o2o /s1

/s1K)s(G

ω+ω+

= (5.26)

where qpqo

qpqo2o KK

K)CRPR1(KK

+++

= ; r

qpqot2 C

K)CRPR1(K ++=ω ;

r

qpqon2 C

KK +=ω

The third block, written out, is

22r

tqpqo

22r

tr

qpqor

qpqor

22r

t3o

CRsAM

)KK(sAM

C

KKsC

KKsC1CRs

AM

1)s(G+++

++=

+++

= (5.27)

which, written on standard form gives

1s/2/s/s1

K)s(Gn3

2n3

2t3

3o3o +ωζ+ωω+

= (5.28)

where 2qpqo

3o CRKK

K+

= ; r

qpqot3 C

KK +=ω ;

tr

rn3 MC

ACR=ω ;

r

t

r

qpqo

CM

ACRKK

21 +

=ζ


Chapter 5

Multiplying the three transfer functions , , and , gives the open loop gain function G . (notice that

)s(G 1o

t3n

)s(G 2o )s(G 3o

)s(o 2 ω=ω )

)1s/2/s(s/s1

K)s(Gn3

2n3

2t2

o +ωζ+ωω+

= (5.29)

where 2ofo

eqpooqe2

f

qpqo

CRVKKK)CRPR1(AK

CRCK)CRPR1(K

Kβ++β

=++

=

oro

eqpooqe

r

qpqot2 VK

KK)CRPR1(AKC

K)CRPR1(K β++β=

++=ω

te

or

Pn3

MVA

β

=ω ; or

te

Po

qpooq

VM

AKKKAK

21 β+

=ζ

From the mathematical description in Equation (5.29), stability and other performance characteristics can be computed. 5.2.4 Stability analysis Stability is the most important performance characteristic of a hydraulic system and often requires some sacrifice in the speed of response. The design of the loop dynamics is usually centred around the requirements for stability. In this chapter the Routh-Hurwitz stability criterion and Bode diagram method for stability analysis will be used. The Routh-Hurwitz criterion is an analytical procedure for determining if all roots of a polynomial have negative real parts. Note, however, that only the stability of a system is determined. From Equation (5.29) we can write the system characteristic equation

0)1s/2/s(s

/s1K1)s(G1n3

2n3

2t2

o =+ως+ω

ω++=+ (5.30)

or

0Ks)K1(s2s

t2

2

n32

n3

3

=+ω

++ω

ς+ω

(5.31)

Writing out the coefficients in Equation (5.31) using the terms defined in Equation (5.29) we can write the characteristic equation in the general form

0asasasa 012

23

3 =+++ (5.32) where ; a tfr3 MCCa = )KK(MC qpqotf2 += a ; )CCRC(A r

2f

2r1 += { }qpqo

2r0 K)CRPR1(KAa ++=

The system in Equation (5.32) is stable if all coefficients ( ) are existing and positive and if

30 aa →

0aaaa 0312 >− (5.33)


Chapter 5

Therefore, for a stable system, we require that

+>

qpqo

qo

r

f

KKK

CRPR

VV

(5.34)

In many mobile system we often have high inertia loads combined with static loads. This means that the pressure sensitivity of the over centre valve will be high and the term in the brackets in Equation (5.34) will tend to unity. A lower pilot ratio will increase stability, but from this simple stability result it is obvious that the stability margin of the overall system will be very small as the hydraulic cylinder is a full stroke component. The enclosed volumes and on the meter-in and the meter-out side, respectively – influence the gain and the damping in the system. Hence, a big and a small leads to increased stability, see Figure 5.4, and vice versa, see Figure 5.5.

fV rV

fV

rV

0 1 2 3 4 0

50

100

150

200

250

Time [sec]

Pres

sure

[Bar

]

Pr

Pf

Fig. 5.4 Pressure response with step in fQ

0 1 2 3 40

50

100

150

200

250

300

350

Time [sec]

Pres

sure

[Bar

]

Pf

Pr

Fig. 5.5 Pressure response with step in fQ

Let us now make the stability analysis using a Bode diagram. The main advantage of the Bode plot over other types of plots for frequency response is that the effects of adding a pole or zero to a transfer function can be seen rather easily. For this reason, Bode diagrams are very useful in designing control systems.


Chapter 5

From the open loop gain function in Equation (5.29), the free s in the denominator indicates an integration, so that the system is type 1 and has zero position error. Variations in the gain constant K, the break frequency in the numerator t2ω , the hydraulic natural frequency n3ω , and especially in the damping ratio ζ occur and cause considerable shifting in the frequency response with different operating points. Hence the frequency response is quite elastic and this must always be kept in mind during design or in viewing test data. Though, the loop gain function is very complicated and some simpler expressions still retaining information essential to stability, is desirable. Examining Equation (5-29) it is possible to make a further reduction. First ζω n32 can be written as

)KKAK(KV

2 oqpoqoor

en3 +

β=ζω (5.35)

and

)KK)CRPR1(AK(KV oqpoq

oor

et2 ++

β=ω (5.36)

Combining Equation (5.35) and Equation (5.36) yields

n3

oqp

oq

oqp

oq

t2

1KKAK

1)CRPR1(KKAK

2 ω+

++ζ=ω (5.37)

The relation is known as the pressure sensitivity, and can be expressed as

Pqpq KK/K =

o

rP x

P2K = (5.38)

In cranes the load is mostly high inertia loads combined with linear steady state loads. This implies that the load pressure can be high, sometimes close to the relief valve setting of the over centre valve. Since the pressure drop over the valve will be high, the opening of the valve must be very small if the load is to be moved slowly. This implies that the pressure sensitivity will be very high, so that Equation (5.38) can be approximated with the following expression

rP

PK

n3t2 )CRPR1(2 ω+ζ=ω (5.39)

This equation indicates that t2ω will be equal or bigger that n3ω , even for a very low damping ratio, and a small pilot ratio. The numerator in Equation (5.29) can then be neglected because the phase from the quadratic factor decay much faster than the added phase from the first order system – so the loop gain function, Equation (5.29), can be approximated by.


Chapter 5

)1s/2/s(sK)s(G

n32

n32o +ωζ+ω

= (5.40)

Looking at Equation (5.40). From a Bode diagram (or Nyquist plot) it is possible to determine closed loop stability. See Figure 5.6.

1

K/

ω

ω

K/2δ

3n

3n1

(j )

K

Go ω

-1 slope

Log scale3n

-3 slope

rad/secω

ω

Fig. 5.6 Bode diagram of the loop gain function.

If the resonant peak of the quadratic term rises above unity gain, then the system becomes unstable because the critical point of the Nyquist diagram would be encircled. From a Bode diagram the crossover frequency is approximately equal to K (velocity constant). The gain level of the asymptotic diagram is n3/K ω at the frequency n3ω . This level is amplified by the factor ζ2/1 , which is the amplification factor of the quadratic, at resonance. Thus, the gain level at the resonant peak is K n32/ ζω , and must be less than unity for stability. Hence, the stability criterion is

12

K

n3

<ζω

(5.41)

From Equation (5.29) an expression for K can be written as

t22of

or

CR1

VV

K ω= (5.42)

and from Equation (5.39) an expression for 2 n3ζω can be found as a function of t2ω , meaning that Equation (5.41) can be expressed purely by hard quantities, as

1)PRCR1(CR

1VV

2of

or <+ (5.43)


Chapter 5

For a given application - both quantities CR and PR are fixed by a cylinder and over centre valve – Equation (5.43) then states that

ofor VV ⋅λ< (5.44) where is some positive constant. From this simple stability result it is obvious that the stability margin of the overall system will be very small as the total displacement of the cylinder is used.

λ

5.3 Load Sensing Directional valves 5.3.1 Introduction The purpose of this section is to make a reduced order model of a pressure compensated mobile proportional valve. Characterising a typical mobile machine, as a difference to typical industrial applications, is that they are operated by a human being and that the normally therefore do not have a well defined working cycle. The demand for controllability and efficiency have led to an increasing use of load sensing (LS) systems. Compared to conventional systems, LS systems are more efficient since flow and pump pressure continuously are matched to the actual need, however an ideal LS-system does not give any damping at all. The valve considered in this section is a Sauer-Danfoss A/S a load-independent proportional valve type PVG 32. A schematic drawing is shown in Figure 5.7.

ABA

BAB

Mechanical actuator PVM

T

Basic valves PVB

Electrical actuation PVE

PPump side module PVP

Fig. 5.7 Schematic drawing of a PVG 32 valve group PVG 32 is a hydraulic load sensing valve. The module system makes it possible to build up a valve group to meet requirements precisely. In Figure 5.7. is shown a valve group with 3 basic modules.


Chapter 5

5.3.2 Function of the PVG 32 valve In Figure 5.8 is shown a sectional schematic drawing of a PVG with two basic modules. The pressure adjustment spool is placed in the pump side module PVP (see Figure 5.7), and the pressure adjustment spool and main spool is placed in the basic modules PVB. When the pump is started and the main spools, in the individual basic modules are in the neutral position, oil flows from the pump across the pressure adjustment spool to tank. The oil flow led across the pressure adjustment spool determines the pump pressure. When the main spools are in neutral this pressure is the stand-by pressure.

Pressure compensator

Pressure adjustment spool

RESERVOIR

Main spool

Fig. 5.8 Sectional schematic drawing of a PVG 32 system When one or more of the main spools are actuated via the mechanical activation PVM or the electrical activation PVE, the highest load pressure is sensed (load sensing!) and fed through the shuttle valve circuit to the spring chamber behind the pressure adjustment spool, and completely or partially closes the connection to tank. Pump pressure is applied to the right-hand side of the pressure adjustment spool. This means that the pump pressure is kept a certain value, defined by the preloaded spring, higher that the highest load pressure. In the basic modules the pressure compensator maintains a constant pressure drop across the main spool – both when the load changes and when a neighbour module with a higher load pressure is actuated. In other words – when several functions are activated simultaneously the highest load pressure controls the pump pressure. Other modules then get higher pressure drops since the difference between pump pressure and those load pressures increases. In a non-pressure compensated valve, without a pressure compensator, this will increase the flow to those loads which will cause interactions. In a pressure compensated valve the increased pressure drop is compensated by the


Chapter 5

pressure compensator upstream the meter-in spool. So the interference between individual modules is small. 5.3.3 Modelling of the PVG 32 valve In modelling the PVG 32 valve we will use the parameters defined in Figure 5.9. We will only consider one basic module, while modelling other basic modules will be using the same approach.

T

RESERVOIR

LS

M

QA

Q

P3 V3

B26V2 2P

A

Pext

5 Q

P

B1

A2

M K

AHK

P

1

PQ

A3

VP

PX

PPP

VQP0 1A

PA

Qext

M M

A4

P

AKV1

VQ

Fig. 5.9 Schematic sectional drawing defining parameters Input to the valve model is a constant pump flow Q , the opening area of the main spool when actuated away from neutral by the electrical activation, the load pressure

P HA


Chapter 5

LSp , an external load pressure and external flow demand from other basic modules. The output will be the flow out of the valve.

extp extQ

VQ

Q

Q 2BQP +

)XP(A3d

)p)p,pACQ 3extLSd2B −= (max(26 ρ

C=

)FP +

p)(P

(FSP

5.3.4 The pump side module (PVP) The pump delivers the constant flow . The flow that is not used by any of the modules will be diverted back to tank by the spool area A . The pump is looking into the volume , which is composed of the volume in the hose and the volume in the valve. Applying the continuity equation we get

P

3

PV

)QQQ(Vdt

dpextVT

P

FP −−−β

= (5.45)

The flow to tank Q is a function of the area A , which again is a function of the spool position X . Using the orifice equation gives

T 3

P

)pp(2CQ TPT −ρ

= (5.46)

Similar with the flow through the damping orifice with the area 2BQ 6A .

(5.47)

Often the flow through damping orifices are small, meaning that the flow easily can be laminar and not turbulent as when described by the orifice equation (2.34), or the flow can shift between the laminar and turbulent regime. One way to solve this problem is to make the discharge coefficient a function of Reynolds number (see Figure 2.9). Applying Newton 2

(Re)C ddnd law to the pressure adjustment spool gives

p,p,X(F)X(A)pp(dtXdM TPPFPpSPP32

P2

P +−= (5.48)

)XP is the spring force and include the preloaded spring force. is

the flow force trying to move the spool in the closing direction. From Equation (2.58) it can be written as

)p,p,X(F TPPFP

)69cos()pX(AC2F o

TP3dFP −= (5.49) To find the pressure in Equation (5.48) we apply the continuity equation to the volume .

3p

3V

−

β=

dtdX

AQVdt

dp pp2B

3

F3 (5.50)


Chapter 5

In the equations (5.45) and (5.50) the volumes and V are considered constant – meaning that the displacement volume of the spool is negligible.

PV 3

If it is assumed that the pressure build up in the volume is very fast and the mass of the spool is negligible the model can be reduced to second order. To summarise the equations describing the dynamics of the pump side module then becomes:

3V

)QQQQQ(Vdt

dpextVT2BP

P

FP −−−+β

= ; P

2BP

AQ

dtdX

=

)pp(2)X(ACQ TPP3dT −ρ

= ; )p)p,p(max(2AC 3extLS6d2B −ρ

=Q

0F)X(FA)pp( FPpSPPP3 =++− ; )69cos()pp)(X(AC2F o

TPP3dFP −=

(5.51)

5.3.5 The pressure compensator and main spool (PVB) Looking at Figure 3.9 the flow Q from the pump side module is entering the basic module – passing through the compensator spool and the main spool and out to the actuator. There is a small orifice in the compensator spool so that the pressure can act on the right hand side of the spool. The pressure p is a combination of the pump pressure and the compensated/reduced pressure . The compensator spool will take a position so that the pressure drop over the metering land will be constant. The flow through the valve is therefore independent of the load pressure, giving a consistent metering.

V

Kp

0p

0

Kp

LSp− HA

Applying the continuity equation to the spool end chamber volumes and , and using the orifice equation for the damping orifices A and gives:

1V 2V

4 5A

−

β= 4AK

K

1

F1 QAdt

dXVdt

dp ; )pp(2ACQ 014d4A −ρ

= (5.52)

−

β= K

K1B

2

F2 Adt

dXQVdt

dp ; )pp(2AC 2LS5d1B −

ρ=Q (5.53)

The volumes and is again assumed constant, neglecting the volume change caused by spool movement.

1V 2V

Force equilibrium for the compensator spool becomes

)X(F)p,p,X(F)p,p,X(FA)pp(dtXdM KSK0PK1FKK0K2FKK122

K2

K +−+−= (5.54)


Chapter 5

)p,p,X(F 0pK1FK and are the flow forces as a result of oil flowing through the orifices and . Again from Equation (2.58) the flow forces can be written as

),,( 02 KKFK ppXF

1A 2A

)69cos()pp)(X(AC2F o

0PK1d1FK −= (5.55)

)69cos()pp)(X(AC2F oK0K2d2FK −= (5.56)

As can be seen from the expressions for the flow forces we need to know the pressures

and . These can of cause be calculated by applying the continuity equation to the volumes between the three orifices A , , and . However, these volumes are very small and the dynamics arising from the small volumes is so fast that it can be neglected. We though still need an expression for the pressures in the volumes. A way to cope with this problem is to consider the three orifices as hydraulic resistors in series. We want to find the equivalent area as shown in Figure 5.10.

0p Kp

1 2A HA

A 1

PP P0

2

Q

A

PK

V

AH

LSP

eqA

Q V

Fig. 5.10 Three orifices in series The flow through all the orifices is the same so we can write up the pressure drops as functions of the areas and the flow.

)pp(2ACQ 0P1dV −ρ

= => 2

d1

V0P CA

Q2

pp

ρ=− (5.57)

)pp(2ACQ K02dV −ρ

= => 2

d2

VK0 CA

Q2

pp

ρ=− (5.58)

)pp(2ACQ LSKHdV −ρ

= => 2

dH

VLSK CA

Q2

pp

ρ=− (5.59)

)pp(2ACQ LSPeqdV −ρ

= => 2

deq

VLSP CA

Q2

pp

ρ=− (5.60)

Now, summation of the pressure drops in Equation (5.57) → (5.60) gives

LSPLSKK11P pp)pp()pp()pp( −=−+−+− (5.61) Using the right hand side of Equations (5.57) → (5.60) in Equation (5.61) yields


Chapter 5

2eq

2

d

V2H

22

21

2

d

V

A1

2CQ

A1

A1

A1

2CQ ρ

=

++ρ

(5.62)

or

2H

22

21

2eq A/1A/1A/1

1A++

= (5.63)

Now, combining Equation (5.57) and Equation (5.60), and using Equation (5.63) the following result is obtained

2CQ

A1pp

2

d

V21

0Pρ

⋅=− ; where )pp(A

2CQ

LSP2eq

2

d

V −=ρ

⇓

)A/1A/1A/1(App

pp 2H

22

21

21

LSP0P ++⋅

−=−

(5.64)

In a similar manner it is possible to express the pressure drops in Equation (5.58) and Equation (5.59) as functions of the areas and the total pressure drop over the basic module

)A/1A/1A/1(App

pp 2H

22

21

22

LSPK0 ++⋅

−=− (5.65)

)A/1A/1A/1(App

pp 2H

22

21

2H

LSPLSK ++⋅

−=− (5.66)

If we assume that the pressure build up in the volumes V and are infinitely fast, the mass of the spool is negligible, and the orifice dominates the damping of the spool (meaning that there is no pressure drop over and ). Then the equations describing the basic module can be represented as a first order system.

1 2V

0p5A

4A 1p =

K

1BK

AQ

dtdX

= ; where )pp(2ACQ 2LS5d1B −ρ

=

( where 0)X(F)p,p,X(F)p,p,X(FA)pp KSK0PK1FKK0K2FKK02 =+−+− ; )69cos()pp)(X(AC2F o

0PK1d1FK −= )69cos()pp)(X(AC2F oK0K2d2FK −=

)A/1A/1A/1(App

pp 2H

22

21

21

LSP0P ++⋅

−=− ;

)A/1A/1A/1(App

p 2H

22

21

22

LSPK0 ++⋅

−=−p

)A/1A/1A/1(A

pppp 2

H22

21

2H

LSPLSK ++⋅

−=−

(5.67)


Chapter 5

5.4 Two Stage Relief Valve This rapport about modelling of pressure control valves, abbreviated PCV, concentrates on that group of valves often indiscriminately named as relief valves, safety valves, safety relief valves, overload protection valves, and so forth. 5.4.1 Performance Characteristics of a PCV Typically the PCV is inserted into a system adjacent to a trifurcation, having one line coming from a supply, another line feeding a load, and the third line conditionally connecting to tank via the PCV, see Figure 5.11a. Let be the load flow at which the pressure response from the load, according to some static load characteristic, see Figure 5.11b, equals P . For Q the PCV is closed, the system pressure is not controlled and equals the load pressure P .

0LQ

LP

ref 0LS Q<

L )Q(P LL=

b)

PL2

PL1

PSPref

QLO1

QLO2

QS, QV , QL

PS

QL , QV , PS

c)

d)

Q

PS

1

δ

LOAD

SUPPLY

a)

Pref

PSQ S

Q V

Q L

PCV

V1

,

e)

QV

O

PS

Q

TS

QV

QV2

t

SP

QSO

QLO

t

Fig. 5.11 Basic characteristics of a system with a PCV For the system pressure should be kept constantly equal to P , whatever the value of the valve flow lower than the maximum rated flow of the PCV.

0LS QQ ≥ SP

SQref

LV QQ −=

This means, that the ideal PCV characteristics are the ones shown in Figures 5.11b and 5.11c. (The chosen graphical representation, using pressure as abscissa and flow as ordinate, anticipates a later encountered advantage in the analytic modelling, as it proves to be easier to express the PCV flow explicitly as a function of pressure while, in reality, flow is considered the independent variable). The static characteristic is ideally a vertical line. The ideal, dynamic performance is characterised by an unaffected system pressure P , despite the sudden change in valve flow because of sudden changes in load characteristics e.g. from to , see Figure 5.11b and Figure 5.11c. Virtually, the characteristics of a PCV are the ones shown in Figure 5.11d and figure 5.11e.

S

1LP 2LP


Chapter 5

The most essential deviations from the ideal characteristics are: The pressure build-up, having some value different from zero and equal to the inverted slope of the static characteristic: δ

V

S

QP

∂∂

=δ (5.68)

The different static characteristic for increasing and decreasing valve flow (due to friction effects). The overshoot ο in the system pressure in case of transient flow steps. Settling-time in the transient response.

SP

ST The magnitudes: , , and plus rated flow and pressure form a minimum set of essential performance-figures of a PCV. However, the dynamic performance figures may as well be expressed in frequency domain terms like resonant frequency

δ ο ST

rν , , bandwidth, etc. peakM m −

5.4.2 Modelling of the two-stage PCV The PCV is of the almost classical design as shown in Figure 5.12. The valve is designed for a maximum rated flow of 200 l/min at a maximum pressure level of 250 bar.

QS

PS

VS

A S

x

IQ

Q

TPQ

ST

QT

IT

P

A I VI

I

y

Figure 5.12 Two-stage pressure control valve

Both the static and the dynamic characteristics for the valve also depend on the connected hydraulic circuits. To obtain a simple situation the return line is simply described by a return pressure , and the upstream conduit is assumed to be a simple fluid volume with negligible friction and mass force effects and with a stiffness determined by the effective bulk modulus of the hydraulic fluid.

rP

SV

An analytical/experimental modelling technique should be utilised on the two-stage PCV, because the total performance of the PCV heavily depends on the interaction between the three flow resistances shown in Figure 5.13, that is the two series connected resistances in the pilot line and the pilot controlled resistance of the main flow path. The pressure-flow-area relationships of these internal resistances are too complicated for a trustworthy theoretical/analytical modelling solely.


Chapter 5

IQ

QS STQ QT

Figure 5.13 Principle of function of the two-stage PCV Furthermore, it shall be attempted to establish a detailed model of the valve, which can be useful on component design level. 5.4.3 A Detailed Dynamic Model (Model I) The first steps of the structural identification process suggest 17 basic equations (Equation (5.69) to Equation (5.86)) by which the type of relief valve and its operation is described. Some of the functional relationships, like in Equation (5.70), (5.76) etc., are given in a general form. Based on elementary fluid mechanics more specific expressions are suggested such as equations (5.71), (5.79) etc. Some of these relationships need experimental verification due to considerable doubtfulness regarding the virtual flow conditions. Through experimental procedures the relationships are determined as characteristics of “black-boxes”, representing sectional phenomena of the valve. By experimental means the input-output relations, often pressure-flow-area relations, are recorded. The finally accepted equations, appear as those which combine reasonable simplicity and acceptable accuracy. In this way vital system parameters are estimated and a total model is established. Finally, the total model is checked through an overall experimental verification. The equations used are, mainly, flow-pressure-area relationships, momentum equations, compressibility equations plus continuity and force balance equations applied to the four main energy reservoirs of the valve (see Figure 5.12):

° The internal volume . IV° The volume of the valve inlet and upstream conduit. SV° The mass/spring - system of the main stage. ° The mass/spring - system of the pilot stage.

(Generally index P refers to parts of the pilot valve while index M refers to parts of the main stage). Continuity equation applied to the volume V I

outdiinci QQQQ −+= (5.69) where

ciQ compression flow to the volume IV


Chapter 5

inQ incoming flow = Q I

IQ flow through the fixed throttle channel of the main spool

diQ main spool displacement flow

outQ outgoing flow = ITQ Flow through the fixed pilot-throttle: IQ

)PP(fQ ISII −≅ (5.70) where is some flow-function. Assuming turbulent orifice flow the flow can be written as

If IQ

)PP(/2ACQ ISOdII −ρ= (5.71)

where dIC discharge coefficient

OA orifice area ρ mass density of oil

Defining ρ= /2ACk OdII Equation (5.71) can be rewritten as

)PP(kQ ISII −= (5.72) The displacement flow Q is given by di

xAdt

dVQ I

Idi &⋅== (5.73)

where IA effective spool area facing the internal chamber

x position of the main spool (= 0, when main-spool is seated) The compression flow is: ciQ

III

ci PxAV

Q &β

−= (5.74)

where

β fluid bulk modulus (husk !) The outlet flow Q is equal to the pilot stage flow . out ITQ Combining the Equations (5.69) to (5.74) yields:

ITIIIII QxAQPxAV

−+=β

−&& (5.75)


Chapter 5

Flow equation for the pilot valve The flow through the pilot valve is:

=>∆

=0y,0

0y,)y,P,P(fQ ref1P

IT (5.76)

where Pf some flow function

PP∆ equal to P TI P−

TP return pressure (in the main return line)

refP preset reference pressure y displacement of the poppet. By definition . 0y ≥

Equation (5.76) contain a ‘hard’ non-linearity. Further it is assumed that the return pressure of the pilot valve is approximately equal to the return pressure P , that the upstream pressure of the pilot valve is approximately equal to the pressure , and that the compressibility and mass effects in the up- and downstream channels of the pilot valve can be neglected.

T

IP

Usually in hydraulic valves a realization of the quadratic flow is attempted to avoid sensitivity to viscosity variations and as a first step the following equation for the flow through the pilot valve is assumed:

d

y

α

Q IT

)PP(/2)y(ACQ TIPdPIT −ρ= (5.77) The opening area A can be calculated as )y(

)cossinysinyd()y(A 22P αα−απ=

← Fig. 5.14 Poppet valve

By defining

ραπ= /2)sin(dCk dPP (5.78) Equation (5.77) can be written as

TIPIT PPykQ −= (5.79)where

Pk flow coefficient of the pilot valve d pilot valve seat diameter

dPC discharge coefficient of the pilot valve α half the conical angle of the poppet


Chapter 5

Equation (5.79) is derived under the assumption that the ratio is small, i.e. fore relative small openings.

d/y

Force balance for the pilot poppet If for and for , Newton’ second law applied to the poppet yields:

0y ≥ 0t = 0y > 0t >

fPPSPPTTPIIPP FFykyBPAPAyM −−−−−= &&& (5.80)where

PM mass of poppet

IPA efficient poppet area affected by the pressure P I

TPA efficient poppet area affected by the pressure P T

PB viscous damping coefficient

SPk spring rate of the pilot spring

PF preloaded spring load on the poppet

fPF flow force Equation (5.80) tacitly assumes, that coulomb friction forces on the poppet can be neglected. Further, the first and the second term on the right hand side can be lumped together into A , where . )PP( TIP − TPIPP AAA ≅=The preloaded spring force on the poppet is equal to:

SP0P kyF = (5.81)where

0y initial compression of the pilot spring The flow force acting on the pilot-poppet, is

)2sin(CdCk;)PP(ykF vPdPfPTIfPfP απ=−= (5.82)where

fPk flow-force coefficient

vPC empirical factor due to viscous friction ( ≈ 1) The preset pressure level of the valve is, by definition the value of at which the pilot valve is on the point to open. By definition when the poppet is on the seat. Consequently, can be found from the steady-state solution to equation (5.80) combined with equation (5.81) - thus . This magnitude is seen to be a characteristic of the valve, while

refP PP∆0y =

P

refP

SP0ref A/kyP =the opening pressure is a characteristic

of both the valve and the actual operating situation of the return line. Tref PP +OP =

When equation (5.80) to (5.82) are taken together, the force balance for the pilot poppet becomes:

[ ] )PPP(Ay)PP(kkyByM refTIPTIfPSPPP −−=−+++ &&& (5.83)


Chapter 5

This simple form of the force balance equation can be used without complications if the valve opens at e.g. and remains open. However, a general validity of the equation must provide for the phenomena occurring in the closure for the valve, that is when

0t =

y = 0 for When this happens, an impact between the poppet and the seat occurs, and and may vary drastically, depending on the nature of this impact. For the actual valve-configuration it is generally reasonable to assume a perfectly non-elastic impact (with a coefficient of restitution equal to 0).

.0y&&

tt 0 ≥=y&

Continuity equation applied to the volume SV

STicsdsin QQQQQ +++= (5.84)where

inQ = = supply flow or load flow SQ

dsQ main spool displacement flow

csQ compression flow to the volume SV

STQ flow through the main stage throttle (from upstream to return line) Consequently:

xAdt

dVQ S

Sds &== ( ) diQ= (5.85)

SS

SSS

cs PV

PxAV

Q &&β

≅β

+= (5.86)

=>−

=0x,0

0x,)PP,x(fQ TSST

ST (5.87)

where STf some flow function characterising the main flow path

Often it is assumed, that:

TSSST PPxkQ −= (5.88)with

ργπ= /2)sin(dCk SdSS (5.89)where

Sk flow coefficient of the main spool

Sd main spool seat diameter

dSC discharge coefficient of the main spool γ half the conical angle of the main spool

Combining Equation (5.72) and Equation (5.84) to (5.88) gives:


Chapter 5

0xfor ;PPkPPxkPVxAQ ISITSSSS

SS ≥−+−+β

+= && (5.90)

Force balance equation for the main spool If the main stage is open, or initially closed and thereafter definitely open, that is for and for , Newton’ second law yields:

0x ≥0t = 0x > 0t >

µ−−−−−−= FFFxkxBAPAPxM fMMMMIISSM &&& (5.91)where

MM mass of the main spool

MB viscous damping coefficient

Mk spring rate of the main stage spring

SA efficient area of the main spool exposed to the pressure P S

IA efficient area of the main spool exposed to the pressure P i

MF preloaded spring force on the spool

fMF flow force acting on the spool

µF ry friction force In the following is considered negligible. Further µF

0MM xkF = (5.92)where

0x initial compression of the spring Through change-of-momentum-considerations the flow force can generally be expressed as

fMF

)2sin(CCdk;)PP(xkF vSdSSfMTSfMfM γπ=−= (5.93)

where fMk flow-force coefficient

vSC empirical factor due to viscous friction ( ≈ 1) Combining Equation (5.91) to (3.93), the force balance equation of the main spool becomes:

[ ] 0MIISSTSfMMMM xkAPAPx)PP(kkxBxM −−=−−++ &&& (5.94) The 26 equations [(5.69) to (3.94)] form an overall dynamic model which, however, must be considered a hypothesis because of approximations. An overall experimental evaluation is required to verify the validity of the model. Flow and force characteristics of the main and pilot poppet The flow coefficients and flow force equations in the above section have been derived theoretically. In general, for example the discharge coefficient of an orifice must be determined experimentally. It will be a function of Reynolds number and orifice


Chapter 5

geometry. Only in some special cases the flow and force characteristics can be predicted analytically. The various and complex valve geometry means that the above results may have a rather limited use or may even be contradictory – hence a further analysis concerning these characteristics may be needed. It an be shown that poppet valves under certain conditions can be “bi-stable”, in that either of two different stable flow patters could be induced under otherwise identical conditions. this could give rise to differences and even a discontinuity in the flow coefficient and the flow force. In Figure 5.15 these results are illustrated. also the construction of the pilot poppet will have an effect of deflecting the jet radially and thereby reducing the flow force influence. Thus in a simulation model the flow force coefficient and the discharge coefficient of the pilot valve C could be implemented as functions of y according to Figure 5.15.

fPk dP

x

x

Flowforce, Discharge Coefficient

Cd

Ff

A B

A

B

A

B

Fig. 5.15 Pilot poppet valve characteristics

In Figure 5.16 is shown three different configuration of the main orifice. The first one (Case 1) shows the standard configuration. In Case 2 the spool has no under cut and in Case 3 the seat has a bigger fly cut. The jet patterns in Case 1 and Case 2 can be verified by fluid dynamics simulation (CFD). In Case 3 the jet is assumed to follow the fact of the poppet, and thereby giving a flow force corresponding to equation 5.93.

x

Flowforce

x

Flowforce

x x

CASE 3CASE 1

x

CASE 2Flowforce

x

Fig. 5.16 Main stage valve characteristics

Representation of model I Above analysis can be summarized into the following 5 fundamental equations.


Chapter 5

Continuity equation, volume : IV

ITIISIIII QxAPPkPxAV −+−=

β−

&& (5.95)

Pilot flow Q : IT

TIPIT PPykQ −= (5.96) Force balance, pilot poppet:

[ ] )PPP(Ay)PP(kkyByM refTIPTIfPSPPP −−=−+++ &&& (5.97) Continuity equation, volume SV :

0xfor ;PPkPPxkPVxAQ ISITSSSS

SS ≥−+−+β

+= && (5.98)

Force balance, main spool:

[ ] 0MIISSTSfMMMM xkAPAPx)PP(kkxBxM −−=−−++ &&& (5.99) A block-diagram representation of this model is shown in Figure 5.18. Because the functions of the valve involves a number of non-linear effects, it has been necessary to extend the linear block-diagram symbols with the special elements as defined in Figure 5.17.

ELEMENT

DIVIDER

SQUAREROOT

MULTI-PLIER

TYPE

NEGATIVECLIPPER 1e

2

2

e e1 0

e 2

e e1 0

e 2

e 1 e 0

ee e=0

e =0

1

e 1

e=e 0 1 e

DESCRIPTION

e

DIAGRAMSYMBOL

e 0 forfor

e 0 = e 1

e 0 = 0 e1

1 0

0

Fig. 5.17 Special block diagram elements The model shown in Figure 5.18 is particular suited for investigations of the ability of a constant pressure valve to respond on flow disturbances. It appears from the block-diagram that the function of the valve corresponds to that of a controller with a fixed, but adjustable reference . The output is the system pressure refP


Chapter 5

SP , the flow is the load, and the return pressure may be considered as a disturbance. Regarding the performance of the valve, the most relevant feature is the response in systems pressure in function of variations in the load flow

SQ TP

SP SQ . Parameters of the valve A number of the parameters characterising the valve in accordance with the model shown in Figure 5.18 are determined by simply measuring the geometry of the valve. Others - e.g. friction coefficients - are simply estimated by rules of thumb. A compilation of all the parameters used in the model is shown in the Table 5.1 on the following page. The way in which the individual parameters has been or should be evaluated is indicated in the column to the right. The need for partial experimental verification of the parameters as well the structural identification of the valve will be discussed later.

SYMBOL VALUE UNIT ORIGIN AI 38 10 4. ⋅ − m2 Geometrical measurement AS 2 84 10 4. ⋅ − m2 Geometrical measurement AO 196 10 7. ⋅ − m2 Geometrical measurement CdI 0.6 Normal practice CdP 0.6 Normal practice CdS 0.6 Normal practice CvP 1 Normal practice CvS 1 Normal practice BM 10 Ns m/ Estimated VI 852 10 6. ⋅ − m3 Geometrical measurement VS 2 0 10 3. ⋅ − m3 Estimated β 7 0 10 8. ⋅ − N m/ 2 Normal practice ρ 870 Kg m/ 3 Estimated α 20 degrees Construction data γ 45 degrees Construction data d 2 3 10 3. ⋅ − m Construction data dS 19 10 3⋅ − m Construction data M M 0.1 Kg Geometrical measurement MP 0.0015 Kg Geometrical measurement k M 31 103. ⋅ N/m Construction data kSP 5886 103. ⋅ N/m Construction data x0 0.0177 m Calculated cf. Eq. (5.87)

Table 5.1 Parameters of the PCV-model


Chapter 5

PA

k SP k fP

P ref

P T

P I

+

_ _

_

_

+

+

+

+

P

__

__

+

_ +

_

_+

AI

x 0

k fM

MkS

Ak M

B M 11

MM

ss

F µ

k S

_

k P

Pilo

t Val

veB P P

Ms

s11

+A

I

V I

A

_

+ +β

k1 s

_+

QS-

Vo l

ume

I

+_ _

S _SVβ

1 s

I-Vol

ume

Mai

n St

age

AS

P S

Fig. 5.18 Total block diagram of the two stage pressure control valve (Model I)


Chapter 5

5.4.4 Simplifications of Model I (⇒ Model II) It is to be expected, that the response speed of the pilot valve is superior to that of the main stage spool both because of their intended relative functions and because of the difference in the mechanical spring stiffness-to-mass ratio of the two elements. However, also the flow-force component of the net spring stiffness should be taken into account. From Equation 5.97 it is seen, that the purely mechanical natural frequency nmPω of the pilot valve is

Hz997 rad/sec6264M/k PSPnmP ≅==ω (5.100) while the mechanical flow force natural frequency nfPω , with a pressure difference

equal to 100 bar becomes TT PP −

Hz1210 rad/sec7603M/))PP(kk( PTIfPPnfP ≅=−+=ω (5.101) For the main spool the mechanical natural frequency nmMω is

Hz28 rad/sec176M/k MMnmM ≅==ω (5.102) while the mechanical-flow force natural frequency nfMω , as derived from Equation (5.9), becomes

Hz 370 rad/sec 2324M/))PP(kk( MTSfMMnfM ≅=−+=ω (5.103) with bar. From these figures it can be found reasonable to establish a second model without the pilot valve dynamics.

150PP TS =−

Through the next chapters only the reduced model II is considered. A block diagram of this model is shown in Figure 5.19. In this model equation (3.97), describing the pilot valve dynamics, is truncated, the position y of the poppet becomes irrelevant, and the former used flow equation (5.79) can be replaced with an expression for the static characteristic of the pilot valve. As such a valve, being per se a single stage PCV, normally has rather linear characteristics, the following linear relation is assumed:

TrefITrefIPIT PPP;)PPP(CQ +≥−−≅ (5.104) in which is the inverse of the pressure-build-up factor of the valve. PC


Chapter 5

F

S

_ +

k AMAI

x 0_

__

+

_

µ

+ P IP

P T_

Pilo

t Valv

e

+

P ref

_C P

AS

Vβ_

_

Q +_

S

Ss

S-Vo

lum

e

1P S

AI+

_+

β

V II-V

olum

e_

+

s1_

A

+k I

M

k

+ _1 M

_

B M

fM

k S

Mk

s1 s

Main

Stag

e

Fig 5.19 Simplified block diagram of a two stage pressure control valve (Model II)


Chapter 5

To get a proper selection of the C -value the flow-pressure function of the pilot valve has been simulated. In the simulation the main

spool was locked against its seat, so that only the pilot flow could pass the valve. Continuous plots of the pilot flow versus the differential pressure at different valve-settings were recorded as shown in Figure 5.20.

P

)P,PP(fQ refTIIT −=

refPTI PP −

0 50 100 150 2000

1

2

3

4

5

6

7

8

9Qpilot [l/min]

PI -PT[bar]

Fig. 5.20 Static characteristics of the pilot valve Reasonable PCV-performance requires, that does not exceed P drastically, i.e.

bar should be a conservative estimate. Hence only the range of the curves below approximately Qpilot < 1 l/min is relevant to the virtual operation of the total valve, and for this range linear approximations according to equation (5.104) with

(l/min)/bar apply fairly well.

IP ref

10PP refI ≤−

075.0CP = To gain a systematically survey over the comprehensive model and to identify significant effects in the overall performance, a linearised model and a transfer function is helpful. This will be the topic of the next section. 5.4.5 Linearisation of the Model (⇒ Model III) Assuming, for simplicity, that the return pressure is negligible and selecting a specific operating point , the equations of the model, that is equations (5.95), (5.104), (5.98), and (5.99), can be linearised and combined to yield the following three linear equations:

TP)Q,P( 0S0S

IPIIS1III pCxA)pp(pxAV

−+−α=β

−&& (5.105)

)pp(pxpV

xAq IS1S32SS

SS −α+α+α+β

+= && (5.106)

IISSS54MMM ApAppx)k(xBxM −=α+α+++ &&& (5.107)


Chapter 5

where 0IS

I1 )PP(

Q−∂

∂=α (5.108a);

0

ST2 x

Q∂

∂=α (5.108b);

0S

ST3 P

Q∂

∂=α (5.108c);

0

fM4 x

F∂

∂=α (5.108d); 0S

fM5 P

F∂∂

=α (5.108e)

From computer simulation with model I, it has been found that spool position and inner pressure P referring to the operating point ( are

approximately , . Next is chosen equal to 2000 as a representative value for many practical systems. Then to can be evaluated.

0xbar)

cm0I

x 0

242,l/min200()Q,P 0S0S =

SV 3

1α 5αmm56.0= bar168P 0I =

It can be shown that . Thus A is neglected in the following. A block diagram representation according to the linear model is shown in Figure 5.21.

II VxA << xI

α

V+

+α1

+2sMM

αAS

_

+

_5

M

1M k(sB +

AI

4α+ )

++β

1

s + (I1α CP )

AI s A S +

kI

Q

+

_+

β

2

S

VS s + 3

1

( α1+α )SP

Fig. 5.21 Block diagram of the linearised model (Model III) After considerable calculations it can be shown that the transfer function has the following form:

)s(q/)s(p SS

012

23

34

4

012

23

3

S

S

bsbsbsbsbasasasa

)s(q)s(p

+++++++

= (5.109)

where each of the coefficients a and are determined by rather complex equations. i ib To manage the numerical calculations it is easiest to derive the transfer function via a state space model. In state space notation the linear model becomes:


Chapter 5

S

SS

I

S

31

S

1

S

2

S

S

I

1

I

1P

I

I

M

5S

M

I

M

4M

M

M

S

I

q

V

000

ppxx

V)(

VVVA

VV)C(0

VA

0001M

AMA

Mk

MB

ppxx

⋅

β+

⋅

α+αβ−

βα−

βα−

β−

βαα+β−

β

α−−

α+−−

=

&

&

&

&

&&

(5.110)

with the output vector [ ] . 0 0 0 1 After having calculated the elements of the system matrix and the distribution matrix, which obviously is much easier than to calculate the coefficients of equation (5.109), the factorised form of the transfer function was found to be:

+

ωζ

+

ω

+

ωζ

+

ω

+

ωζ

+

ω

τ+

=

1s2s1s2s

1s2s)s1(G

)s(Q)s(P

d

d

2

dc

c

2

c

b

b

2

ba

S

S (5.111)

where

up) build pressure(l/minbar107.5G 2 =δ=⋅= −

Hz12s

rad7.75rads0132.0 aaa =ω⇒=ω⇒=τ

=ζ

=ν⇒=ω

068.0

Hz1799s

rad11304

b

bb

=ζ

=ν⇒=ω

343.0

Hz40s

rad251

c

cc

=ζ

=ν⇒=ω

065.0

Hz1800s

rad11309

d

dd

Neglecting the static gain, a Bode-plot of Equation (5.111) is shown in Figure 5.22. It appears from the figures above that the two second order systems indicated by b and d in the numerator and denominator, respectively, are very much alike numerically. However, a reduction is, unfortunately, not generally applicable.


Chapter 5

100 101 102 103 104-60

-40

-20

0

Frequency (rad/sec)

Gain dB

100 101 102 103 104

-30

-60

-90

0

30

Frequency (rad/sec)

Phase deg

Linearized model

Nonlinear model

Linear model

Nonlinear model

Fig. 5.22 Frequency response of a two stage PCV To check the accuracy of the linear model against that of model I, block diagram Figure 5.18, a curve derived from computer simulations with model I has been drawn on Figure 5.22 together with the linear frequency transfer function. The two curves exhibit a reasonable agreement. They describe the ability of an open, constant pressure valve to respond on flow disturbances. The high-frequency asymptote of the frequency characteristics are -20 dB/dec and -90 degrees, respectively, because, at higher frequencies the system is essentially an integrator, integrating the input flow in the volume to yield the output-pressure .

SQ

SV SP Characteristic subsystems of the PCV To get a physical understanding of the parameters of the transfer function it might be useful to compare with the specific subsystems of the valve, as they appear in the block diagram, Figure 5.21. The i-volume can be characterised by a time constant of

Hz176s

rad1110rads100.9

)C(V

II4

P1

II =ν⇒=ω⇒⋅≅

+αβ=τ −

The s-volume can be characterised by a time constant of

Hz5.5s

rad35rads029.0

)(V

II31

SI =ν⇒=ω⇒≅

α+αβ=τ

The main-spool has a mechanical natural frequency which earlier (Equation (5.102)) has been found to be:

Hz28 nmM ≅ν

Taking into account the flow forces on the main spool, the natural frequency will increase with increasing pressure, while the damping ration will decrease.


Chapter 5

The mechanical-hydraulic natural frequency of the main spool is

Hz1733s

rad10892VM

Anhm

Im

2I

nhm =ν⇒=β

=ω

Comparing these figures with the numerical values of the parameters of the transfer function, Equation (5.111), it seems reasonable to conclude, that the time constant of the numerator stems from the s-volume, and that the two numerically identical second order terms, appearing both in the numerator and the denominator are identical with the mechanical-hydraulic second order term of the main spool. The other second order term of the denominator might correspond with the mechanical natural frequency and damping ratio of the main spool.

aτ

The resonance peak, characterized by a resonance frequency rν and the maximum amplitude ratio , is a characteristic feature of the dynamic performance of the PCV. mM

[dB][Hz]

00

20

15

10

5

mM rν

60

30

15

45

37.5 75 112.5 150

V = 1000 cms3

Q = 150 l/mins

Pref

[bar]

mM

νr

[dB][Hz]

0 1000

10

20

30

40

50

60

70

80

0

20

15

10

5

Q = 200 l/min

refP = 150 bar

s

mM

mM rν

V = 1000 cm

100

10 30

00

5 15 P = 150 barref

50

s3

[Hz]

15 45

[dB]20

mM rν

60

2000

rν

3000 4000

V

[cm ]

s

3

150 [l/min]

Q

200

s

νr

mM

Fig. 5.23 Displacement of the resonant peak in function of variations in , , and SV refP SQ

For the specific operating situation referred to in Figure 5.22 it appears, that flow rate oscillations in the 30-40 Hz range will result in pressure oscillations, amplified by

or a factor of 5, compared with the value determined by the static pressure build-up coefficient solely. Based on computer simulations, the curves in Figure 5.23 gives a survey over the variations of

dB 15M m ≈

rν and with some of the most important operating-state parameters like the upstream volume , the preset reference pressure level , and the amplitude of the sinusoidal input flow rate.

mM

SV

refP 0sq----- oo0oo -----


Appendix A

A Flow Force

Compensation Fluid Power Systems (Hydraulisk Komponentanalyse) AaU ~ Forår 2003

A1 A2

Introduction...................................................................... Flow Force Compensation Methods.........………….…… Radial hole orifices • Angled hole orifices • Pressure drop compensation • Recirculation lands • Profiled spool lands • Scalloping techniques

1

3

A1 Introduction Orifices is a basic means in control valves. It is a sudden restriction of short length in a flow passage and in valves it has a variable area. Two types of flow regime exist depending on whether inertia or viscous forces dominate. The flow velocity through an orifice must increase above that in the upstream region to satisfy the law of continuity. At high Reynolds numbers, the pressure drop across the orifice is caused by the acceleration of the fluid particles from the upstream velocity to the higher jet velocity. At low Reynolds numbers, the pressure drop is caused by the internal shear forces resulting from fluid viscosity (see Figure A1).

1 2

0A

2A

3

Fig. A1 Flow through an orifice (turbulent) Since most orifice flows occur at high Reynolds numbers, this region is of greater importance. The flow between points 1 and 2 in Figure A1 is potential flow and experience justifies the use of Bernoulli’s equation in this region. Applying Bernoulli’s equation between points 1 and 2 gives the well known equation

)pp(2ACQ 210d −ρ

= (A.1)


Appendix A

where is called the discharge coefficient. Although the analysis leading to Equation (A.1) is not valid at low Reynolds numbers, the equation can be extended to the laminar region by plotting discharge coefficient as a function of Reynolds number.

dC

Another important factor in control valves is the axial force necessary to operate the spool. This force has three principal components: inertial force, friction force, and a usually far larger force which is produced by the flow through the metering orifices. With the assumption that the flow is two-dimensional, irrotational, non-viscous, and incompressible, the solution for this last mentioned force has been found by von Mises, for the configuration shown in Figure A2.

Valve body

g

fe d

x

h

Spool

c

b θa

i

Fig. A2 Square-Land Configuration

The axial force on the piston equals the axial component of the net rate of efflux of momentum through the boundary a-b-c-d-c-f-g-h-i-a, where a-b is the vena contracta of the jet. In an actual valve the area of a-b is much smaller than that of d-e where the fluid enters the upstream chamber. Since the velocities are inversely proportional to the areas, the influx of momentum through d-e is negligibly small compared with the efflux at a-b, which is equal to QVρ , where is the total rate of flow, the velocity of jet at vena contracta, and

Q Vρ the density of fluid. The net axial force can then be written as

)cos(QVFFF hifg θρ=−= (A.2)

By use of Bernoulli’s equation, Equation (A.2) can be transformed into the more useful form

)cos(pwxC2F d θ∆= (A.3) where is the peripheral width of orifice, x the axial length of orifice, and the pressure difference between the upstream and downstream chamber.

w p∆

Equation (A.1) and Equation (A.3) are the most used equations today when estimating the flow through valves and the actuation force, but the effect of geometrical deviations from the ideal ones in Figure A1 and Figure A2 is difficult to calculate theoretically. In relation to the global variables some general methods though have been developed to influence the flow-force and avoid cavitation, as discussed in the next sections. A2 Flow Force Compensation Methods


Appendix A

Over the past forty years, a considerable amount of effort has been spent on creating and developing compensation methods. Consequently an extensive range of techniques have been proposed and documented, some of which are more effective than others. However, the majority of methods tend to use one of three general principles to achieve compensation, as identified by Backé. These are as follows:

1. Influence the jet angle such that all mass flows entering and exiting the spool chamber control volume are as perpendicular as possible to the valve axis. This minimises the net change in momentum of the fluid in an axial direction and hence the reactionary axial forces acting on the valve spool.

2. Generate a pressure drop within the valve to counteract the reactive force of the

flow. 3. Balance the axial momentum components of the flow entering and exiting the

spool chamber control volume. This generally requires alteration of the influx and/or efflux jet angles and can involve partial reversal of the flow by means of flow recirculation within the valve chamber.

In the following text, a review is made of current flow-force compensation methods. Each method is discussed in terms of its effectiveness in influencing the flow-force characteristic. A2.1 Radial hole orifices This method utilises holes drilled radially into the valve sleeve to meter the flow from the valve chamber, as shown in the upper Figure A3. Another configuration is to drill the holes in a hollow spool. The holes drilled in the sleeve is placed in diametrically opposite positions. This ensures that there is no lateral force on the spool, i.e. the momentum components ρQVsinθ cancel out. The method can be applied equally effectively to both meter-in and meter-out configurations. The flow-force characteristic of a single hole was investigated experimentally by Clark and is shown in the lower Figure A3. Initially as the holes are being uncovered the jet angle is 69 deg, but as the opening increases the angle increases until at full opening it becomes 90 deg. At this point the net axial rate of efflux is zero and there is no flow force. However, Clark points out that the abrupt change in the force gradient can lead to stability problems. The problems with the abrupt change of the force gradient and the limited compensation of the overall flow force can be eliminated/improved by replacing the individual holes by a series of smaller holes, as illustrated in the upper Figure A4. Arranging the holes such that the overlap, s, between successive holes is held to , an approximately linear area/displacement characteristic is created with a corresponding flow force, as shown in the lower Figure A4. The choice of hole sizes and distribution is, of course, dependent upon the metering characteristics required and the permissible pressure losses.

d18.0 ⋅


Appendix A

Percent Diameter Orifice Open

With radial holeUncompensated Orific

e

0 20 40

Flow

-forc

e

Hole Completely uncovered

60 80 100

Exhaust

Sleeve Valve body

Small diameter hole through sleeve

Inlet

Uncompensated Orifice

Total force

Flow

-forc

eFlow

Force from each hole

Spool

s

Sleeve

Holes drilled through sleevearranged in a spiral with

overlap "s"

d

Fig. A3 Radial hole compensation Fig. A4 Staggered radial holes A2.2 Angled hole orifices Flow forces exist where there is a net rate of efflux of momentum. If the inlet is angled as shown in Figure A5, an axial component of inlet momentum will exist and this will reduce the net rate of efflux momentum. The flow force is then represented by the following formula:

))cos(V)cos(V(QF 12f α−θρ= (A.4) where α is the inlet angle, and θ the exit angle, both relative to the spool axis. To obtain maximum advantage from this method the inlet angle α must be as small as possible and the inlet velocity must be as high as possible. These requirements create problems, of which the inlet velocity is the most significant since it implies increased pressure loss at the inlet, and controlling dimensions, which determine the inlet and outlet velocities, become critical. The inlet design conditions must be carefully determined, the implication here being that for maximum advantage and performance, spools will have to be designed for limited flow and pressure range.

1V

If the first hole in the stagged radial hole compensation method, shown in Fig. A.6, is angled, a further reduction in flow-force magnitude can be realised. The efflux jet impinge upon the spool land, generating an opening reacting force. The qualitative effect of this configuration is shown in the lower Fig. A.6, and tends to reduce the initial “hump” in the flow-force curve.


Appendix A

Flow

Flow

-forc

e


Increasing α

Inlet

Sleeve

V1

α

Valve body VExhaust2

Radial hole compensation

Radial hole + angled hole compensation


Flow

-forc

eFlow

Exhaust

Valve bodySleeve

Angled holes

Inlet

Fig. A5 Angled hole compensation Fig A6 Angled + Staggered radial holes A2.3 Pressure drop compensation In the analysis of momentum effects frictional forces in the valve chamber were neglected. However, if the annular space between spool and sleeve is small, velocity in the valve chamber will be high and a friction loss, proportional to the square of the fluid velocity in the valve chamber, will occur. Referring to Figure A7 this will result in the pressure in the inlet side being greater than that of the exhaust side.

Exhaust

Sleeve Valve body

V

Inlet

Flow-

forc

e

uncompensated valve

Flow

decreasing spoolshank diameter

Fig. A7 Thick spool flow-force compensation

This gives an opening force which opposes the closing force. The smaller the area between spool and sleeve, the greater the opening force generated. However, this cannot be carried too far since it causes choking of the flow. Typical flow-force curves for this type of valve is shown to the right in Figure A7. Pressure drops associated with the thick spool method are large. Furthermore, perfect flow-force compensation cannot be achieved using this scheme as the steady-state flow-force varies linearily with the flow-rate, while the force used to compensate it varies with the square of the flow-rate. For


Appendix A

low capacity valves the pressure loss will be a linear law and the transition to the square law may be unpredictable. A2.4 Recirculation lands The general principle of recirculation lands is to shape the valve chamber so as to control the direction of the efflux jet and use it to modify the flow-force characteristic of the valve. If the fluid that leaves a valve chamber can be directed back on the spool, then an opening force will exist which compensates the steady state flow force. One method, presented by Clark, is illustrated in the upper Figure A8. Here the fluid that is metered across the orifice is directed in a curved path and circulated through a cavity in the land. A reaction force is generated on the spool as the fluid leaves the cavity at angle β. An exact calculation of the reaction force is virtually impossible. The method of compensation is more effective at higher flows because considerable fluid is recirculated, making the net force versus stroke characteristic quite non-linear, as seen in the lower Figure A8. Flow force tests is presented by Clark for spools with different exhaust angles, the results of which are presented in the lower Figure A8. For and , the valve is overcompensated. The “hump” at low flow conditions, should diminish with reduced radial clearance and sharper metering edges.

030=β 050=β

f

Uncompensated valve

Flow

-forc

e

0

β = 30

Flow

Flow

-forc

e

=β 50

70=β 0

β

Exhaust

chamber

Curvedexhaust

Increasing radialclearance

Displacement x

e2θ

gα 2

α 3

Exhaust

θ1θ

d

cα1

3

b

a

x

h

Inlet

Fig. A8 Recirculation land Fig. A9 Negative force compensation Another compensation method, presented as negative force port, were proposed by Lee and Blackburn. A schematic representation of the complete valve chamber profile used is shown in the upper Figure A9. The stem is shaped to look like a turbine bucket and the recess in the sleeve makes the down-stream chamber larger. The smooth profile ensures minimal loss in the momentum of the fluid in its passage through the valve chamber. The jet


Appendix A

enters at angle and leaves at angle and hence if the cross section of the flow remains constant, the closing force would be

1θ 2θ

))cos()(cos(QVF 21 θ−θρ= (A.5)

Obvious, if , i.e. , the force will tend to open the valve. In addition to this an eddy is formed in the chamber causing an additional flow to enter the chamber and angle . This also tends to open the valve. The presence of the eddy makes the total force almost impossible to predict analytically. Experimental results, shown qualitatively in the lower Figure A9, verifies the predicted negative characteristic but also reveals the presence of small positive “hump” at low flows, the magnitude dependent on valve radial clearance, and the metering edge radius. Unfortunately there is no method of calculating the dimensions of the various features illustrated in the figure.

)cos()cos( 12 θ>θ

3θ

12 θ<θ

A variation of the compensation method proposed by Clark has been proposed by Nakano et al. The geometry of the compensation method is shown below to the left in Figure A10. They made an experimentally study of the effect of the geometric parameters, shown in Figure A10, on the magnitude and trend of the flow force characteristic. From their work the following results were drawn:

β

D1

Dd1

B

w2

3R R

4 β

β

R Rβ 2 1

w1

Flow

-forc

e

Uncompensateed

valve

d

Flow

Fig. A10 Positive force compensation The presence of blend radii, R, have a beneficial effect upon compensation. Rounded corners reduced the magnitude of the flow force with 35 %. For the range , compensation is not significantly affected. 0

10 9069 ≤≤ β

Chamber width, 2ω , made less than sleeve chamber width, 1ω , has a negative effect upon compensation. Reduction in improves the compensation effect, but may lead to instability if carried to far. Increasing gives small improvement in compensation.

1D

1dNo clear guidelines were given on the influence of the other parameters. Flow force characteristic for the most effective and stable choice of parameters is shown to the right in Figure A10. The flow force is approximately reduced to 28 % of that measured on an equivalent sized conventional square-edged valve. A2.5 Profiled spool lands


Appendix A

This arrangement differs from those previously considered in that the metering characteristic is generated between the spool and sleeve edge, as shown in upper Figure A11. The throttling zone is formed tangentially between the two components and intrudes into the spool chamber. Under meter-in conditions, as well as the flow force trying to shut the valve there exist an additional force component, acting in an opening directing. This is a result of a pressure imbalance on the spool. The portion of the spool land from a-b is located in the sleeve chamber and so is subjected to approximately supply pressure conditions. The downstream land, d-e, is exposed merely to the return pressure. The portion, b-c, is acted upon by a pressure which is a function of the flow velocity. These pressures combine to generate a compensating force. This compensation technique was studied by Lang and Fassbender, who produced a single metering edge device with a profile similar to that discussed. The effectiveness of the compensating force present in the meter-in configuration is demonstrated in lower Figure A11. The meter-out characteristics represent purely the reactionary force to the flow as no compensation force is generated under this condition. A modification of this method (Lang and Fassbender) is illustrated in the upper Figure A12. In addition to the flow-force and pressure force described above, flow between the closely adjacent spool land and sleeve gives rise to a friction force acting on both components. With a meter-in configuration, this force acts to open the valve.

Meter-out

Flow

- forc

e

Meter-in

Flow

Sleeve

cd

e

Spool

b

a

Flow

Meter-out

0

Flow

-forc

e

Meter-in

Sleeve

Spool

Fig. A11 Profiled spool land Fig. A12 Profiled spool land Utilisation of friction and pressure forces allows the flow force to be balanced more evenly, since they do not depend on the same variables. Experimental results for the profile, shown in the lower Figure A12 demonstrate effective compensation. A2.6 Scalloping Technique Scalloping refers to the technique of cutting notches into the spool shoulder used for metering the flow, as shown in Figure A13.


Appendix A

Fig. A13 Scalloped spool

As partial, rather than full annulus opening is provided, all metered flow is guided through the scallop. By careful design of the scallop geometry, the direction of the metered jet can be controlled and the resultant flow-force characteristic influenced. In addition, scallops give the designer control over the flow versus spool displacement characteristic and increase valve stroke which reduces the criticality of alignment between the spool to sleeve metering edges. Notching the spool shoulder also gives rise to slower metering effect (i.e. a reduced valve gain) and provides a smoother action for the valve. An added consequence is that valve leakage around the null position is greatly reduced, due to the greater overlap of the spool in the sleeve. In the development of scalloping techniques, several distinct types of notch geometry have emerged. These are as follows: Those in the form of a shallow cylindrical pocket, formed by plunging an end mill cutter radially inward, and at a slight angle to the spool axis, as shown in Figure A14 (Junck and incorporated in various other US patents). This notch form is effective at providing flow force compensation when applied to a meter-out configuration. Those formed by a cutter mounted in a horizontal milling machine (see Figure A15), as proposed by Junck and Schexnayder. These slots provide effective compensation in a meter-in configuration but generate rather high flow-force in a meter-out case. Those formed by plunging a ball-ended cutter radially inward, and possibly at a slight angle with respect to the spool axis, s proposed by Melocik and Latimer. This type of scallop can be applied effectively to both meter-in and meter-out configurations if slight adjustment is made to the cutter axis, as shown in Figure A16 and Figure A17, respectively. It is claimed that these notches, defined solely by curved surfaces, result in a minimum 50 % reduction in flow-force, s compared with the previously described scalloping geometries. In the case of meter-in it is theorised that the curved surfaces of the notch minimise flow-force by their tendency to first direct flow generally through a central focal zone and second, to cause it to spread out. This action directs fluid to form impingement upon any immediate adjacent surface of either the sleeve or the spool. In the case of meter-out, compensation is attributed primarily to the effect of guiding the efflux jet radially outward.

Exhaust

Inlet

Exhaust

Inlet

Fig. A14 End mill cutter Fig. A15 Horizontal milling


Appendix A

Fig. A16 Meter-in Fig. A17 Meter-out A compensation scheme combining the practical benefits of partial circumferential metering with the flexibility of recirculation lands was devised by Seamone. Here scallops in the form of “wedge cuts” produced by grinding symmetrical flats across the spool shoulder, shown in the left Figure A18, were used to meter flow into a zero force port. Similar to that described earlier in section A2.4.

ExhaustInlet

γFl

ow-fo

rce

Flow

γ=30

0 γ=78

γ=71

Fig. A18 Scallops in form of “wedge cuts” As illustrated in the right Figure A18, the flow exit angle, γ , had a significant effect on flow-force if made less than 70°.

----- oo 0 oo -----


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