MODELLING AND CONTROL OF A SAUER DANFOSS PVG-32 VALVE

Transcription

1 MODELLING AND CONTROL OF A SAUER DANFOSS PVG-32 VALVE 5 semester project Group MCE5-522 Department of Energy Technology Aalborg university Autumn/winter 29

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3 Title: Semester: Semester theme: Modelling and control of a Sauer-Danfoss PVG 32 valve 5th Mechatronic System Analysis Project period: to ECTS: 23 Supervisor: Project group: Henrik Clemmensen Pedersen MCE5-522 Thomas Schmidt Mikael Højen Cemre Yigen Morten Hyldgaard Sørensen SYNOPSIS: The purpose of this project has been to thoroughly understand the principles behind the Sauer- Danfoss PVG 32 proportional valve. To achieve this result, a mathematical dynamic model of the valve has been developed, and implemented in Matlab Simulink. Also, a linear model have been developed, for the purpose to gain understanding of the interaction between the different parameters and also to get an estimation of how fast the system behaves, which is a usefull information when designing a controller for a system applying this valve. Furthermore, several tests have been carried out for the purpose of validating the dynamic Simulink model. After the data was analysed, and the uncertain parameters of the model was alteret, the validation was achieved. Also the results from the linear model have proven to be a good estimation of how fast the system behaves. The main purpose of the project is therefore achieved. Copies: 6 Pages, total: 135 Appendix: 3 Supplements: 1 CD By signing this document, each member of the group confirms that all participated in the project work and thereby all members are collectively liable for the content of the report.

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5 I Preface The report is written by project group MCE5-522, attending the 5. semester of Mechatronic Energy Technology at Aalborg University. It is addressed to students on a further scientific education. The theme for this semester is Mechatronic System Analysis. Study Guidance Formulas in the report are numbered by chapter followed be a dot and the equation number. Sections are numbered to sub sections on the form (Chapter.Section.Subsection). Figures and tabels are numbered the same way as equations. All references to formulas, sections and figures are referenced by number and page. Sources are cited by the Harvard-style, [ Lastname, year ], as a reference to the bibliography in the back of the report. Appendix are stated with A in front of the pagenumber, Enclosures with a B, and the contents of the enclosed CD with a C. Symbols and their units are described in the section Nomenclature in appendix. Appendix D consist of a foldout page. This is meant as a help when keeping track of the different parts and symbols of the PVG 32. Along with this and the Nomenclature it should allways be possiple to find the description of any symbol without having to leave the page concerned. Contents of the CD The Matlab Simulink model. The report.

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7 Table of contents 1 Introduction Initial problem Problem analysis The directional control valve The internal valves and circuit diagram of the PVG The total structure and the modules of the PVG Exploring the PVG 32 under action Conclusion on the Problem Analysis Problem statement Problem statement Problem solving Introduction Modeling the pumpside module Modelling the compensator spool Modeling the main spool Modelling the PVE module Modeling the load Validation of unlinear advanced model Linear model Controlling the valve Conclusion 91 6 Future work 93 Bibliography B1 I Appendix B3 A Govering Equations B5 A.1 Orifice equation B5 A.2 The orifice equation in models B7 A.3 Translatoric friction B8 A.4 Flow force B9 A.5 Continuity equation B1 B Test Reports B15 B.1 Compensator spool in PVP module spring B15 III

8 IV TABLE OF CONTENTS B.2 Compensator spool in PVB Module spring B17 B.3 LVDT Spring B19 B.4 Main Spool Spring B2 B.5 Cavities of Main Spool B22 B.6 Volumes of the Hoses and Pipes B23 B.7 Application in Hydraulic System B25 C Transfer functions B29 C.1 Notes when deriving transfer functions, used in tuning the model B29 D Foldout page with overview of the PVG 32 B33

9 Chapter 1 Introduction Today, hydraulics is used in various applications, from the simple hydraulic cylinder known from: truck mounted cranes, personal lifts, wheel diggers and many other systems where a powerfull linear force is needed, to more advanced systems, like: Power steering or hydrostatic transmissions. Some alternatives to hydraulics are pneumatic or electrical systems. The main-problem with pneumatics, compared to hydraulics, is that air is much more compressible, which makes it difficult to control and in some applications very dangerous to use. Electrical motors and linear actuators is often a cheap alternative to hydraulics, but it is difficult to achieve the same high linear force and torque that a hydraulic system can deliver. (a) Mobilecrane.[Danfoss, 26] (b) Rendegraver.[Danfoss, 26] Figure 1.1: Examples on mobile machinery where the PVG-valve is being used. Some disadvantages exists when comparing hydraulics to electric components. The efficiency for instance. If the hydraulic pump delivers flow to different hydraulic components, the total efficiency of the system will drop radically if the load pressure at the various components varies alot. Hydraulics is nevertheless very popular because of the extreme forces it can transmit. Whether it is a linear force or a torque that is needed, hydraulics can deliver it with a very compact unit at the point of action. Usually a large pump, cooler and filter system is needed. Sauer-Danfoss, is a Danish company, who produces a wide range of hydraulic components. Among these are the PVG Load independent valves. The PVG valves are very flexible, because they can be controlled both manually, or by an electrical 1

10 2 CHAPTER 1. INTRODUCTION input. Figure 1.2: [Danfoss, 26] In this project, the group is provided with a PVG 32 valve from Sauer-Danfoss to be thoroughly analysed. The PVG 32 valve is the smallest in the PVG-series, which also contains a PVG 1 and a PVG 12. The PVG 32 can operate at the same pressures as it has bigger counterparts, but it can t handle the same amount of flow. This particular valve is used on a wide range of mobile-systems like; wheel-diggers, forestry machines, garbage trucks, tractors and truck-mounted cranes, because of its load-sensing principle which reduces wear and tear on the other components on the machine[danfoss, 26]. Since the PVG 32 valve is widely used, it intensify the need for understanding the valve, which leads us to the Initial problem. 1.1 Initial problem In the light of the introduction the following initiating problem can be outlined: How does the PVG32 proportional valve work?

11 Chapter 2 Problem analysis In the following chapter, the initial problem will be investigated. And therefore the general functioning of the PVG valve will be explained. This section of the report will only explain the functions in Steady-state. The valve will be explained dynamically later when the problem statement has been outlined. First a general description of how the valve works will be given. Then each individual module of the valve will be specified. The startup situation of the valve is described, in order to give an example of how the different parts interact. 2.1 The directional control valve The PVG 32 is a directional control valve which is used to control the direction, and the amount, of fluid flow from a pump to its load. The key part in a directional control valve is the main spool, which is moving inside the housing. The movement of the spool will open and close for the oil flow to the different ports. The ports can be connected to different loads, sources and reservoirs, depending on the purpose of the valve in the hydraulic system. The PVG 32 is a 4/3 valve, which means that it has 4 ports connected to the valve, and 3 different positions the spool can switch between. The direction of the fluid between the ports is determined by the position of the spool. Figure 2.1 contains the hydraulic symbol of the mentioned valve. B A T P Figure 2.1: The figure displays a 4/3-way directional control valve in the middle position centered by the springs. At this position the flow between the ports are blocked. The valve in the figure is electro-hydraulic controlled. 2.2 The internal valves and circuit diagram of the PVG 32 A look inside the PVG 32 directional control valve reveals that three internal valves are used to create the directional valve and to fulfill other features of the PVG 32. The internal valves are respectively a 3

12 4 CHAPTER 2. PROBLEM ANALYSIS direction valve called main valve, and two pressure compensator valves. The modules of the PVG 32 each has a certain task to insure proper function of the valve. The hydraulic circuit diagram of the PVG 32 with the division into the three modules is seen in Figure 2.2. A similar figure can also be seen in Appendix D in the back of the report. Q L Q V Side A P LS Side B LVDT P bp M M P p P 5 Q B1 A 5 P 4 To tank A H A H To tank X M PVE module P K A 2 P ext M K A 4 Q B3 A K V 1 P 1 A6 Q B2 V 2 P 2 P A 1 Q v X k PVB module P P M p A P A 3 P 3 V 3 X P Q T Q P V P RESERVOIR PVP module Figure 2.2: The figure illustrates the different parts of the PVG 32 with all the interior and definitions of areas, fluid flows and pressures. The main valve 2 The main valve is a vital part of the PVG 32, since this is the part that control the direction of flow in and out of the PVG valve. The whole structure is build upon this main spool, which can be controlled remotely, either using a mechanical lever or an electrical control signal. The electrical control system is managing the position of the spool, inside the main valve. The valve itself, is a housing where the spool can move freely inside, even though the space between the housing and the spool at some parts are as small as a few micrometers. This valve and its control units are illustrated in Figure 2.3. The electro hydraulic control system is the part on the right side of the spool, and the mechanical lever is on the opposite side.

13 2.2. THE INTERNAL VALVES AND CIRCUIT DIAGRAM OF THE PVG 32 5 Mechanical lever Q L Q V Electric control P p M M A H To tank A H To tank Figure 2.3: This figure illustrates the main spool of the PVG 32, which acts as a remote controlled valve. The electrical control unit uses an input signal, to displace the main spool, through a secondary hydraulic system, with electromagnetic valves. Even if there is a change in pressure inside the valve under operation, the position of the spool is kept constant for the same input signal. In reality the position will never be constant, but the electrical control unit will continuously make the spool converge to the desired position. This is done by a secondary hydraulic system, which is managed through four electromagnetic valves. The electromagnetic valves are working in pairs with a switching frequency of 1-1 Hz[Danfoss, 2]; when two valves are open, two will be closed. By using an uneven up-time, down-time period, one side of the spool will have a higher pressure than the other. Because of the high switching frequency, the pressure changes will happen at a very short time. This will cause the spool to move to the desired position. A hydraulic circuit diagram of the secondary hydraulic system is given in Figure 2.4. Electromagnetic valve NC 1 NC 2 Pump Main spool NO 1 Checkball valves NO 2 2 Reservoir Figure 2.4: Basic hydraulic circiut diagram of the secondary hydraulic system used to move the mainspool. As it is evident in the basic hydraulic circuit diagram, the electromagnetic valves are basically on-off valves. All the valves has an electromagnetic component to manage the state by a duty cycle, and the pump side valves also has a spring attached. The function of the springs, are to change the initial position, when there are no input signals. While the pump side valves initially are blocking for fluid passage (off), the reservoir side valves are open for passage (on). The electromagnetic valves can be divided in pairs. The valves on the left-hand-side, NC 1 and N 1, are connected in series, with

14 6 CHAPTER 2. PROBLEM ANALYSIS the pump in one end and the tank in the other. Between the valves one side of the main spool is connected. All the valves are working in pairs. The valves on each side of the main spool gets the same signal, but they are working inversely, since one of the valves are normally open, and the other is normally closed, hence, whenever one is open for fluid passage, the other will block Danfoss [2] The pressure compensator valves The two pressure relief valves are controlled internally, which means the pilot pressure that controls the compensator spools runs internally in the PVG. If the pressure at the load rises, the rising pressure will be led to the compensator spool and the two internally controlled valves will compensate whenever the pressure at the load changes. To implement this feature, the compensator valves are using a feedback system of the load pressure and a preloaded spring attached to the spool. The existence of the preloaded spring forces F spr i ng means that the spool will yield resistance when trying to open for the fluid flow. The spools are initially blocking for passage, and only when a force F is present, which is opposite and greater than the initial spring force, the spool will move and open for flow. This is illustrated in Figure 2.5. The mentioned force F, is generated by the pump-side pressure P pump working on the area of the spool A spool. The force F, has to overcome an initial spring force called F cr ack, this will move the spool a distance of x cr ack. This displacement x cr ack serves to secure some pressure drop and to block for leakage. The longer the x cr ack distance is, the less flow will run through the cavities between the spool and the valve-cylinder. F spring F crack F spring M spool X A spool F X crack X Q v P pump Figure 2.5: This figure illustrates the principle idea behind a preloaded spring attached to a spool inside a directional control valve. The graph on the left shows that in order to compress the spring so that fluid can pass, a force, high enough to do so, is needed. The main reason why the preloaded springforces initially are blocking for fluid passage, is not to block the fluidflow, but to force the pressure inside the valve to rise in order to overcome the springforces. This serves to block the fluidflow when the pressure inside the valve is too low, and open up again when the pressure has increased. The pressure needed to establish equilibrium between springforce and pressure force, for the compensator spools, are for the one in the PVB module 5 B ar and for the one in the PVP module 15 B ar [Pedersen, 29] higher than the payload pressure. The payload pressure is implemented in the force-equilibrium of the spool, by leading a small amount of fluid at the payload to the spring side of the compensator valve, so the pressure P LS works together with the spring. This approach is called feedback, and this is the reason why the internal pressure is continiously adjusted according to the payload pressure. An equilibrium diagram of the forces observed by the spool is shown in Figure 2.6.

15 2.3. THE TOTAL STRUCTURE AND THE MODULES OF THE PVG 32 7 x A spool F spring P feedback P pump Figure 2.6: This figure illustrates the forces and pressures observed by the two compensator spools. The pressure on both sides have an effect on an area of the same size 2.3 The total structure and the modules of the PVG 32 Sauer Danfoss has chosen a modular based architecture in order to build the PVG 32, by respecting the basic principles behind the valve and to keep a flexible structure. This gives the possibility to connect multiple PVG 32 in parallel to drive multiple actuators, typically pistons. The modules in the PVG 32 are respectively PVE (Electro hydraulic module), PVP (Pump side module) and PVB ( Compensatorand Main spool module) modules. Each module is treated separately to give an idea of their basic features. PVE module The primary function of the PVE module is to move the main spool, and thereby changing the passage area and direction of the fluid. By changing the passage area, simultaniously, the flow through the PVG 32 to the payload is controlled. In order to move the main spool, the PVE module transforms an input signal to a fixed spool displacement through a secondary hydrodynamic amplifier system. To assure that the position of the spool has reached the desired position, the PVE module constantly measures the position of the spool. If needed the PVE module will compensate for a misposition through its internal control system. This feature of the PVE module is based on a closed loop feedback system, as seen in Figure 2.7. The input signal is constantly compared to the measured signal from the sensor. Whenever they are equal, the system will balance the forces on both sides of the valve, to stop further displacement of the spool. refference + - Measured error Controller System input System System output Measured output Sensor Figure 2.7: The figure illustrates a simple closed loop feedback system.wikipedia [n.d.a] PVP module The PVP module has three ports. The pump is connected to one of the ports, the reservoir is connected to another and the last port is letting the fluid flow through to the PVB module.

16 8 CHAPTER 2. PROBLEM ANALYSIS The main purpose of the PVP module is to adjust the pump pressure to a level of about 15bar above the load pressure of the highest consumer. Because the motor driving the pump often has a fixed operating point, the flow into the valve will be constant. If this flow where just dropped to tank at the highest possible working pressure, the effect loss would be large. Therefore the PVP module lowers the pressure so that it is at the minimum needed level for the most demanding actuator. If for example two cylinders where to be controlled, the one working at the highest pressure will be the one determing the pump pressure. It is therefore very ineffective to controle an actuator with a high pressure demand together with an actuator with a high flow demand, or for example controlling an actuator with a high pressure demand using very little flow over long time, because the rest of the flow will then be lead to tank at a high pressure, thus wasting energy. PVB module The last module is the PVB module. This module contains the main spool and a compensator spool. The function of the main spool is to direct the correct amount of flow to an actuator. Defined by the spool position. In order to do so it changes the area of the passage leading to the two output ports, while it moves. The flow is assumed turbulent and therefore given by the orifice equation (Equation 2.1). 2 Q = C d A P (2.1) ρ The purpose of the compensator spool is to keep the pressure drop across the main spool constant so that the position of the spool, and hereby the area, is the only thing determing the flow. 2.4 Exploring the PVG 32 under action The position of the main spool either widens or narrows the passage area for the fluid. The fluid is typically oil. In order to understand how the valve works, it is necessary to describe the process from a working condition where the PVG 32 doesn t lead any fluid through, to a working condition where the fluid passage through the PVG 32 is open. This is illustrated in Figure 2.8. Initially the main spool is blocking for fluid-passage, this is equivalent with an OFF-stage. The off stage of the valve means, there is no fluid passage from one end of the valve to the other. The main spool is, simply, blocking for fluid passage. Even when the passage is blocked, the pump will keep running and pumping fluid. The fluid will fill the chambers until it meets resistance at the main spool, and build up pressure. As the pressure rises, eventually, the spring connected to the pressure adjustment spool will give in and open the passage to the reservoir. Because the spring is preloaded with a force, the pressure has to rise untill the spring starts to give in. Whenever the pressure exceeds this level, the fluid will start flowing into the reservoir. This is analogous to a motor running idle; all the work done by the pump on the fluid is going to waste. The picture changes when the main spool is moved so that it opens for fluid passage. This is equivalent with an ON-stage. Immediately after this process, the pressure will fall inside the valve, and the pressure adjustment spool will automatically narrow the passage to the reservoir. The fluid runs into the actuator and it is now possible to transfer some power through the fluid. If the feedback pressure makes the spools block for passage, the pump side pressure will just rise untill the forces is balanced and the spool once again is in equilibrium.

17 2.5. CONCLUSION ON THE PROBLEM ANALYSIS 9 Q V M M M M ON-stage OFF-stage Figure 2.8: The picture illustrates the difference in the ON and OFF stage, which is determined by the position of the spool. 2.5 Conclusion on the Problem Analysis The initial problem was: How does the PVG32 proportional valve work? Due to the problem analysis it is now possible to, partly, answer the question. Partly; because only the steady state operation of the valve has been analysed, although some dynamics have been introduced in order to describe the steady state behaviour, a more extensive analysis needs to be carried out in order to fully answer the question. To help focussing on the right part of the problem a problem statement has been defined. This problem statement narrows down the initial problem, so that the time and effort now only concerns the problem statement and not the broader initial problem. The statement is shown in the succeeding chapter. 2

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19 Chapter 3 Problem statement 3.1 Problem statement Main problem How to model and understand the dynamics of a Sauer Danfoss PVG 32 valve To answer the Problem statement, the following items will be determined: Make a describtion of the valve which also takes the dynamics into account. Create a simulink model which simulate the real nature of a PVG 32 valve. Perform test to use for validation of the model. Validation of the dynamic model. The relative error of the response are not allowed to exceed 3% in order for the model to be considered valid at the given point of operation. If the relative error is less than 1% the model are describing the reality in a good and reliable manner, at that specific set of parameters and test conditions. Simplify the advanced dynamical model to a linear model. This is considered achieved if the model can be used as a tool to better understand the valve, and the control of it. Understand how to use the developed tools to optimize the valve. This criteria is considered as fullfilled if the two models can be used in the development of a controller. 11

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21 Chapter 4 Problem solving 4.1 Introduction This part of the report are focussing on deriving models describing the valve, and at the end validating them, so they may be used af a tool in understanding af the pvg works. This way the problem statement will be fullfilled. A unlinear dynamic model describing allmost all parts of the PVG are developed and validated, also linear models of subsystems in the PVG are developed. Including a linear model discribing the dynamics from the controle input to the position of the main spool. First the unlinear dynamic model will be discribed, starting with the PVP-Module. 4.2 Modeling the pumpside module Ap Q V Pp Q B2 A 6 k c M p Pp V p θ A 3 V 3 P 3 Xp Q p P T Q T Figure 4.1: Illustration of the pump side module with the pressure adjustment spool The PVP-module, or pump side module, is usually connected to a pump in a hydraulic system. The function of this module is to adjust the pressure difference through the PVB module to a pre-defined pressure level. As illustrated in Figure 4.1, the spool will relief the pressure if it exceeds a certain level 13

22 14 CHAPTER 4. PROBLEM SOLVING defined by the spring. What happes is that the pressure will rise in chamber V 3, the spool will move to the right and lead the fluid to tank. As seen in the figure, there are several parameters that needs to be determined in order to establish a dynamic model. The parameters can be split into measureable quantities and variables. The measureable quantities that needs to be determined are listed in Table 4.1. A 3 Area of the opening A 3 [ m 2 ] V 3 Volume of spring chamber V 3 [ m 3 ] V p Volume of covering from the pump piping to compensator spool chamber [ m 3 ] A 6 Cross-sectional area of the orifice A 6 [ m 2 ] A p k p Cross-sectional area of the pump side spool Spring constant of the spring located on the pump side spool Table 4.1: Measureable quantities to be used in the pump side module [ m 2 ] [ N m ] The area of the spool-opening A 3 needs to be determined as a function of spool-displacement x p. This function is derived later in subsection Initially the spool displacement is defined as x p = mm. This means that the spool is sticked to the left wall. The volume of the chamber V 3 is calculated based on the biggest possible chamber size. This value is estimated to be V 3 = 8cm 3. Because the volume depends on the spool displacement, the volume will decrease as seen in Equation 4.1. V 3 (x p ) = V 3 A p x p (4.1) The orifice in the pump side spool is neglected, because it is assumed that the pressure buid up in such a small chamber will happen very fast. The volume of V p represents the whole volume covered by the fluid from the pump to the compensator spool, including piping and spool chambers as seen partly in Figure 4.1 and Figure 4.7. The size of this volume are given in equation 4.2 and how the quantity is found are explained in section B.6. V p = 2.7L (4.2) The volume V p dependens, in reality, on the travel of the spool, and should increase with a positive spool displacement x p, but is neglected as it yields a very small change compared to size of the volume. The mass of pump side spool M p was determined using a scale. The mass of the pump side spool was determined to be: M p = 7 g m (4.3) The pump side spring constant k p, is found in an experiment, which is described in section B.1. The value of the spring constant is repeated here: k p = 15 N m (4.4)

23 4.2. MODELING THE PUMPSIDE MODULE 15 The pressure in the tank P T is determined to be B ar relative or 1B ar absolute, because the tank is open to the atmosphere. Cross sectional area of the pump side spool A p is determined from the diameter of the pump side spool, d p = 18mm. Thus the cross sectional area of the pump side spool is. A p = π r 2 = 2.54cm 2 (4.5) Determination of the PVP spool opening area The pump side compensator spool opening area is designed as a cone with different slopes as indicated in 4.2a. The area function of the opening area is dependent on the displacement x p. The shape of the spool yields a function of the resulting area. This function can be derived by measuring the diameter of the spool at the points where the magnitude of the slope changes. The slope φ 6 has an inclination of and φ The diameter of the slopes at the different inclinations are; r 1 = 9.mm, r 2 = 8.84mm and r 3 = 8.15mm. These values have been used to make two 6 point vectors of the resulting area function. These vectors can be seen as a plot in 4.2b. 1.8 x φ 6 φ 7 Area [m 3 ] PVP spool.4.2 r1 r2 r Displacement [m] (a) PVP spool slopes. (b) Plot of the opening area function of the PVP module spool. Figure Force equilibrium on the pump side spool To determine the motion of the spool, Newton s 2. law will be utilized. The forces that acts on the spool is shown in Figure 4.3.

24 16 CHAPTER 4. PROBLEM SOLVING F FR-P F Pp M p F S F P3 F FL3 x p A 3 Figure 4.3: Forces acting on pump side spool From figur 4.3, Newtons 2. can be utilized to describe the motion and the forces acting on the spool. M p ẍ p = F P p F P3 F SP F F L3 F F R P (4.6) ẍ p Resulting acceleration of the mass [ m s 2 ] F SP Force of the spring at the pumpside spool [N ] F F L3 Flow force due to the opening A 3 to tank [N ] F P p Force due to the pump pressure on the left side [N ] F P3 Force due to the pressure on the right side [N ] F F R P Friction force between the spool and the cylinder [N ] As explained previously, the small orifice seperating the flow to the left chamber of the pumpside spool is neglected. By doing that, it is assumed that there won t be any pressure build-up in this chamber. The pump pressure P p will therefore be observed here as instantaneously. As seen from equation 4.6, there are only two forces as a result of the pressure that needs to be determined, respectively F P p and F P3. F P p = P p A 3 (4.7) F P3 = P 3 A 3 (4.8) The pump pressure P p and the spring-side pressure P 3 To be able to express the pump pressure P p in known terms, the flows in and out of the chamber needs to be taken into account. The relationship is given in the continuity equation. An explanation of the continuity equation can be found in Equation A.7).

25 4.2. MODELING THE PUMPSIDE MODULE 17 Q i n Q out = dv d t + V β dp d t Q p Q v Q T = dv d t + V β dp d t Q p Q v Q T +Q b2 = V β dp d t (4.9) (4.1) There are three flows through V p. The flow from the pump Q P is dependent on the size and type of the pump. According to the datasheet, this PVG 32 can handle flows up to 13l/mi n[danfoss, 2]. The flow to the PVB module Q v, is dependent on the main spool position. This will be explored in the next chapter about modelling the main spool. A look on Figure 4.1 reveals that a movement of the spool will give cause to a change in volume as stated in equation 4.9. This change in volume will have an effect on both the left and right chamber of the spool. In order to take this volume change into account, the change in volume is replaced with the flow Q B2, which is the flow that runs through the pilot channel into the spring chamber. This is visible in equation 4.1. This replacement assumes that the flow into the spring chamber equals the volume change of V p, which is an assumptions which require the pressure build up in the spring chamber to be much faster than the pressure build up in the pump chamber. The flow Q B2 is determined using the orifice equation. Q B2 = C d A 6 (x p ) 2 ρ (P 2 P 3 ) (4.11) Here ρ is the density of the oil. The density is dependent on the temperature and normally it varies from kg, but is assumed to be a constant at 9 kg. The flow discharge coefficient is defined m 3 m 3 to be C d =.6. P 2 is the pilot pressure in the spring chamber of the PVB modul. This will be evident in the next chapter about the compensator spool modulation. The last unknown term in equation 4.1 is the Q T, which is the flow to the tank. This is, once again, calculated using the orifice equation: Q T = C d A 3 2 ρ (P p P T ) (4.12) Now that the flows in- and out of the spool is determined, an expression for the pressure P p can be derived by solving the continuity equation with respect to dp p d t : dp p d t = (Q p Q v Q T +Q B2 ) β V p (4.13) Where β is the Bulk Modulus, which is determined to be 7B ar as stated in section A.5. To determine the pressure in the spring chamber P 3, the continuity equation is used, and in the same way as it was when finding P p : Q B2 = A p ẋ p + V 3 A p x p β dp 3 d t (4.14) By solving equation 4.14 with respect to dp 3 d t dp 3 d t the following expression is found: β = (Q B2 + A p ẋ p ) (4.15) V 3 A p x p

26 18 CHAPTER 4. PROBLEM SOLVING The flow and spring forces at the pumpspool The flow force F F L3 is a variable force that will always work in the direction closing the valve. The flow force is determined by Equation F F L3 = 2 C d A 3 (P p P T ) cosθ (4.16) Where θ is the angle of the flow which is shown in figur 4.1 on page 13. Since there are no slot s in the spool, the angle of the flow is determined to be 69 [Andersen & Hansen, 27b]. The force of the spring F SP is a function of the spring constant k p and the displacement x p. The spring is always decreased in length, which means there will be a constant spring force, even when pump spool hasn t been moved and x p is zero. The constant spring force F pr eload p is calculated in section B.1 with a distance x pr eload p = 2mm to be: F pr eload p = 3N (4.17) By adding the preload to the spring force, a function for the overall spring force is obtained: F SP = k p x p + F pr eload p (4.18) F SP = N x p + 3N (4.19) The viscous and coulomb friction forces at the pump spool In section A.3 the friction force is explained in details. This chapter divides the friction force into three different friction forces; striction, coulomb and viscous friction. Measuring the friction forces is complicated, and the division in the three types of friction force is advanced. Available data in the litterature divides the friction force in a simpler model with two quantities, respectively a static friction force and a kinetic friction force. Both of these quantities coefficients are usually given as a constant. Therefore these quantities has similarities to the more advanced model of the friction forces. The static friction force corresponds to striction and coulomb combined, and the kinetic friction corresponds to the coulomb friction force. The static friction force will only have an effect between the spool and the cylinder, when the spool is at rest, and because the spool is almost never at rest, the static friction force will be neglected. The kinetic friction force will therefore be the same as the coulomb friction force. In order to look up the kinetic k k friction coefficients, the shape and material needs to be examined. It is assumed that the spools and the cylinders are made of steel, and there are some leakage oil in between the objects. For this reason the friction coefficients are found for lubricated steel. The kinetic friction coefficient differ, depending on the source. The sources agree on a value from.5 to.1 for the kinetic friction coefficient, Based on these observations the coefficient for kinetic friction between two lubricated steel plates are: k k =.5 (4.2) The friction force depending on this coefficient yields a constant value to the total friction force. This friction force is dependent on the normal force between the cylinder and the spool. Because it is assumed that the spool is placed horizontal, the normal force for a spool in motion will be the grativy force multiplied with the mass. Eventhough there is high pressure fluid inside, these won t have an effect on the kinetic friction, because the pressure will be observed on both sides of the spool as seen in figure 4.4.

27 4.2. MODELING THE PUMPSIDE MODULE 19 F t P p F sk Figure 4.4: The kinetic friction force between the spool and cylinder The kinetic friction force are of the following size: F sk = k k F t = k k g m p = m.7kg =.344N (4.21) s2 The edge effect on the spools Another important contribute to the friction force is, the fact that the edge of the spool will hit the cylinder as seen in figure 4.5 and the fact that there aren t much fluid around these edges. However, this is just an illustration and these conditions are heavily exaggerated. F t P p F se Figure 4.5: The kinetic friction force between the spool and cylinder because of the edges This will cause the oil in between the cylinder and the housing to be minimized, which makes a kinetic friction coefficient for dry steel probable. This friction coefficient, once again, vary with the source and is defined to be: µ ke =.57 The normal force in this case will be given by the pressure observed by the cylinder. Because each opposite edge of the spool hits opposite sides of the cylinder, only the pressure working on one half of the cylinder with the length l 1/2 will contribute to the normal force for each edge. A non uniform shape of the spool with respectively a diameter of D 1 = 18mm and D 2 = 9mm yields the following total surface area of the spool. A SP = D 1 2 π l 1 + D 2 2 π l 2 = mm 2 (4.22)

28 2 CHAPTER 4. PROBLEM SOLVING D 1 A active D 2 Figure 4.6: The active area of the cylinder which on which the pressure works as the normal force Noticing that only the contribution from the pressure working on the upper part of the cylinder will have an effect, the active area reduces likewise as seen in the previous figure to one half again and the active area are calculated based on this equation: A acti ve = 1 4 A Sur f ace The normal force given by a pressure working on the active area, when the gravity force is neglected is given by: F n = P f f A acti ve As an example, for a pressure of P f f = 1B ar and an active area of A acti ve =.1m 2 yields the following result, which makes it obvious why the gravity force and the kinetic friction force for lubricated steel calculated in Equation 4.21 are neglected: F n = P f f A acti ve = Pa m 2 = 1kN It should be noticed that this way of finding the friction force, will only give a very rough starting point for the model, it could be far from the friction seen experienced in reality. The friction force based on the edges will work on both sides of the cylinder. In order to take this into account the force needs to be multiplied by 2. The friction force, because of this effect, will give a constant contribution to the total friction force: F f c = 2 k ke F n = 2 k ke A acti ve P f f (4.23) This parameter can vary with the pressure and the actual active area, but because it is much larger than the friction force caused by the gravity, the total coulomb friction force will be determined by this edge effect. The pressure inside the pump module is assumed to be the pump pressure P p. However, besides the variables that has an impact on this edge effect, there is another important and unknown factor; How often will this edge effect occur? This can not be known, and the coulomb friction will be a parameter which is adjusted in order to tune the model. The viscous friction coefficient The viscous friction force is another contribution to the total friction, which is a term in the advanced friction model as explained previoulsy. This friction is dependent on the viscosity of oil, the distance between the spool and cylinder, and the velocity of the spool, as seen in the shear stress equation: F f v = τ oi l = µ du d y (4.24) According to the datasheet of the PVG 32 the kinematic viscosity of oil is given to be in the range from ν oi l = 12 75mm 2 /s. Using a value of ν oi l = 32mm 2 /s to find the shear stress, yields a transformation

29 4.2. MODELING THE PUMPSIDE MODULE 21 to a dynamic viscosity given by equation The velocity gradient can be calculated in the manner of the velocity of the spool and the smallest perpendicular distance between the housing walls and the spool as seen in equation Due to the precision of the materials, this perpendicular length is assumed to be l = 5µm. The velocity is zero at the housing, and at the spool it is the velocity of the spool. µ oi l = ν oi l ρ oi l = =.29 (4.25) du 1 = d y ẋ p = ẋ p (4.26) The total friction force is given as the coulomb and the viscous force contribution as seen in Equation Because the values in this equation is based on some parameters that can vary a lot, a more general form of the friction force, with a coulomb and a viscoius friction coefficient, is derived. F F R P = F f v + F f c F F R P = ν oi l ρ oi l ẋ p + 2 k ke P f f A acti ve d y F F R P = c v ẋ p + c c (4.27) The coefficients to the viscous friction force and the coefficient for the coloumb friction force is summarized to viscous coefficient c v and a coloumb coefficient c c as stated in Equation c v = ν oi l ρ oi l d y c c = 2 k ke P f f A acti ve (4.28) Inserting the value calculated in this chapter yields the values of the friction coefficient for the pumpside spool. Both the viscous friction coefficient c v and the coulomb friction coefficient c cp can be adjusted because of numerous unknown parameters. c v = 576 [ kg s ] c cp = P P [N] Conclusion The dynamics in the PVP module have been described mathematically, and Newton s second law have been utilized to described the motion of the spool. By applying the forces found in the previous chapter, Newton s second law, decribing the motion of the spool, is as decribed in Equation 4.29 The final model with all the terms derived of the main spool dynamic and these expressions are inserted in equation M p ẍ p = (P p P 3 ) A p k x p 2 C d A 3 (P p P T ) cosθ ( c v ẋ p + c cp ) (4.29) To model the dynamics of the PVP module the following assumptions have been made:

30 22 CHAPTER 4. PROBLEM SOLVING Constant flow from the pump Q p. Constant flow angle θ. Instantaneous pressure P p in the pressure chamber, hence neglecting the orifice. Pressure in the reservoir is kept constant at bar relative. The coulomb friction is neclected for now. 4.3 Modelling the compensator spool The compensator spool, is located in between the main spool and the pressure compensator spool. The function of this spool is to keep the pressure a certain amount above the load pressure. This way the spool will compensate for a fall or rise in pressure when ever needed, in order to keep a constant pressure drop from the compensator spool to the load. Another feature of the compensator spool is to block backwards fluid flow, if the pressure from the pump should fall below the load pressure. P K Q B1 A 5 Q v A k A 2 k c1 k c2 V 1 A 4 M k Q A B2 6 P 1 V p A 1 φ 1 P V 2 P 2 x k P P Q v Figure 4.7: A sketch of the compensator spool. In Figure 4.7, the constants and variables that needs to be determined to model the compensator spool, can be seen. Before determining equations for the variables, the constants will be determined. Table 4.2 shows as list of the constants and variables that will be determined in this section. Determination of the volume of the spring-chamber V 2 and the volume of the pressure chamber V 1 is changing the volume of the cavities before movement of the spool, to the volume emerged or contracted as an impact of change in x k. The original volumes are determined based on a zero point when compensator spool is at the most left position, as seen by the x k mark in Figure 4.7. The initial volume V 1 is actually, but in order to make the model work, a small volume has been addded.

31 4.3. MODELLING THE COMPENSATOR SPOOL 23 V 1 Volume of pressure chamber V 1 [ m 3 ] V 2 Volume of spring chamber V 2 [ m 3 ] A 1 Area of the spool opening A 1 [ m 2 ] A 2 Area of the spool opening A 2 [ m 2 ] M k Mass of compensator spool [ ] kg A k Cross-sectional area of the compensator spool [ m 2 ] k c1 Spring constant of the spring located on the compensator spool [ N ] m k c2 Spring constant of the spring located on the compensator spool [ N ] m A 4 Cross-sectional area of the orifice A 4 [ m 2 ] A 5 Cross-sectional area of the orifice A 5 [ m 2 ] A 6 Cross-sectional area of the orifice A 6 [ m 2 ] Table 4.2: Measurable constants, necessary to model the compensator spool V 1 ( xp ) V 2 ( xp ) =.5 cm 3 + A k x k (4.3) = 1 cm 3 A k x k (4.31) Cross sectional area of the compensator spool A k is determined from the diameter of the compensator spool, d = m. Thus the cross sectional area is: A k = π r 2 = π (9 mm) 2 = 2.54 cm 2 (4.32) The mass of the compensator spool was determined by using a scale, which revealed a mass of m k = 66g (4.33) There are two springs in the compensator spool, which leaves two spring constants to find. There is a thorough description on how the spring constants was determined in section B.2. k c1 = 9945 N m k c2 = 35 N m (4.34) (4.35) The longest spring that is active all the time, has the smallest spring constant k c1, while the short spring has a larger spring constant. This comes in sight when looking at the combined spring constant k c2. The reason for this is to avoid that the spool travels too far in the positive direction x k The areas in the compensator module The orifice A 4, that is located in the compensator spool, which leads fluid to the pressure chamber V 2 on the left side of the spool is determined to have a diameter D 4 = 1.5mm which yield a cross sectional

32 24 CHAPTER 4. PROBLEM SOLVING area as stated in equation The orifice areas A 5 and A 6, which are located in the pilot-lines has the same diameter D 5 6 = 1mm and this yields a cross sectional area given in equation A 4 = 3.5 mm 2 (4.36) A 5 = A 6 = 1 mm 2 (4.37) There are two discharge areas between the compensator spool and the housing from which the fluid flow Q v can pass through. These discharge areas are created two different places in the spool. The first discharge area A 1 will decrease with a positive spool displacement x k as a result of a higher pressure drop between the pump and load pressure. Contrary, the opening area A 2 will increase with a positive displacement x k. The spool opening that creates the discharge area A 2 has a cylinder shape with a uniform diameter D A2 = 18mm. Whenever there is a gap between the spool and the housing, fluid is free to flow through this gap.the gap determines the discharge area for A 2. This disharge area is calculated in equation 4.38 as a function of the displacement x k. A 2 = π D x k = π 18 mm x k (4.38) The part of the spool that creates the discharge area A 1 is cylinder shaped as well. This part of the spool is not uniform, instead numerous cavities are drilled in it at the edges. With these cavities there doesn t have to be a gap between the spool and the housing before fluid can flow. Whenever the spool is near the edge, the fluid is free to flow through these cavities. Because of this, the discharge area is no longer cylinder shaped. The shape of these cavities are based on two different slots, and there are four of each these types of slots at the edge of the spool. This arrangement of the slots makes the discharge area non-uniform. Using the surfaces of a cylinder, but only where the slots are placed will give one result, while using the perpendicular area of the slots will give another. The discharge area for a fluid through an orifice, is based on an opening from which it is possible for the fluid to flow through. Because of this fact, if there are two discharge ares, the fluid flow will depend mostly on the smallest. Using the program AMOC [Power, 1999], it is possible to determine the smallest cross sectional discharge area A 1 as a function of the spool displacement x k. The slot arrangement and the shape of the 2 different cavities in the compensator spool are illustrated in Figure 4.8: (a) Slot arrangement (b) Disc 1. (c) Disc 2. Figure 4.8: Slot arrangement and slot geometry

33 4.3. MODELLING THE COMPENSATOR SPOOL 25 As seen in 4.8a, there are 8 slots; 4 smaller and 4 bigger slots. Disc 1 is the name of the bigger slots which is illustrated in 4.8b, while Disc 2 are the smaller slot and is illustrated in 4.8c. (a) Horizontal discharge area. (b) Vertical discharge area. (c) Discharge area Figure 4.9: Discharge area In 4.9c, graphs of the discharge area as a function of the spool displacement x k is shown. The figure contains plots for the discharge area from respectively the four disc 1 slots and four disc 2 slots and a total discharge area as a combination of all of the slots. It can be seen, that the discharge area for Disc1 is allmost constant when the spool displacement exceed approximatly 4.5mm, which is because the smallest discharge area changes from a horizontal surface area as shown in 4.9a, to be the vertical cross area as shown in 4.9b. After defining the slots in AMOC, the program generates a [2x1] matrix, that contains a vector describing the total opening area in the compensator spool, and a vector describing the spool travel. To get a better resolution, Linear interpolation is used to achieve 1 points in the table instead of Force equilibrium on the compensator spool In order to determine the dynamics of the compensator spool, Newtons 2nd law will be utilized. The forces acting on the spool are shown in Figure 4.1:

34 26 CHAPTER 4. PROBLEM SOLVING F FF2 F FR A 2 F P1 M k F SC F P2 A 1 F FF1 X X K Figure 4.1: Forces acting on the compensator spool After determining the forces that acts on the compensator spool shown in Figure 4.1, Newton s 2nd. law can be utilized for determining the motion of the compensator spool: M k ẍ k = F P1 + F F F 1 F P2 F F F 2 F F RC F SC (4.39) ẍ k Resulting acceleration of the mass [ m s 2 ] F P1 Force due to the pressure in the left chamber [N ] F P2 Force due to the pressure in the right spring chamber [N ] F F RC Friction force between the spool and the cylinder [N ] F SC Force of the spring [N ] F F F 1 Flow force due to the opening A 1 [N ] F F F 2 Flow force due to the opening A 2 [N ] In order to calculate the forces involved in this equation, the pressures having an effect on the compensator spool needs to be found. The pressures in each end of the spool, respectively F P1 and F P2 have a direct effect on the spool, while the pressures effecting the flow has an impact on the flow forces. These quantities will be derived in the following section Pressures inside the compensator spool The pressure P p is already determined in connection with the pumpside module by Equation As it was evident in the previous section, it is only needed to derive an expression for the pressures P 1, P 2, P and P k in order to determine the pressure forces and flow forces. To determine pressure drops through the compensator spool, it is necessary to determine the flow through the compensator spool Q V, which is dependent on the load pressure P LS. So in order to achieve this, the main spool discharge area A H needs to be brought into the equation (the derivation of A H will be discussed in section 4.4). By utilizing the orifice equation, a series of equations that

35 4.3. MODELLING THE COMPENSATOR SPOOL 27 describes the flow Q V can be set up. Q v = C d A 1 2 ρ (P p P ) P p P = Q v = C d A 2 2 ρ (P P k ) P P k = Q v = C d A H 2 ρ (P k P l s ) P k P l s = ( Qv C d ( Qv C d ( Qv C d ) 2 ) 2 ) 2 1 A A A 2 H ρ 2 ρ 2 ρ 2 (4.4) (4.41) (4.42) For simplification these equations is merged to one equation with all the orifice areas merged to an equivalent area. Summarizing all the pressure drops and putting them equal to the pressure drop of the equivalent orifice yields: ( ) Qv 2 ( ) ( ) ρ C d Qv 2 ρ = A 2 1 A 2 2 A 2 C H d = 1 A 2 1 A 2 2 A 2 H A 2 eq A eq = 1 A A A 2 H 1 A 2 eq Lastly the orifice equation for the equivalent orifice: Q v = C d A eq 2 ρ (P p P l s ) P p P l s = ( Qv C d ) 2 1 A 2 eq ρ 2 (4.43) For this equivalent orifice equation to be valid, the volumes between the orifices must be small and the flow in each orifice must be the same. Allthough a very small amount of fluid will leak to the spring chamber due to the pressure difference and a tiny opening between the spool and the surrounding walls in the housing, this leakage can be neclected, because the cavity in the spring chamber is very small and the larger fluid flow contribution to this spring chamber from the pilot line. It is therefore legit to assume that the same flow will occur in all the mentioned orifices. As regards to the volumes in between the orifices, not only are the chambers very small, the fluid flow Q v is large during operation. If these chambers wasn t neglected, the pressure build up in these chambers would happen allmost instantaniously because of the large flow. Another important fact to point out, is that if all the orifices observe the same flow, the flow in and out of all these cavities will be zero. In addition, if the volume can t change, there wont be any change in pressures as a result of the fluid flow. Therefore it is also reasonable to neglect the volumes in between the mentioned orifices. The impact of these assumptions, is that a change in pump pressure, will be faster for the main spool, than it would be in reality. Solving Equation 4.4, 4.41, 4.42 and 4.43 for P P, P, P k and P l s respectively, gives these equations:

36 28 CHAPTER 4. PROBLEM SOLVING P p = P + P = P k + P k = P l s + P l s = P p ( Qv C d ( Qv C d ( Qv C d ( Qv C d ) 2 1 ρ A ) 2 1 ρ A ) 2 1 ρ A 2 2 H ) 2 1 ρ 2 A 2 eq (4.44) (4.45) (4.46) (4.47) The flow Q B3 through the orifice with the discharge area A 4 needs to be found in order to determine the pressure P 1. By utilizing the orifice equation, the following occur: Q B3 = C d A 4 To determine the pressure P 1, the continuity equation is utilized: 2 ρ (P P 1 ) (4.48) Q B3 = A K ẋ c + V 1 + A K x k β dp 1 d t (4.49) By solving Equation 4.49 with respect to dp 1 d t : dp 1 d t β = (Q B3 A K ẋ c ) (4.5) V 1 + A K x k The pressure P 2 in the spring chamber V 2, is determined by the flow into the chamber (Q B1 -Q B2 ). Q B2 was determined by Equation 4.11, so whats left to be found is Q B 1: Q B1 = C d A 5 2 ρ (P LS P 2 ) (4.51) Now that there is derived an expression for both Q B1 and Q B2, the pressure in the spring chamber P 2 can be determined by utilizing the continuity equation for the spring chamber: Q B1 Q B2 = A K ẋ k + V 2 A K x k β dp 2 d t (4.52) By solving equationequation 4.52 with respect to dp 2 d t : dp 2 d t β = (Q B1 Q B2 A k ẋ k ) (4.53) V 2 A k x k As a result of the areas and the found pressures the pressure forces observed at both ends of the compensator spool, can be calculated using the relationship between force, pressure and area: F P1 = P 1 A k F P2 = P 2 A k

37 4.3. MODELLING THE COMPENSATOR SPOOL 29 Because the area on both ends of the compensator spool are the same, these forces can be combined to one equivalent equation where the direction of the force are taken into account, as seen in Equation F P12 = F P1 F P1 = (P 1 P 2 ) A k (4.54) Flow forces in the compensator spool As seen from Figure 4.1, there are two flow forces acting on the spool F F F 1 and F F F 2. Flow forces will always work in the direction to close the opening where the flow occurs. This means that the two flow forces involved in the compensator spool works in opposite directions. The flow force F F F 1 that occures at the spool opening A 1 can be described as shown in Equation 4.55: F F F 1 = 2 C d A 1 (P p P ) cos(φ 1 ) (4.55) φ 1 The angle of the flow as shown in Figure 4.7. [r ad] P p The pressure on the pumpside of the compensator spool [Pa] P The pressure between the two openings A 1 and A 2. [Pa] A 1 Discharge area as a function of the displacement x k. [ m 2 ] The flow force F F F 2 occurs at the spool opening A 2, and is given by Equation 4.56: F F F 2 = 2 C d A 2 (P P k ) cos(φ 2 ) (4.56) φ 2 The angle of the flow through the orifice with the area A 2 [r ad] P k The pressure in the chamber between the main and compensator spool [Pa] A 2 Discharge area as a function of the displacement x k. [ m 2 ] The spring forces in the compensator spool There are two springs in the compensator spool which are being turned in to a single equivalent spring force F SC. The first of the springs has a preload, and the other spring isn t activated before the spool displacement reaches 5.2mm. This force depends on the spring constant k c1 and the displacement x k. Because the spring is preloaded, the spring is allready loaded at the chosen zero point for the displacement vector x p as seen in Figure 4.7. This initial deflection of the spring is estimated to be x spr i ng = 1.57mm. Using this value, the preload of this spring F pr eload is calculated in section B.2 and repeated here: F pr eload = 15.12N (4.57) The size of x spr i ng is the minimum compression of the spring, which also is the distance where x k is defined to have it s zero point x k =. By adding the preload to the spring force, a function F SC 1 for the spring that is always active is found:

38 3 CHAPTER 4. PROBLEM SOLVING F SC 1 = k c1 x k + F pr eload k F SC 1 = 9945 N m x k N (4.58) When the spool displacement exeeds 5.2mm, both springs will be compressed. The spring force will then be a function of a new preload F pr eload k2 to secure that in the transition both functions will have the same value. The spool displacement x k and a combination of the two springs which is defined to be k c 2 will also be a part of the function, which is expressed by: F SC 2 = k c2 x k + F pr eload k2 At a deflection of x k = 5.2mm the spring force F SC 2 out to have the same value as F SC 1. The spring force F SC 1 for the longest spring at the given point is: F SC 1 (5.2mm) = 9945 N 5 mm + 15N = 155N m Based on this observation, the following calculations, will determine the second preload value: F SC 2 (5.2mm) = F SC 1 (5.2mm) N m m + F pr eload k2 = 155N F pr eload k2 = 155N 35 N m 5.2 mm = 155N 175N (4.59) = 2N (4.6) This yields a spring force model for the part where both the springs are active, which is expressed in Equation 4.61 F SC 2 = k c2 x k + F pr eload k2 F SC 2 = 35 N m x k 2N (4.61) The two expressions determined for the springs is combined in one equation, which is given in Equation F SC = 9945 N m x k + 15N if mm < x k 5.2mm 35 N m x k 2N if x k > 5.2mm The friction forces in the compensator spool The friction-force F F RC depends on the velocity of the spool, and will always work against the moving direction of the spool. The friction force for the compensator spool is in aggrement with the friction force derived for the pressure adjustment spool in Equation The coefficients for the viscous

39 4.3. MODELLING THE COMPENSATOR SPOOL 31 friction coefficient c v will be the same, if assumed the proporty of oil doesn t change from one module to the other. The coulomb friction coefficient c c is given in Equation As the active area and the pressure are the only values that are different, calculating these will yield a value for the coloumb friction coefficient. The pressure inside this module is different for different parts of the spool, but is assumed to have the pump pressure P F F = P p. The spool has a diameter of D 1 = 18mm and D 2 = 9mm, which yields a surface area of: A SC = D 1 2 π l 1 + D 2 2 π l 2 = π π 3.7 = 13cm 2 (4.62) And the active area will become a quarter of this size as explained in the previous deriviation. A acti ve c = 1 4 A SC = 3.3cm 2 (4.63) The coulomb friction coefficient will now be calculated. As stated previously the friction coefficient c v can be adjusted because of numerous unknown parameters. c v = 576 [ kg s ] c cc = P P [N] Conclusion After inserting all the terms that are derived previously in this chapter, the model for the compensator spool will be as stated in equation 4.64 for a spool travel x k 5.2mm: M k ẍ k = (P 1 P 2 ) A k + 2 C d A 1 (P p P ) cos(φ 1 ) (k c1 x k + F pr eload k ) (c v ẋ p + c cc ) 2 C d A 2 (P P k ) cos(φ 2 ) (4.64) And in equation 4.65 for a spool travel x k 5.2mm: M k ẍ k = (P 1 P 2 ) A k + 2 C d A 1 (P p P ) cos(φ 1 ) (k c2 x k + F pr eload k2 ) (c v ẋ p + c cc ) 2 C d A 2 (P P k ) cos(φ 2 ) (4.65) This is all the forces acting on the compensator spool. To model the dynamics of the compensator spool the following assumptions have been made: Constant flow angles Instantaneous pressure build up in the chambers containing P k and P. The coulomb friction is neclected for now. This concludes the modelling of the compensator spool, and the focus will now be directed to the main spool.

40 32 CHAPTER 4. PROBLEM SOLVING 4.4 Modeling the main spool Side A Q L Q V Side B LVDT P bp P LS M M φ 3 φ 3 P 5 A 5 A H P 4 To tank A H P K To tank x M Figure 4.11: Main spool Calculating the pressures in the main spool module The pressure drop over the main spool are calculated as: P k P l s = ( Qv ) 2 1 C d A 2 H ρ 2 (4.66) The pressures P 4 and P 5 is determined by the electro-hydraulic actuation and this will be described in details in section 4.5. The model of the main spool consist of a number of measurable constant. The mass of the main spool was determined, using a scale to be: M m = 185g (4.67) Calculating the areas in the main spool module The area A H, of the main spool opening is determined as a function of the displacement x m. x m is when the main spool is in the middle position, i.e. when no oil is flowing. In 4.14a a plot of the opening area of the main spool in terms of spool displacement can be seen. It is clear that there is a deadband of 1.5mm in both directions. At maxmimum displacement x m = 7mm the opening area is mm 2. It is also seen that the area function is symmetric about the point x m =. The displacement of the main spool is controlled either by the manual lever connected directly to the spool or by the electro-hydraulic actuation as mentioned in section 2.3. The opening area in the mainspool A H as a function of x m is, 2like for the compensator spool, determined using the program AMOC. 4.12a shows the arrangement of the slots. There is a total of 4 slots at the edge of the main spool, and in figure 4.12c on the facing page, the geometry of one of the slots is illustrated, seen from above.

41 4.4. MODELING THE MAIN SPOOL 33 (a) Slot arrangement (b) Angle of the slot. (c) Slot geometry. Figure 4.12: Slot arrangement and slot geometry As seen in 4.12b, unlike the slots in the compensator spool, the slots in the main spool is angled which will have an impact on the flow angle, which will be determined later on. As stated earlier, the main spool has 4 slots per edge, which all have the same shape. The geometry of the slots yields a total discharge area as a function of the spool displacement, as shown in 4.13b (a) Angle of discharge area. (b) Total discharge. Figure 4.13: Total discharge The discharge area is always the smallest possible cross sectional area, 4.13a shows how the area is determined. Unlike the function of the discharge area for the compensator spool, the discharge area

42 34 CHAPTER 4. PROBLEM SOLVING for the main spool, as shown in 4.13b is more uniform, because the smallest cross sectional area follows the angle of the slots. 7 x Discharge area [m 2 ] l 1 φ3 h Main spool Spool displacement [m] x 1-3 (a) Opening area in terms of spool displacement. (b) Slope of slot in main spool Figure 4.14 As explained previously in section 4.2 AMOC creates a matrix, describing the total discharge area of the mainspool at a given displacement x m. The discharge area of the main spool opening A H is given in 4.14b. The main spool can travel in both directions, changing the flow of the fluid to two different ports, which means the model has to compensate for the change in flow direction. If x m is positive the main spool in figure Figure 4.11 is displaced to the right, which changes fluid flow from port A to port B. In order to keep consistency in the terms for the pressure, P l s follows the flow to port B, and reversed for P bs. Solving this problem can be done by having the model monitor the value of x m and changing the direction of the fluid flow through the load. In praxis this would have an impact in the resistance of the load as it is less likely that the resistance would be the same for both directions. As seen in 4.12b, the slots in the main spool are angled, which will make that the flow follow the slope of the slots. The angle of the flow φ 3 through the spool opening A H is dependent on the angle of the slope in the slots. The slope is illustrated in 4.13b, and the angle φ 3 is determined by Equation 4.68 φ 3 = tan 1 ( h l ) (4.68) From 4.12b the height of the slot h is determined to be (3.115mm.15mm) and the lenght of the slot is determined to be 5.64mm which yields Equation 4.69 ( ) 3.115mm.15mm φ 3 = tan 1 = (4.69) 5.64mm The angle of the slot needs to be described in [r ad] to be implemented in simulink, which yields Equation 4.7:

43 4.4. MODELING THE MAIN SPOOL 35 ( π ) φ 3 = =.48[r ad] (4.7) 18 A m is the cross-sectioal area of the main spool. The Diameter D m of the main spool was measured with a precise vernier caliper and determined to be D m = 18mm. The calculation of A m is as follows. A m = π D2 m 4 = π = 2.54 cm 2 (4.71) The area of the orifice A 5 placed in the pilot line, was not to be found in any datasheets. As such, it was necessary to determine the area based upon common sence. Making the area too small would mean that the pressure build up would be too slow. Knowing that the flow in the pilot line is very small the area of the orifice can be set equally small. For this it is estimated that an orifice cross-sectional area of the following size is reasonable: A 5 = 1mm 2 (4.72) The spring forces in the main spool module In section B.4 the experiment to determine the spring constant of the main spool k m can be seen. The spring constant is defined to be: k m = 3695 N m (4.73) In the PVE module their is a small spring attached to the LVDT sensor k l vd t yielding work on the main spool, as section B.3 outlines. The size of the spring constant is: k l vd t = 19 N m (4.74) The spring constant for the LVDT sensor in comparison with the spring constant attached to the main spool is very small, thus the spring connected to the LVDT can be neglected as it will have a very small effect on the dynamics of the main spool Forces acting on the main spool Newton s 2nd law will be utilized to determine the motion of the main spool. The dynamics behind the motion of the main spool will be explored by utilizing newtons 2. law. The forces acting on the spool for a negative spool displacement are shown in figur Figure 4.15.

44 36 CHAPTER 4. PROBLEM SOLVING x M x F P 4 F P 5 F SM M M F F R M F F F H2 F F F H1 Figure 4.15: Forces acting on the main spool After determing the forces that acts on the compensator spool shown in Figure 4.1, Newton s 2nd. law can be utilized in determining the motion of the compensator spool as seen in Equation ẍ m M m ẍ m = F P4 F P5 F SM F F R F F F H1 F F F H2 (4.75) Resulting acceleration of the mass [ m s 2 ] F F F H1 Flow force 1 at the area A H from port B to load [N ] F F F H2 Flow force 2 at the area A H from load to port A [N ] F F R M The friction force between the spool and the walls of the housing [N ] F P4 The pressure force in the left chamber working on the spool [N ] F P5 The pressure force in the right chamber working on the spool [N ] F SM Double acting spring force for the main spool [N ] Inertia of the Lever The main spool is directly coupled to a mechanical lever. As this is exerting a force to the main spool whenever it is subject to an acceleration, it is necessary to determine the mass of it. A complete model of the lever was designed in Solidworks, see Figure After that, the "mass properties" tool was used to determine the moment of inertia, of the lever, around the axis of revoulution, ie. where the lever is atached to the PVG 32. The moment of inertia is [ g mm 2]. As this is a rotational term and the mainspool is moving along a linear axis, the rotational force has to be transformed into linear force. As seen in Figure 4.17 the lever is coupled to the main spool via a 2mm long rod welded to the axel wich the lever rotates about. The equation for the moment of inertia is a follows: τ = I α (4.76)

45 4.4. MODELING THE MAIN SPOOL 37 Figure 4.16: An illustration of the lever that controls the main spool. 7mm 7mm M M Main spool 1mm Displacing the main spool 1mm 2mm Axis of revolution φ 2mm Figure 4.17: Working principle of the lever. The lever is coupled to the mainspoll via a 2mm long rod. Where τ is the torque, I is the moment of inertia and α is the angular acceleration. These components are the rotational inertia analogues to linear inertia equation: F = m a (4.77) Here, F is the force, m is mass and a is the acceleration. To convert the rotational inertia in the lever to translatoric inertia in the main spool, it is necessary to convert the linear acceleration of the main spool, ẍ M to rotational acceleration. This is done by expressing a main spool displacement as a corresponding rotation around the axis of revolution. So, if the main spool is displaced by 1mm the corresponding rotation of the lever is: ( ) 1 sin 1 =.521 r ad (4.78) 2 Furthermore, a displacement of 1m of the main spool corresponds to a rotation of 5.21 r ad. Thus

46 38 CHAPTER 4. PROBLEM SOLVING [ ] an acceleration of the main spool of y m can be converted to a rotational acceleration by this ratio: s 2 [ ] [ m r ad y = 5.21 y s 2 s 2 ] (4.79) Inserting this, and the value of I into the moment of inertia equation. τ = [ g mm 2] [ ] r ad 5.21 a s 2 (4.8) However, for this expression to be valid for a rotational equivalent to the linear, it is necassary to convert the torque [N m] produced by the lever to a force [N ]. F = τ [ g mm 2] [ ] 5.21 a r ad m = s 2 (4.81) 2[mm] Here m is the distance from the axis of revolution to the end of the rod. The final calculations are carried out and the expression is as follows. F = a (4.82) Where a is the acceleration of the main spool. This expression is a way of expressing how the rotational torque from the lever exerts a linear force on the main spool, and is the equivalent of the "rotational motion mass" transformed into "linear motion mass" and has the unit of [ kg ]. These calculations is only valid for small φ 4. Because in reality, if the main spool is displaced an infinite long distance x m i n f i nte, then φ 4 goes towards 9. For this expression to be valid, it is assumed that φ 4 is linearly dependent on x m and as such, φ 4 will grow as x m grows. However, the maximum displacement of the main spool is x m max = 7mm, and thus the maximum error is.5%. Therefore the unlinear behavior of φ 4 is neglected. As Equation 4.82 is an equivalent to the mass of the main spool this can just be added to the original mass of the main spool The flow forces at the main spool The flow forces F F F H1 and F F F H2 acting on the main spool is located at the left hand side of the main spool if the displacement x m is negative as seen in figure F F F H1 is allways located in the middle of the main spool and F F F H2 is either on the left or the right side of F F F H1 depending on how the main spool is displaced. However, F F F H1 and F F F H are always acting in the same direction. F F F H1 and F F F H2 are calculated using the flow force equation. The first flow force F F F H1 depends on the area A H, the pressure drop over the main spool from P K to P LS, while the second flow force depends on the area A H and the pressure drop from the returning load pressure P BP to the reservoir pressure P t. These flow forces are derived in equation F F F H1 = 2 C d A H (P K P LS ) cos ( φ 3 ) F F F H2 = 2 C d A H (P BP P T ) cos ( φ 3 ) (4.83) (4.84) In steady state when the external load is in balance, the flow out of Port B to the load will be the same as the flow from the load to the resevoir through Port A, when the displacement to the spool is negative in figure Figure Because both ports have the same discharge area A H, the orifice equation requires the pressure drop to be the same over both orifices. These observations leaves the

47 4.4. MODELING THE MAIN SPOOL 39 flow force to be modelled as one, with twice the magnitude, which is given in Equation This is also valid for a positive spool displacement, because both flow forces will work in the opposite direction. F F F H = 4 C d A H (P K P LS ) cos ( φ 3 ) (4.85) The pressure forces and spring forces The force acting on the main spool in terms of pressure difference from the electro-hydraulic actuation is calculated by the pressure difference between P 4 and P 5. F P45 = F P4 F P5 = (P 4 P 5 ) A M (4.86) The spring in the main spool is a double acting spring preloaded with a predefined force. This is done to prevent the spool from oscillating when small pressure differences occur. Preloading the spring means the pressure difference has to build up to overcome the preloaded force and hence move the main spool. According to the experiment in section B.4 the spring is initialy deflected with x pr eload m 18mm. The calculations shows that this deflection adds a preload to the main spool of the following size: F pr eload m = 65.5N (4.87) Further calculation of the resulting spring force are as follows. F SM = k m x M + F pr eload m = 3596N X M N (4.88) The friction forces in the main spool The friction-force F F RM is given by Equation 4.27 with a coloumb friction coefficient c c and a viscous friction coefficient c v. The viscous effect is dependent on the viscosity of the oil, and because this doesn t change, the size of the coefficient is the same as for previous spools. The viscous friction coefficient c v depends on the active area which is given by the surface area. The dimensions of the spool are seen in Figure ,48 18 Figure 4.18: The dimensions of the main spool used to calculate the surface area

48 4 CHAPTER 4. PROBLEM SOLVING The main spool have been fully modeled in Solidworks. The "measure" tool was used to determine the surface area to be 5173mm 2 The biggest diameter is D 1 = 18mm and has a length of l 1 = 14mm, while the smallest diameter is D 2 = 12mm and has a length of l 2 = 7mm. This gives a surface area of: A SM = D 1 2 π l 1 + D 2 2 π l 2 = π π.7 = 31cm 2 (4.89) The active area based on the explanation following Figure 4.4 becomes: A acti ve m = 1 4 A SM = 12.93mm 2 (4.9) Inserting this into the friction coefficient equations will yield a value for the friction forces of the following size. Remark that the friction coefficients can be adjusted because of numerous unknown parameters as explained previously. c v = 576 [ kg s ] c cm = P P [N] The final model of the main spool M m ẍ m = (P 4 P 5 ) A m k m x m (c v ẋ m + c cm ) 4 C d A H (P K P LS ) cos ( φ 3 ) (4.91) This concludes the modeling of the main spool. To be able to establish a dynamical model it was necessary to determine all of the forces acting on the spool. Furthermore it was assumed that the their was the same flow rate in both orifices allthough this is only somehow true in steady state. Also the coulomb fiction is neglected. The assumptions are listed in Equation To model the dynamics of the main spool the following assumptions have been made: Constant flow angles The same flow force exists in both the output port and the input port of the main spool. The coulomb friction is neclected for now. Next, the modeling of the PVE module will be covered. 4.5 Modelling the PVE module This section will derive and explain a dynamic model of the PVE module. The PVE module contains an electric control unit and a hydraulic system. These units are ment to displace the main spool, as explained in section 2.3. A model of the PVE module will be derived by examining the governing equations. With the purpose of deriving a model, all the measured constants for this module will be studied and used as the proper coefficients. Those quantities are listed in Table 4.3.

49 4.5. MODELLING THE PVE MODULE 41 A checkval ve A dr ai n A suppl y V 4 V 5 Area of the checkvalves A and B The area of orifice leading to the reservoir The area of the supply orifice Volume of the Port A chamber Volume of the Port B chamber [ m 2 ] [ m 2 ] [ m 2 ] [ m 3 ] [ m 3 ] P pi l ot The pilot pressure at the supply line [Pa] P r eser voi r The reservoir pressure at the drain line [Pa] Table 4.3: Measureable quantities to be used in the PVE module. Because the PVE module contains both an electric and a hydraulic part, these parts will be treated seperatly. First, the hydraulic circuit with the diagram shown in Figure 4.19, will be studied. P pilot Q supply P upper NC1 V upper NC3 Port B Port A P 5 P 4 V 5 V 4 NO2 b A checkball a NO4 V lower P lower P tank Q drain Figure 4.19: Hydraulic circuit diagram in the PVE module to displace the main spool The areas of the orifices The two, spring loaded checkball valves has the same diameter D seat = 1.7mm. Whenever there is a partial vacuum in port A or Port B, the fluid will flow through the checkball valve from the lower cavity to the Port through a discharge area A checkval ve. A checkval ve = π D2 seat 4 = 2.3mm 2 P upper

50 42 CHAPTER 4. PROBLEM SOLVING The drain orifice Q dr ai n, is the orifice between the lower cavity and the tank. It has a diameter of D dr ai n = 1mm, and this gives the following area: A dr ai n = π D2 dr ai n 4 =.79mm 2 The supply orifice leading the fluid from the pilot relief valve to the upper cavity is determined to have a diameter of D suppl y = 4mm which yields: A suppl y = π D2 NC 1 4 = 12.6mm Chambers and piping between the components A positive displacement of the main spool will change the volume in Port A and port B. The volume in Port A will increase and port B will decrease. The size of the main spool determines the change in volume in the chambers. The spool, which is cylinder formed, has a diameter of D = 18mm and therefore a cross sectional area of: A m = 254mm 2 (4.92) The position of the spool determines the volume of the chambers at port A and port B. In an experiment which is described in section B.5 the size of the initial cavities are determined to be: V 4 = 13cm 3 V 5 = 27cm 3 The volume of these cavities changes with the displacement of the spool. In order to account for this, the direction of the spool displacement x m yields the following connection: V 4 (x m ) = V 4 + x m A m = 13cm 3 + x m 254mm 2 (4.93) V 5 (x m ) = V 5 x m A m = 28cm 3 x m 254mm 2 (4.94) To model the fluid in the pipes between the relief pilot valve and the solenoid valves, a cavity with the name upper cavity is added. Likewise there is added a cavity with the name lower cavity at the other end. These volumes are assumed to be: V lower = 1mm 3 (4.95) V upper = 2mm 3 (4.96) Pressures inside the PVE module Inside the PVE module there are four nodes which contains a volume. To model the dynamics of main spool movement, the pressures inside these chambers P l ower, P upper, P 5 and P 4 needs to be determined. The flows running in and out of each node is used in the continuity equation to calculate the pressures. These continuity equations are given in equation 4.1. Notice that the change in volume is neglected in the upper and lower chamber, because these only contains piping.

51 4.5. MODELLING THE PVE MODULE 43 Q suppl y Q NC 1 Q NC 3 = V upper β dp upper d t Q NC 1 +Q check B Q NO2 = A m ẋ m + V 5 x m A m β Q NC 2 +Q check A Q NO4 = A m ẋ m + V 4 x m A m β Q NO2 +Q NO4 Q check A Q check B Q dr ai n = V l ower β dp lower d t dp 5 d t dp 4 d t (4.97) (4.98) (4.99) (4.1) The pressure drop over the main spool causes the spool to move. The continuity equations will be solved with respect to the pressures in order to determine the pressure drop across the main spool: P upper = P 5 = P 4 = P lower = β V upper (Q suppl y Q NC 1 Q NC 3 ) d t (4.11) β (Q NC 1 +Q check A Q NO2 + A m ẋ m )d t V 5 x m A m (4.12) β (Q NC 3 +Q check B Q NO4 A m ẋ m )d t V 4 + x m A m (4.13) β (Q NO2 +Q NO4 Q check A Q check B Q dr ai n )d t V l ower (4.14) The pressures inside the ports depends on the position of the spool. This is because the volume changes with the movement of the spool in Port A and Port B. The pressure inside Port B is given in equation 4.12 and for Port A in equation These continuity equations dependens on flow through different components. The flow through the checkball valve is modelled as a rectified orifice. This means that the fluid only can flow in one direction. The area of this equivalent orifice is still the value found for A checkval ve, and using this in the orifice equation yields: C d A checkval ve 2 ρ Q check B = (P lower P 5 ) if P l ower > P 5 if P l ower P 5 Q check B A checkval ve Flow to Port B from the lower cavity Area of checkball-seat equivalent to a orifice area [ m 3 s ] [ m 2 ] P 5 Pressure in port B [Pa] P lower Pressure in the lower cavity [Pa] The drain flow to the tank, and the supply flow through the pilot relief valve are dependent on the pilot pressure and the pressure in the tank. The pilot pressure depends on the flow through the pilot line and the amount of electric actuation. For continuous electric actuation this flow is measured to be Q pi l ot = [.7 : 1.]l/mi n. As seen in Figure 4.2 the pilot pressure for this flow is approximately P pi l ot = [13 : 14]B ar [Danfoss, 2].

52 44 CHAPTER 4. PROBLEM SOLVING bar Max (Elect Act) Min. (Elect Act) l/min Figure 4.2: The pilot line pressure as a function of the flow through the pilot line.[danfoss, 2] Assuming a tank pressure of P t ank = B ar and a pilot pressure of P pi l ot = 13.5B ar, the values can be used to calculate the flow through the drain and supply orifice, given in the following equation. 2 Q dr ai n = C d A dr ai n ρ (P lower P r eser voi r ) (4.15) 2 Q suppl y = C d A suppl y ρ (P ) pi l ot P upper (4.16) The flow through the solenoid valves NC and NO are determined by a 4D model explained in the succeeding subsection, which covers all the flows through the hydraulic circuit in the PVE module Flow through the solenoid valves; NO and NC The solenoid valves are used to build up a variable pressure inside port A and port B. To implement this, there are used two types of solenoid valves, respectively a normally open NO and a normally closed NC valve. The NO valves are connected to the tank side, while the NC to the pilot relief side. The flow through each solenoid valve depends on the dutycycle, which controls the on-off dynamic of the valves. To take the dynamic properties of the fluid and the on-off dynamics of the solenoid valve into account, a 4D model is created to determine the flow as a function of 4 parameters [Danfoss, n.d.a]. This is seen in T Tfluid T Vs D T Orifice 1 u1 u2 u3 4D-model 4-D T(u) NC1, NC3, NO2 & NO4 NO2 Fow_ripple Q P valve u4 Figure 4.21: The 4D models of the solenoid valves; One for the normally closed and one for the normally open valves.

53 4.5. MODELLING THE PVE MODULE 45 As seen in the figure, four different parameteres determines the value of the flow, and these parameteres are listed here. An explanation of why these parameters have an effect on the flow through the solenoid valves is convenient for understanding the model. These are included in the table The temperature of the fluid: The temperature of the oil has an effect, as it changes the viscosity. The effects of the viscosity will be explained later on. The supply voltage to the PVE module: The supply voltage has an effect on the dynamic proporties of the solenoid valve. A higher supply voltage, will increase the magnetic field causing the solenoid valve to have a faster reaction time. The duty cycle to the solenoid valve in question: The duty cycle describes how long, during a period T p the valve has to switch between on and off. A higher dutycycle for a NC valve means, that longer time during a period, the valve will be ON and fluid can flow through it. During on time, the inductor in solenoid valves will have a voltage drop enforcing the current to rise. This inductor convert electric energy to mechanical through a magnetic field, by moving the valve connected to the inductor and opening or closing for flow. The pressure drop across the valve: The pressure difference has a direct effect on how much flow that can pass through the valve. The higher pressure difference, the more flow. Changing the viscosity yields a change in the density of the fluid, which will force the pressure to change. This is seen in Equation All the other parameters in this equation are constant [Andersen & Hansen, 27a]. ν = µ ρ e( T ) P (4.17) ν Kinematic viscosity of the fluid µ Dynamic viscosity at atmospheric pressure ρ The density of the fluid [ m 2 ] s [ N s ] m 2 [ kg m 3 ] T The temperature of the fluid [ C ] P The pressure of the fluid [B ar ] Changing the density has an effect on the stiffness, also known as the compressibility of the fluid κ, which is the inverse of the Bulk modulus. β F = 1 = 1 κ F 1 ρ δρ δp β Bulk modulus [B ar ] κ Compresibility [ B ar 1 ] δρ The density change of the fluid [ kg /m 3 ] δp The pressure change of the fluid [B ar ] = ρ δp δρ (4.18) Bulk modulus is used in the continuity equation and is more thoroughly explained in Equation A.7. Changing the temperature and thereby indirectly changing the Bulk modulus, effects the importance of the pressure changes inside the continuity equation [Andersen & Hansen, 27a]. The 4D-model is based on various different conditions, by changing the voltage, the fluid temperature, the duty cycle and the pressure drop across the valve. To plot a graph of how the fluid changes

54 46 CHAPTER 4. PROBLEM SOLVING with varying duty cycle, some of the parameters must be held constant. If assumed that the fluid temperature is T = 5 C, and the supply voltage is V s = 13V, a graph of the flow as a function of the duty cycle can be plottet for different pressuredrops. The graph at figure 4.22 the fluid flow through orifice NC 1 and NO2 as a function of the duty cycle. Flow Q [m 3 /s] 3 x Bar 4 Bar 6 Bar 8 Bar 1 Bar 12 Bar 14 Bar 16 Bar Flow Q [m 3 /s] 5 x Bar 4 Bar 6 Bar 8 Bar 1 Bar 12 Bar 14 Bar 16 Bar Duty cycle DT [%] Duty cycle DT [%] Figure 4.22: The graph in the left represents the flow for a NC 1 valve as a function of the duty cycle for different pressures, while the graph in the right represents the flow for NO2. The one with the biggest flow is measured at 16B ar, and the lowest flow at 2B ar The flow through a NC valve increases with the duty cycle, while the flow through an NO valve will decrease. Furthermore the NC valve has a more linear characteristic, while the NO valve has a larger constant region, causing the flow through each valve to be different for the samme presure drop. A pressure drop P = 2B ar over both valves, yields an actual flow to each chamber as illustrated in Figure When the duty cycle is DT = 8%, this will yield more flow into than out off port B. Besides, this yields a lower duty cycle to port A seen in Figure 4.28, resulting in more flow out of Port A than in. This consequently will cause a pressure difference between the chambers in both ports, resulting in a force on the main spool. 1 x Bar.5 Flow Q in Q out [m 3 /s] Duty cycle DT [%] Figure 4.23: The actual flow in and out of the chamber at the ports.

55 4.5. MODELLING THE PVE MODULE The model of the electric actuator The electro-hydraulic unit serves to move the spool according to a desired position. In order to assure, that the spool has moved to the desired location, the electric unit uses a feedback system. This system is build using the principle of closed loop feedback control theory as illustrated in Figure The error in position turns into a duty cycle which defines a different flow pattern for each port through the electromagnectic valves. The two electromagnetic valves connected to each port builds up a pressure both at Port A and Port B. The pressure inside these ports are dependent of the spool velocity, position and the flow pattern of the connecting valves. The pressure difference at each port are forcing the main spool to move. Spool postion Spool velocity X set V rel Defined spool position + - Error in pos DT NC1 X err DT NO2 X err DT NC3 X err QNC1 DT NC1 QNO2 DT NO2 QNC3 DT NC Port B P 5Q5 Port A P 4Q4 + - Spool System Actual spool position DT NO4 X err QNO4 DT NO4 Measured spool position LVDT sensor Figure 4.24: The closed loop feedback control of the pve module. To sum up and explore the entire process from an input signal to a displacement of the spool, the process needs to be broken into parts. The parts that are involved in this process, includes a sensor to measure the actual position of the spool, an input signal that defines the position of the spool and an output signal to each electromagnetic valve. The difference between the input and the measured signal is used to calculate two different PWM signals; one for port A cavity and one for port B cavity. The input signal defining the position The input signal defines the position of the main spool. How this is implemented in the PVE module is basically using the voltage source as a full scale resolution, and the input signal as a varying parameter used to compare with the voltage source. This is done by dividing the input voltage V i nput with the supply voltage V s, and using this relative value to specify a main spool displacement value x m. V r el = V i nput V s (4.19) Because the spool can travel ±7mm, a relative value of V r el =.5 is defined to be the mid position of spool travel x m = mm. Furthermore, the maximum displacement in both directions is defined to be V r el =.25 in negative travel direction, and V r el =.75 in positive. This gives a linear relationship between the spool displacement and the relative input signal as shown in Figure 4.25.

56 48 CHAPTER 4. PROBLEM SOLVING 8 6 Main spool position x set [mm] Relative input voltage V ref [.] Figure 4.25: The ouput for the transfer function. The relation between the relative input signal V r el and the defined settling point for spool travels is given by Equation This equation is used to transfer the input signal to a position, so it is possible to compare with the measured position. x set =.28 V r el.14 (4.11) Position Sensor There are two common categories of position sensors; A linear sensor to measure a linear movement, and an angular sensor to measure an angular movement. The one inside the PVG 32 is a linear sensor of the type linear variable differential transformer (LVDT). Figure 4.26 illustrates the principle idea behind a LVDT sensor. V in Moveable core Primary coil L s L p L s + L p V in L s V out Secondary coil L p - + V out - Figure 4.26: The figure on the left side shows the actual composition of an LVDT sensor, while the figure on the right side shows an electric circuit diagram of a LVDT sensor The advantage of this sensor is the low maintenance requirements because the sensor doesn t have any contact with the moving parts. This provides a longer lifetime of the sensor because of less wear on the components. The moving part of the sensor is an iron core, which is directly connected to the main spool, so a spool movement yields a movement of the iron core used in the sensor. An electric conductor is wound around the iron core and connected to an AC voltage source. This is called the primary winding. The windings will create a magnetic field, this field will be magnified in a direction perpendicular to the windings as shown in Figure 4.27.

57 4.5. MODELLING THE PVE MODULE 49 V in + - iron core I in O primary winding Figure 4.27: Illustration of how the flux is magnified through the iron core using an alternating current. By placing an iron core inside the windings, nearly all the generated flux is passed through the iron core and not the air. This happens because of the high relative permability µ r of the iron core relative to the low relative permability of air at µ r = 1. A high relative permability gives a low magnetic reluctance, which is the equivalent of electric resistance. The relation is given by Equation This equation shows that the magnetic reluctance of the iron is much smaller than the air reluctance, and therefore the main part of the magnetic flux will pass through the iron core. l R = µ µ r A (4.111) R Magnetic Reluctance [ At ] W b l Lenght of the element [m] µ Permability of free space µ r Relative permability of the material A The cross sectional area of the magnetic circuit [ H ] m 1 [ H ] m 1 [ m 2 ] Henvisninger Two other sets of windings, called the secondary windings, is connected to the same iron core as seen in Figure Depending on the spool position, one of the secondary windings will cover the iron core more than the other. As a result of this, the mutual inductance between the primary inductor and the two secondary inductors will vary as a function of the position of the iron core [Wikipedia, n.d.b]. The total flux going through the iron core is given by the flux density which is dependent on the current I i n, and the number of turns of the windings n. This is given in the following Equation φ = n I i n (4.112) Depending on the size of the iron core, and the number of turns in each inductor, the relation between the input voltage and the output voltage differ [Wikipedia, n.d.c]. According to the model made by Sauer Danfoss [Danfoss, 2], the transfer function for the LVDT sensor in the PVEH module is given in Equation 4.113, where the x m variable is the actual position and the x m variable is the measured. x m x m = s + 1 (4.113) This first order transfer function of the LVDT sensor has a time constant of τ =.59s. Additionally, it is seen that the steady state gain is lower than 1.

58 5 CHAPTER 4. PROBLEM SOLVING The duty cycle for each port depending on position error The error in position defines the duty cycle of each port, and the duty cycle defines how much flow the electromagnetic valves pass through to each port. The error in position is calculated taking the difference between the defined spool position and the actual spool position as seen in Equation x er r or = x set x m (4.114) The two electrocmagnetic valve types NC and NO are both recieving the same duty cycle. The only difference of the dutycycle for these valves are in the implementation. The pilot side valves NC 1 and NC 3 has a time delay of 2ms and the reservoir side valves has a time delay of 3ms. The duty cycle signals for both valves in each port has an in-between time delay of 17ms. The function of the time delay is to imitate the physical proporties of a solenoid valve. This time delay is implemented in the control. The reason for this is to avoid a shortcut, which means that the valves connected to the same port won t open for passage at the same time. Whenever there is a hightime, first the NO valve will close and 17ms later the NC valve will open. This also serves to imitate the physical behaviour of the valves. Since the NC valves work against the pressure they are somehow slower. The error in position defines a duty-cycle which can vary between and 1 percent. Figure 4.28 illustrates the dutycycle as a function of the error in position. 1 9 Port B Port A 8 7 Duty cycle DT [%] Error in position [mm] Figure 4.28: The duty cycle output for the port A cavity and port B cavity side valves as a function of the error in position. The duty-cycle as a function of the error in position for both port A and port B valves are given at Equation There is an outer boundary for the output values in these equation ranging from % to 1%. DT A = 7 x er r or DT B = 7 x er r or (4.115)

59 4.6. MODELING THE LOAD Modeling the load In order to validate the model of the PVG 32, the valve has to be added to a hydraulic circuit with an external load. In addition, the load will have flowmeters and pressure gauges around the load, so it is possible to compare the measurements with the simulations. This load has the function to switch the load between using one piping line or two. One of the piping lines will allways be active, while the other can be opened if desired. Consequently this means that the flow Q v has to be split up into the two lines and therefore several equations for determination of flows and pressure have to be established Flows and pressures in the load P f Pressure in the front of the load [Pa] P m Pressure in the mid section of the load [Pa] P b Pressure in the back of the load [Pa] A L1 Area of the orifice A L1 [ m 2 ] A L2 Area of the orifice A L2 [ m 2 ] A L3 Area of the orifice A L3 [ m 2 ] Table 4.4: Constants necessary to model the load LOAD A L1 Q L1 V m P m Q L3 Q L2 V b P b Q b A L3 Q f V f P f Q L Q V P bp P LS Figure 4.29: The working load of the system

60 52 CHAPTER 4. PROBLEM SOLVING The pressure in the load are calculated utilizing the continuity equation wich takes the following form. Q f Q L2 Q L3 = V f β P f β P f = d t (4.116) t (Q v Q L3 Q L2 ) V f Q L2 Q L1 = V m β P m t Q L1 +Q L3 Q b = V b β P b t P m = P b = β (Q L2 Q L1 ) V m d t (4.117) β (Q L1 Q L3 Q b ) V b d t (4.118) Now the Flows throughout the load i.e. the flow through the orifices Q L1, Q L2 and Q L3 can be calculated using the orifice equation. 2 Q 1 = C d A L1 ρ (P m P b ) (4.119) 2 Q 2 = C d A L1 ρ (P ) f P m (4.12) 2 Q 3 = C d A L1 ρ (P ) f P b (4.121) The orifice equation assumes that the flow is turbulent, which will be investigated further when the test data is analysed. 4.7 Validation of unlinear advanced model This section takes results from the test section B.7, and the model in order to verify the model. To do so, the area of the load orifices are to be determined Introduction to the test In the laboratory test the PVG32 was controlling the flow through two orifices of variable contraction area. Refer to section B.7 for more details about the test. Beside the two orifices the test arrangement also included a solenoid valve which would block the flow to one of the orifices. The system was tested with three parameters varied. The parameters where, the contraction area of the two load orifices, the magnet valve, and the control voltage on the PVG, which in steady state leads to a certain flow. The solenoid valve was either opening or closing during the tests. The classification of the test series is illustrated in Figure 4.3. Each section of the block symbolizes a single test and each test has a unique set of parameters. Opening Closing Small area Large area Small flow Large flow Figure 4.3: Classification of test series

61 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 53 As described in the test report, the things being measured includes, the steady state flow into the load Qv, and the flow in the branch containing the solenoid valve L2 and the orifice L1. The flow in this is denoted Q L2. An illustration of the load is seen on Appendix D. Beside flows, also pressures was measured. Three pressure transducers were applied to the the load, measuring the input pressure P f, the intermediate pressure P m, and the output pressure P b. In contrast to the flow measurements, the pressure measurements were logged over time, and not just one single steady state value Calculating areas of the load In order to compare the results from the model with the test results, it is necessary to find the areas of the load during the test. As no direct measurement of the orifices have been carried out, it is only possible to find the areas through calculations, based on pressure and flow meaurements. The test was carried out with two different opening areas for each of the two orifices, one large denoted b and one small denoted s. The relationship between flow, pressure and area of an orifice is described in the orifice equation. For more details, see Equation A.1. This orifice equation solved for area is shown in Equation Q A = C 2 d ρ (P i n P out ) (4.122) Using this equation it is possible to set up the following equations describing the areas of the load orifices from the measured data. A L1 = A L2 = A L3 = Q L2 C d 2 ρ (P m P b ) Q L2 C 2 d ρ (P f P m ) Q f Q L2 C 2 d ρ (P f P b ) (4.123) (4.124) (4.125) Applying these expressions to each of the 12 tests, returns the areas given in Table 4.5.

62 54 CHAPTER 4. PROBLEM SOLVING Test no. A L1 [mm 2 ] A L2 [mm 2 ] A L3 [mm 2 ] Table 4.5: Areas of the load orifices based on data from the specific test As you can see, the results yields a great amount of variety, so in order to find the best suitable area the Reynolds numbers are calculated. The reynolds number is an indicator of which forces is dominant in the current flow. If the inertia forces are dominant the Reynolds number is high and the flow is turbulent. On the other hand, if the Reynolds number is low the viscous forces are dominant and this kind of flow are called laminar. For the flows in this particular test, Equation is used to calculate the Reynolds number. The equation is for a smooth pipe, but it s assumed that the parameters of the equation yields the same effect to the flow through an orifice. The transition from laminar to turbulent will however happen at a lower Reynolds number because of the more disturbing geometry of an orifice. Re = V D ν (4.126) Re Reynolds number [ ] ] V Velocity of the fluid [ m s D Diameter of the pipe [m] [ ] ν Kinematic Viscosity m 2 s The kinematic viscosity is unknown, but the group has been informed that it was a common mineral oil that where used in the tests. The kinematic viscosity for some common mineral oils are listed in Figure A.2 in section A.1. As there is no further information available the kinematic viscosity is assumed to be ν = 32cSt. The velocity and the diameter, used in the formula, are calculated using the following two equations.

63 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 55 4 D = π A (4.127) The reynolds number for each flow are listed in Table 4.6 V = Q A (4.128) Test no. L1 L2 L Table 4.6: Reynolds number of the flows in the load orifices The transition point from laminar to turbulent flow isn t well defined, but as a rule of thumb it happens at 23 for hydraulic hoses and pipes [Andersen & Hansen, 27a]. In this case it s an orifice and because of the disturbing effects of the rough geometry it will likely happen at a much lower value. Based only on the Reynolds numbers all the tabulated flows seems turbulent, some of course more than other. The flow which will determine the area used in the orifice equation of the load, is the one with the highest Reynolds number, hence this flow is the most turbulent. The elected areas are tabulated in Table 4.7. A L1 [mm 2 ] A L2 [mm 2 ] A L3 [mm 2 ] Small Large Table 4.7: Areas of the load orifices. Asuming turbulent flow. Turbulent flow through an orifice is determined by the orifice equation, which states that the flow rate is related to the square root of the pressure drop. In contrast; laminar flow has a linear relationship determined by Equation 4.129, the so called Hagen-Poiseuille equation.

64 56 CHAPTER 4. PROBLEM SOLVING P Pressure drop [Pa] µ Dynamic viscosity [ N s ] m 2 D Diameter of the pipe [m] [ ] Q Flow rate m 3 L Length of the pipe [m] s P = Which can also be written as in Equation µ L Q π D 4 (4.129) A 2 Q = P (4.13) π 8 µ L 17.5 Flow characteristics of orifice A L3 in load 15 Flow through orifice [l/min] Small opening GR522 test 2.5 Small opening GR52 test Large opening GR522 test Large opening GR52 test Pressure drop across orifice [Bar] Figure 4.31: Plot of steady state flow and pressure for four different areas of the same variable orrifice A L3 in the load Remember that the areas calculated with the orifice equation yields great difference. Therefore it could be interesting to see if the flows really form the graph of a square root if plotted. To assist a conclusion, data from another test series performed by another group of students, with the exact same equipment, three days before the measurements done by our group, is also included. The figure showing the four plots and their corresponding linear fits are shown in Figure The linear fits are made so they contains origo, because when there is no pressure drop there is no flow. The figure shows the magnitude of the flow rate, as the pressure drop across the orrifice A L 3 is varied. All 4 plots shows linear behaviour. This is definitely a surprising result. One should expect to see a

65 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 57 square root characteristics because of the high pressure drop, but this is not the case. Let s consider the sources of error. As stated in section B.7, the tolerance of the flow meter as low as 1 l mi n, and for the pressure transducers it can differ as much as 1B ar [Pedersen, 29], so when pressure drop is considered it will be up to 2B ar total. The first graph to consider is the one which would have the highest Reynolds number, since if this is a laminar flow, it is reasonable to assume the other flows are aswell. The one considered is the lowest graph, "Small opening GR522 test", wich is the blue one. Imagine that each of the 3 points making up the blue graph is allowed to move in a square 2 bars wide and 1 l/min high, will it be possible to fit this graph to a square root function? Even if the sources of errors are considered to be all in the favour for a turbulent flow, it is simply impossible to fit these data to a square root function, as the curvature of the plot are all greater or equal to zero. If it are to reach zero flow at zero pressure it can t have a negative curvature in all of it s regime, which all real square root functions must have (see Equation 4.131). For the graphs to have a negative curvature and still reach zero it must have it s highest point below the linear fit, and the lowest point above. The curvature must also be largest at the smallest pressures, which require that the slope between the start point and the midpoint should be significant larger than the slope after the midpoint to the endpoint, which is impossible to achieve while still reaching origo, even though the sources of error are considered. The same is, by the way, also true for all of the 4 graphs in the figure. So it is reasonable to assume the flows in all of the orifices are laminar, as the flow through the other two orifices of the load happens at a much lower Reynolds number. d 2 1 x = d x 2 4 x 1.5 (4.131) In order to still use the data to test the model, it s necessary to change the load block of the model, to assume linear flow (see Equations 4.132). Q L1 = K L1 (P m P b ) Q L2 = K L2 (P f P m ) Q L3 = K L3 (P f P b ) (4.132) The constants in the expressions is the slope of the linear fits of the measured data. In order to find the constants for the last to expressions two more plots are needed. The plot of the solenoid valve and its linear fit is seen in Figure As the area is the same for the solenoid valve, the measurements for both groups can be utilized to form the graph. The error tolerances are assumed to be the same absolute value. This means that the measuring error compared to the magnitude of the measurements can be very large, but as this graph has 8 data points, it is reasonable to assume that the fit on the graph would explain the relationship between flow and pressure for the solenoid valve.

66 58 CHAPTER 4. PROBLEM SOLVING 7.5 Flow characteristics of orifice A L2 in load 6 Flow through orifice [l/min] Solenoid valve GR522 and GR Pressure drop across orifice [Bar] Figure 4.32: Plot of steady state flow and pressure for the solenoid valve A L2 in the load 7.5 Flow characteristics of orifice A L1 in load Flow through orifice [l/min] Small opening GR522 test Small opening GR52 test Large opening GR522 test Large opening GR52 test Pressure drop across orifice [Bar] Figure 4.33: Plot of steady state flow and pressure for four different areas of the same variable orrifice A L1 in the load

67 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 59 The constants for the last orifice is the thoughest one to capture, as there is very little data for disposal. Only two data points for each orifice area besides zero. The linear fits are made so they fit zero and the data point at the highest pressure, and the fits actually allmost corresponds to the midpoint aswell, eventhough there is relative high errors in this domain. The slopes of all the orifices, based on the data that this group collected, are tabulated in Table 4.8. K L1 [ L B ar mi n ] K L2 [ L B ar mi n ] K L3 [ L B ar mi n Small Large Table 4.8: Flow constants of the load orifices. Asuming laminar flow. ] At this point all the necessary parameters of the load are known, so it is now possible to compare the test results with the results from the model Comparison The model are to be verified. This is done by comparing the results with those from the abovementioned test, but before this is possible it is necesary to define some uncertain parameters, which are to be alteret, based on the results from the comparison to the test results. The parameters and their initial values are defined in Table 4.9. Parameter Symbol Value Unit Flow rate of the pump Q v 17 L/min Supply voltage bat_vol 12 V Bulk modulus β 7 Bar Viscous Fric. PVP comp. c v p 9 Ns/m Viscous Fric. PVB comp. c vc 9 Ns/m Viscous Fric. Main. c vm 25 Ns/m Coulomb Fric. PVP comp. c cp N Coulomb Fric. PVB comp. c cc N Table 4.9: Uncertain parameters. Will be alteret to fit test results. Steady State Flow The first thing to do, will be to find out whether the model has the right steady state respons. To evaluate this, the results from the second part of the test described in section B.7 will be used. The model is provided with the same input voltage as measured in the test, and the steady state value of Q v is logged. The result is seen in Figure 4.34.

68 6 CHAPTER 4. PROBLEM SOLVING Steady State Qv Comparison Simulation Test Flow rate [L/min] Input voltage [V] Steady State Qv Comparison Simulation Test Flow rate [L/min] Spool travel [mm] Figure 4.34: Plot of steady state flow with varied input voltage, and corresponding calculated spool travel The two responses show the same tendencies, but the simulated result is higher in magnitude. The simulated graph clearly saturate at a point between the two inputs 4.45V and 4.21V. The test are showing the same, but it is not fully saturated until somewhere after the input has reached If the graph is assumed to follow a linear slope until it saturates, the lowest input value can be captured as the point where the line reaches the maximum value, this is also the flow rate of the pump, which is read to be 16.3L/mi n. A small zoom of the plot, illustrating the procedure, can be seen in Figure The minimum input value, before saturation, is determined to be U mi n = 4.19V. The actual flow will follow a quadratic slope but with such a steep slope at such a short range, it could be approximated by a straight line. By changing the flow rate of the pump, the system is changed, but it is assumed that it will only affect the maximum value of the two last datapoints, so there is no need to re-simulate all 7 situations. As the flow out of the valve is determined only by two parameters; the opening area of the main spool

69 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 61 and the pressure drop across it, the error must be found here. The pressure drop across the main spool is graphed in Figure Steady State Qv Comparison Simulation Test Flow rate [L/min] Input voltage [V] Figure 4.35: Procedure to determine minimum input voltage before saturation, and pump flow during the test 14 Steady State P m Comparison 12 1 Pressure [Bar] Voltage [V] Figure 4.36: Plot of steady state pressure drop, in the main spool, with varied input voltage The whole purpose of a proportional valve is the ability to control the flow out of the two ports pre-

70 62 CHAPTER 4. PROBLEM SOLVING cisely. This is achieved through implementation of the compensator spool in the PVG. It is seen from the graph that the spool can t keep the pressure drop constant when the system saturates, nor when the main spool blocks the flow. In all other situations the compensator is doing what it should; keeping the pressure drop constant. This is also supported by more simulations than presented, at this point, in the report. The only way that the pressure drop could lead to errors, is, if the pressure drop across the main spool was less than the 5B ar in the test. This value is mainly determined by the preload of the spring in the compensator spool, and the area characteristics of the A 1 opening of the compensator spool. Even though this can affect the flow, the change from 5bar to i.e. 4B ar, only changes the output 1.6% because when the flow is calculated it s the square root of the pressure that enters. So this can t be the only reason to the difference in flow. The reason have to be the area of the mainspool, either directly or by an error in the input voltage, which will have a great effect in the flow. If the measurements where.3v too low, the model actually imitate the flow allmost perfectly according to Figure The effect of the input voltage U on the spool travel, is also highly dependent on the supply voltage, as the actual input to the modulator is the ratio of the input voltage and the supply voltage, not the input voltage directly. If the area characteristics of the main spool corresponds to the flow characteristics from the data sheet, the difference in the flow, can only be explained by measurement errors and errors in the supply voltage. Area A H [mm 2 ] Comparison of the Area Characteristics of Main Spool LABTEST1 LABTEST2 AMOC DATA SHEET E DATA SHEET D DATA SHEET C Spool travel [mm] Figure 4.37: Area of main spool opening, based on data from data sheet, geometric measurements and labtests. The Figure 4.37 contains 6 plots, all describing the area A H as a function of spool travel x m. Three of them are based on flow and pressure data from the data sheet Danfoss [2], and two of them are based on data from the test in section B.7. All graphs are formed using the orifice equation and the assumption of a pressure drop of 5bar across the main spool. The last graph is made solely on

71 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 63 geometric measurements, using a vernier caliper, and the computer program AMOC Power [1999], and this is the one used in the model. Judging from the figure, there is no doubt that the spool of which the measurements where done, was a spool of type E. The data from the two tests doesn t fit another type of spool. Therefore it is reasonable to assume that the model calculates the flow correctly, and the reason why the tests doesn t correspond to the output from the model, are errors in the measurements and in the adjustment of the supply voltage. To further investigate the effect of a small change in the supply voltage the following calculations are made (see Equation 4.133). ( ) U E(U ) =.5 U assumed 14mm (.5V ( 1 E(U ) = 28 U assumed U U actual.5 1 U actual ) 14mm.5V ) U (4.133) E(U ) Steady state error in the position of the main spool. [mm] The number illustrates how much more the spool travel than it actually does. The sign convention follows that of xm. U assumed The assumed supply voltage [V ] U actual The actual supply voltage [V ] U The input voltage [V ] The effect of having a supply voltage of 11.5V instead of 12V is plottet in Figure The graph illustrates the error in spool travel, as a function of the input voltage. The error is larger at the higher signals, and smaller at the lower, an average error of about.42mm is present in the interval from 3.5V..5V, which corresponds to the plot which contains the comparison of the two steady states flows, see Figure The simulation is done 7 times again, and each time, the steady state value of the flow is tabulated. The result is shown in Figure E( U) Figure 4.38: The error in main spool travel x m by assuming 12V power supply, and actually having 11.5V U

72 64 CHAPTER 4. PROBLEM SOLVING Steady State Qv Comparison Simulation Test Flow rate [L/min] Input voltage [V] Steady State Qv Comparison Simulation Test Flow rate [L/min] Spool travel [mm] Figure 4.39: The steady state flow rate Q v from the model compared to the test results. The model is simulated at at supply voltage of 11.5V as this is assumed to be the actual voltage for test aswell. The primary error are considered to be the error in the supply voltage, as the data now completely corresponds to the original. The only change from Figure 4.34 to Figure 4.39 is the battery voltage, and the flow rate of the pump flow. In order to use the data to verify the rest of the model, the supply voltage of the model is kept at bat_vol = 11.5V, and the flow rate of the pump is changed to Q v = 16.3L/mi n. Before proceeding to test other parameters, the steady state flow of the first part of the test, described in section B.7, will be compared to the simulation. The result is seen in figure Figure 4.4. The data that is now plotted is different in the way that, the solenoid valve has been activated, either closing og opening, in all of the tests. A total of 8 steady state values are captured from both the simulation and the test. Each input voltage have corresponding flows. One for the opening of the solenoid valve, resulting in a steady state measurement of the flow, where the oil flows in both branches of the load, and a closing where the oil only flows in one of them, when the steady state measurement is done.

73 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL Steady State Qv Comparison Simulation Test Flow rate [L/min] Input voltage [V] Figure 4.4: The steady state flow of the first part of the test described in section B.7. Only the tests where the solenoid valve is activated are included in the graph, which make up a total of 8 test and 8 simulation points. If compared to the graph of the other data series (Figure 4.39) this graph looks a bit surprising. The values still lies inside the tolerances of the measurement equipment, but shows a larger difference in the datapoint at the top. This could mean that the valve is a bit more saturated in reality than the model assumes. The data plottet at the top point is for the test series numbered 2 and 3, where 2 is the upper one. Besides beeing about 1 l/min of, the data shows the same strange jump in flow. The flow is a bit higher when the solenoid valve is open than it is when it s closed, but only at this point. At the other points the flow is exactly the same, both in the test and in the simulation separately, which means that in steady state, a input signal gives the same output flow independent of the load. It could be interesting to analyse wheter the reason for the error at the top point is a result of saturation or simply error in measurements. It s most likely the former, because if the reson is the error in measurements it isn t very likely that the only place where the error should occur is at that point, all the other points, from both parts of the test, have much smaller errors. As the model describe the flow in the other points sufficient, the text will move on and consider the steady state pressures instead. Steady State Pressure The steady state pressure is only affected by the flow into the load, Q v, and the areas of the orifices in the load, or in this case the equivalent linear constants, K L1, K L2 and K L3, listed in Table 4.8. As the steady state flow is validated, and therefore fit the test data, the comparison of steady state pressure will investigate those constants. First the constant K L3 is considered and therefore only the tests where the solenoid valve is closed in the end of the test, are considered.

74 66 CHAPTER 4. PROBLEM SOLVING Test no. P(SI M)[B ar ] P(T EST )[B ar ] Q v (SI M)[L/mi n] Q v (T EST )[L/mi n] Table 4.1: Steady state pressure drop and flow through load orifice L 3 In Table 4.1 the simulated and the tested pressure drop over the orifice are tabulated, along with the corresponding flow rate. Only one data point for each input is available, that makes up a total of 4 datapoints to describe 2 different areas, which is 2 per area, there is only one data point to describe the small area of the orifice, because the flow isn t calculated correctly at one of the inputs. It would be possible to use the data from all the tests, but as it wouldn t change the result the time and effort are placed elsewhere. Test no. K L3 (SI M) [ ] L B ar mi n K L3 (T EST ) [ L B ar mi n Table 4.11: Flow constants of the load orifice L 3 ] The pressure drop from the simulation is allmost the same as the one from the test. It varies as much as 3bar at test no. 12, but as the flow in that point is equaly high, the values of K L3 are considered to be correct. The value of K L3 for each test and simulation are tabulated in Table 4.11, and calculated using Equation The most reliable values are the one where the error in flow is smallest, which will be test no. 9, and as there is only one value for the small orifice, the result from test no. 6 is used for comparison. And the values corresponds to the test values. K = Q P (4.134) The same procedure are utilized in the evaluation of the other two constants K L1 and K L2, this time of course other tests are used. The ones considered appears together with the pressure drop in Table Test no. P1(SI M)[B ar ] P1(T EST )[B ar ] P2(SI M)[B ar ] P2(T EST )[B ar ] Table 4.12: Steady state pressure drop and flow through load orifice L 1 and L 2

75 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 67 The pressure drop over the two orifices also look much alike. Of course such small measurements will contain errors, so even though there is a relative high error in P2, it doesn t mean that the constant is wrong in the model. Test no. K L1 (SI M) [ ] L B ar mi n K L1 (T EST ) [ L B ar mi n ] K L2 (SI M) [ L B ar mi n ] K L2 (T EST ) [ Table 4.13: Flow constants of the load orifices L 1 and L 2 L B ar mi n ] As for L 3 the constants are calculated and listed. They can be seen in Table Test no. 8 yields great results for both orifices, test no. 11 has an error in L 2, but that could be due to measurement errors, test no. 5 doesn t yield as good results as the others. As the values of K is relatively far from each other in some of the tests, it can be necessary to alter the values to fit that particular test series while other parameters are considered. The steady state response has now been analysed and documented, it is thus time to concentrate on the dynamics of the model. As a small summary, the supply voltage was changed from 12V to 11.5V and the pump flow was changed from 17l/mi n to 16.3l/mi n. The error in the pressure drops and the flows varies for the different tests, but are all tolerable. The steady state flow response for test no. 2 and 3 are too high when the solenoid valve is closed, hence also the pressure drop over L 3, this should be kept in mind when analysing the dynmaics. The result of test no. 2 and 3 will only be used as a help in determining the right parameters, not as a goal for the simulation to fit. Dynamics The analysis of the dynamics starts with a comparison of all the tests to the equivalent simulation. The compared values are the pressure drop across the load orifice L 3. The results are seen in Figure 4.41.

76 68 CHAPTER 4. PROBLEM SOLVING 8 Test number 2 8 Test number 3 Pressure [Bar] Pressure [Bar] Time [s] Test number Time [s] Test number 6 8 Pressure [Bar] Pressure [Bar] Time [s] Test number Time [s] Test number 9 8 Pressure [Bar] Pressure [Bar] Time [s] Test number Time [s] Test number 12 8 Pressure [Bar] Pressure [Bar] Time [s] Time [s] Figure 4.41: The simulated (GREEN) and tested (BLUE), uncorrected, pressure drop over load orrifce L 3, at 8 different settings The graphs have all been edited to fit the step time of the solenoid valve. The axes are the same in all of the plots in order to illustrate the absolute error. The tests in the left column are the tests where the solenoid valve opens, and the ones to the right are when it closes. The four plots in the top are for the small orifice openings, and the four at the bottom are for large orifice openings. All the graphs fits more or less in steady state, except the two in the top and the reason why is explained in section If the dynamics of the graphs are considered, they all seem to have a downward slope that is to steep compared to the test. The effect is seen very clearly in the graph of test no. 2. What this means is that when the solenoid valve opens, the pressure at the output of the valve (P l s ) drops too much before it is limitted, compared to how much it drops in the test. What happens is that the pressure drops at

77 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 69 the output, thus there also is a pressure drop at P k, P and P p aswell. When the pressure P p drops in the simulation, a direct drop in force will take place aswell, since the pressure in the spring chamber of the pump side spool is damped and the pressure in the pressure chamber is not. This results in a closing motion of the spool leading more of the contracted fluid to the load instead of the tank. This could infact be a good solution in reality, but the main purpose of the pump side spool is to make sure that the pressure P p is about 15 bar higher than P l s, and therefore it should in fact open to the tank when P l s drops. The reson why it doesn t is that the small orifice in the pump side spool has been neglected. The orifice and the pressure build up in the pressure chamber is inserted in the model, the same way as it was for the compensator spool, and the simulations are made again. The orifice is assumed to be A 7 =.3mm in diameter. Figure 4.42 shows the result. 8 Test number 2 8 Test number Pressure [Bar] Pressure [Bar] Time [s] Time [s] 8 Test number 5 8 Test number Pressure [Bar] Pressure [Bar] Time [s] Time [s] Figure 4.42: The simulated (GREEN) and tested (BLUE), pressure drop over load orrifce L 3. model has been changed to take care of the pressure build up in the pressure chamber The The figure now shows the right motion in test no. 2, and the velocity of the main spool are opening as it should. But the motion is much more oscilatory. It is especially clear in test no. 5 and 6. As the coulomb friction were neglected at the first approach, this is now introduced. When one look at damping in linear systems, the constant added friction of a coulomb friction, doesn t add more damping, but in reality it will, since it will pull energy out of the motion that is not restored. The Figure 4.43 shows the same simulations, but this time with an added coulomb friction of 1N for

78 7 CHAPTER 4. PROBLEM SOLVING each compensator spool. 8 Test number 2 8 Test number Pressure [Bar] Pressure [Bar] Time [s] Time [s] 8 Test number 5 8 Test number Pressure [Bar] Pressure [Bar] Time [s] Time [s] Figure 4.43: The simulated (GREEN) and tested (BLUE), pressure drop over load orrifce L 3. The model has been changed to take care of the pressure build up in the pressure chamber and the coulomb friction. Now the system behaves more like it should, at least the oscillatory motion that was too big is gone, but the dynamics of the system still isn t satisfying. The steady state error at the two top graphs have increased, and the frequency of which test 5 and 6 oscilliates is also wrong. These coulomb frictions therefore have an effect on these parameters. As the system is very unlinear, it is hard to say exactly what a change in a parameter will result in. Therefore some linearizations is done and some methodical trial and error. This will help in the understanding of the system. First the coulomb friction is tuned. As the steady state error have gotten bigger and the dominant natural frequency of the system also have increased, the tuning have been focused on lowering the coulomb friction until the system gets unstable, and then increasing it again, in order to decrease the coulomb friction as much as possible. Both the coulomb friction of the pump side spool and the PVB spool have been considered, and the system is most sensitive to the coulomb friction of the PVB spool, so this one is kept at 5N, and the coulomb friction of the pump side spool is removed again. The process also involved the viscous friction. It was considered if the coulomb friction would be unnecessary if the viscous friction was

79 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 71 just high enough. That isn t the case, the coulomb friction is mandatory, at least for the compensator spool, in order to avoid the oscilation at the current areas of the different damping orifices. The orifice in the newly installed orifice in the pump side compensator was also changed.2mm in both directions, resulting in worse results in either cases. Changing the viscous friction of the two spools results in a strange behaviour compared to what the transfer function of the system predicts. The various transfer functions derivation appear in section C.1. The change in the viscous friction, changes the natural frequency of the response. The viscous friction doesn t directly affect the natural frequency of any of the spools according to the linear transfer functions. But the change in damping will change the filter effect of the spools, so that it could amplify a broader or a more narrow spectrum. Maybe the frequency of another system like the pressure build up frequency of P p and P l s will be amplified, at least this is the only way the linear model can explain the phenomenom the unlinear model shows. The effect is seen in Figure Test number 6 c vk =1 c vk =5 6 Pressure [Bar] Time [s] Figure 4.44: Two simulations of the system. Both displaying the pressure drop across the load orifice, but with two different viscous friction coefficients. Raising the viscous friction, changes the natural frequency, and lowering the coulomb friction yields a faster pressure build up. Which is seen in test no. 6 particullary. The optimized result is seen in Figure 4.45.

80 72 CHAPTER 4. PROBLEM SOLVING 8 Test number 2 8 Test number Pressure [Bar] Pressure [Bar] Time [s] Time [s] 8 Test number 5 8 Test number Pressure [Bar] Pressure [Bar] Time [s] Time [s] Figure 4.45: The simulated (GREEN) and tested (BLUE), pressure drop across load orrifce L 3. The model has been changed to take care of the pressure build up in the pressure chamber and the coulomb friction. The viscous friction is also changed in order to optimize the result. The tests; 5 and 6, now fits quite well. The slope in test 5 is still a bit to steep, but it looks like the frequency fits. The test number 3 also looks okay. Keep in mind that the steady state flow and therefore also the pressure will be higher in the simulation than in the reality (see section 4.7.3). If they where to hit the same pressure level, it looks like they could have the same settling time. Test number two also looks good, except for the drop in pressure where the test actually shows the opposite. The spool travel of the pump side spool (see Figure 4.46), reveals the reason. Because of the deadband of the pump side area, it will first open to tank when it reaches 2mm. When the unwanted pressure drop occur, the spool has just reached the 2mm point, and is moving downward again as the pressure drops. But because it has some stored kinetic energy, it will take some time. This effect is not seen when the system isn t saturated, because in that case the spool will not have to travel that far before it get s damped by the falling pressure, hence also containing less kinetic

81 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL 73 energy and therefore stops faster. If the damping was changed, it would be possible to make it fit the curve of the test, but it will ruin the dynamics of the other tests. The reason is, that it is supposed to hit the 2mm point a bit faster, thus lowering the pressure drop just like the test result. As there is saturation involved this is just noticed and not changed, because the test and the simulation isn t exposed to the same flow Q v and therefore the other tests are considered to be more accurate and therefore also more relevant. To prove that the lowering of the deadband will result in an elimination of the bump, two simulations are made, and the result are given in Figure x 1-3 Time Series Plot of x_p Spool travel x p Time (seconds) Figure 4.46: The spool travel of pump side spool during test no Test number 2 A 3.5mm deadband A 3 2 mm deadband 6 Pressure [Bar] Time [s] Figure 4.47: Two simulations, describing what a change in the deadband of the pump spool will look like in test number two. All the graphs still shows very steep drop in pressure at the time of the step. This could be do to the fact that the model assumes a laminar flow in the orifices, and maybe the flow at that point is turbulent. The result have also gotten more oscillatory and therefore the settling time has increased, and this is the reason for the inclination on the slope at the beginning of test no. 2 and 4. The rest of the tests are plottet using the new found friction values c v p = 35N s/m, c vc = 2N s/m, c cp = N and c cc = 5N, the result is seen in Figure 4.48.

82 74 CHAPTER 4. PROBLEM SOLVING Pressure [Bar] Pressure [Bar] Pressure [Bar] Pressure [Bar] Test number Time [s] Test number Time [s] Test number Time [s] Test number Time [s] Pressure [Bar] Pressure [Bar] Pressure [Bar] Pressure [Bar] Test number Time [s] Test number Time [s] Test number Time [s] Test number Time [s] Figure 4.48: The simulated (GREEN) and tested (BLUE), pressure drop over load orrifice L 3. All 8 tests compared with the simulation from a model, with updated frictions including the orifice in the pump spool. Parameter Symbol Value Unit Flow rate of the pump Q v 16.3 L/min Supply voltage bat_vol 11.5 V Viscous Fric. PVP comp. c v p 35 Ns/m Viscous Fric. PVB comp. c vc 2 Ns/m Viscous Fric. Main. c vm 25 Ns/m Coulomb Fric. PVP comp. c cp N Coulomb Fric. PVB comp. c cc 5 N Diameter orifice PVP comp. D 7.3 mm Table 4.14: Alteret parameters of the model

83 4.7. VALIDATION OF UNLINEAR ADVANCED MODEL Conclusion The model could be optimized in many ways if there were put some more time into it, but as for now the model describes the reality sufficient, so that it can be used as a tool to understand and optimize the valve. Therefore the time and effort will be spent on this instead. The final result using the parameters listed in Table 4.14 is shown in Figure 4.48, and it is seen that the model fits all the tests but 2. It fits test 8 and 9 particullary well. To show that the model fullfill the requirement from the problem statement, stating; no more than 3% relative error, and 1% will be considered as a good result, the relative error of each test are plottet in Figure Test number 2 3 Test number Relative error % Relative error % Relative error % Relative error % Relative error % Time [s] Test number Time [s] Time [s] Test number 8 Test number Time [s] Relative error % Relative error % Relative error % Time [s] Test number Time [s] Test number Time [s] Test number Time [s] Figure 4.49: The error of the 8 different simulations relative to the corresponding tests.

84 76 CHAPTER 4. PROBLEM SOLVING The data are smooth using moving average filter, in order to remove the little peak in error, due to the small misalignment at the time where the step occur. 4.8 Linear model To get a deeper understanding between the parameters interacting when trying to control the system, a linear model will be developed. This model will lack some accuracy compared to the unlinear model, but the simplicity makes it easier to tune a controller, and also helps in the understanding of the system. Therefore it is evaluated to be within the boarders of the problem statement. Because the linear model is an approximation around a given point of work, the model can t give better results than the accuracy of the linearizations Initial thoughts and assumptions The unlinear, advanced model is validated in the previous section. This is therefore a usefull tool when considering simplifications and assumptions when designing the linear model. The compensator spool is designet to keep a constant pressure drop over the main spool. In order to reveal this, the advanced model is simulated. There is a step in the simulation in the main spool position at time t =.1s and a step in the external load at t =.8s. The plot seen in Figure 4.5 shows the same characteristic for the load pressure P l s, the pressure before the main spool P k and the pump pressure P p in the volume V p. This must therefore result in a constant pressure drop over both the compensator spool and the main spool. 1 x Pressure [Pa] P p P k P ls Time t [s] Figure 4.5: The pressure measured at the load P l s, before the main spool P k and at the pump side volume P p This observation makes it possible to model the pressure drop as a constant given in Equation 4.135, making the flow rate controlable by the position of the main spool. K Pkl s = P k P l s (4.135) As the linear model will involve a force equilibrium of the main spool, the flowforces are investigated, to see whether they can be ignored. To assist in this decision the unlinear model are utilized. All the

85 4.8. LINEAR MODEL 77 simulated flowforces of the valve are illustrated along with the force due to pressure, of the pump side spool, in Figure As the flow forces are much smaller than the pressure force, they are all neglectable. The reason is the small flow demand in all the tests. If the valve was exposed to a higher flow, the flow force will increase quadratic. The different functions in the model has to be linearized in order to derive a transfer function. The first order taylor series is used to expand the functions about a working point. Depending on the order of the functions, in worst case scenario, merely a small interval around the working point x is represented by the linear model. The linearization of the PVE module follows..2.1 Multiple Time Series A2 Flow force A1 flow force AH Flow force 2 1 A3 flow force 1 5 Force due to Pump pressure Time (seconds) Figure 4.51: The simulated flowforces in the valve during a simulation with parameters equivalent to test number 2 explained in section B The PVE module The model of the PVE module is derived in section 4.5. The block diagram in figure 4.52 illustrates the starting point of the PVE module. The input signal is transfered to a position and the feedback signal, through the LVDT is subtracted. The difference between these positions gives the error in position.

86 78 CHAPTER 4. PROBLEM SOLVING V rel X m X set V rel + - LVDT X error Figure 4.52: Block diagram for the starting point of the PVE module, from an input signal to an error in position. The block diagram illustrates how the error signal is calculated (see equation 4.136). x er r or = x set x m (4.136) The input signal to a defined position x set is calculated in Equation using taylor expansion. x set x + x set V r el (V r el V ) = x + K i nput (V r el V ) (4.137) V The working point is defined as V. From Equation it is seen that there is no dynamics in this part of the model, hence only a constant gain will be the transfer function. x set x = K i nput (V r el V ) x set = K i nput V r el x set V r el = K i nput (4.138) The LVDT transfer function is of first order, and has dynamics represented by a time constant τ l vd t. Furthermore there is a steady state gain between the actual position x m and the measured x m which is represented by K l vd t. The LVDT sensor model is derived in Equation and repeated in equation x m x m = K l vd t τ l vd t s + 1 = G l vd t (4.139) To compensate for the error in position x er r or, the solenoid valves will try to build up pressure and move the spool in the correct direction. Ultimately, the NC and NO valves will open for fluid flow as seen in Figure X err DT NC1 X err DT NO2 X err QNC1 DT NC1 QNO2 DT NO2 + - Port B system Figure 4.53: Block diagram from the error in position to a flow through one port chamber at the PVE module.

87 4.8. LINEAR MODEL 79 To reach a given flow, first the error in position is transformed into a duty cycle, which are different for each port as seen in Figure The first order transfer function describing this proces is given in equation Apparent from previous calculations, a taylor expansion yields the result given in 4.14 for port B and in for port A. DT 5 x er r or = K DT 5 (4.14) DT 4 x er r or = K DT 4 (4.141) The saturation of the duty cycles in the unlinear model are not implemented, in these linear equations, hence the duty cycle for both ports will be the same with opposite signs, leading to same fluid flow to both chambers. To avoid this, the linear model will focus on one equivalent port obliging this problem. When the duty cycle is negativ, the pressure in the chamber will be negative, and hereby modelling a negative pressure drop as seen illustrated in Figure The transfer function for the duty cycles will hereby change to the ones given in equation P pilot DT 4 x er r or = K DT (4.142) NC1 Q supply V upper P upper NC3 The pressure drop over the NO and NC valves will change as the pressure inside the chamber increase and decrease as a resultport of the B variations in the flow as seen in figure Initially, whenpthe 5 chambers starts filling up, the pressure over a NC valve can reach the pilot V pressure. The linearization of the flow through the NC valve will 5 be calculated based on a pressure drop at P pi l ot, while NO2for the NO valve will be calculated based on pressure less than half than P pi l ot. The b tank side valves needs to close before the pilot side valves open in order to avoid shortcut. Here the linear model falls short in implementing the time constants. Implementing this in the linear lower V model a first order system with a time constant for both P valves are introduced, as seen in equation P lower tank A checkball Q drain Port A V 4 a P 4 NO4 Q NC = K NC DT τ NOC s+1 (4.143) NC Q NO = K NO DT τ NOC s+1 (4.144) NO Figure 4.54: The figure illustrates which part of the PVE modules hydraulic that are modelled. The pilot pressure is previously defined to be P pi l ot = 13.5B ar, hence the linearization of the flow as a function of the dutycyle, will be based on a pressure drop over the NC valve at P NC = 13.5B ar and for the NO valve P NO = 4B ar. A linear approximation of these graphs are given in figure P upper

88 8 CHAPTER 4. PROBLEM SOLVING Flow Q in Q out [m 3 /s] 3 x Bar 13.5 Bar Linfit 13.5 Linfit 2 y = 3e 7*x 3.7e 6 y = 1.1e 7*x 2.4e 6 Flow Q in Q out [m 3 /s] 2 x 1 6 y = 1.8e 7*x + 1.6e y = 1.7e 7*x + 1.3e 5 2 Bar Linfit2 4 Bar Linfit Duty cycle DT NC [%] Duty cycle DT NO [%] Figure 4.55: Linear fit of the flow through an NC valve with a pressure drop of 13.5B ar and a NO valve at 6B ar As the pressure increases in the the port, the pressure drop over the upper valve, NC, will decrease, while the lower valve NO will experience a higher pressure drop. This will effect the flow through each valve. Because a transfer function with a single input is wanted, the change in pressure can only be implemented using feedback. The transfer fucntion and the linearization is calculated for a duty cycle at D = 7% in Figure 4.22 for both valves. These linearizations are given in figure 4.56 with equations. 2 x y = 1.2e 11*x + 1.5e Q (P B ) for DT=7% Linfit 12 x 1 6 y = 8.3e 12*x 1.9e 6 1 Q(P b ) for D=8% Linfit Q NC [m 3 /s] Q NO [m 3 /s] Pressure P x 1 5 B [Pa] Pressure P B [Pa] x 1 5 Figure 4.56: Linearization of the change in flow as a function of the pressure inside the chamber P 4 for the NC valve in the left and the NO valve in the right side graph This transfer function is given in equation with the gain as the slope of the graphs. dq NC P 4 = K NC QP (4.145) dq NO P 4 = K NO QP (4.146) The flows through each valve, including the change in flow, will be summarized with the direction of flow in mind. The reason why the change in flow for the NO valve is added, is because a higher pressure inside the equivalent chamber results in a higher pressure drop over the NO valve which results in a higher flow rate. ( Q 4 = Q NC dq ) ( NC Q NO + dq ) NO P 4 P 4

89 4.8. LINEAR MODEL 81 The flows yields a pressure build up in P 4. The block diagram showing the step from flow through the valves to a pressure difference over the main spool are seen in figure Q 5 Q 4 Port B P 5Q5 Port A P 4Q4 + - Spool System Actual spool position Figure 4.57: Block diagram from the flow through each valve to a pressure build up in the chamber The continuity equations describing the pressure build up inside the chamber are given in equation 4.13 for the equivalent port. Notice that the change in volume are neglected, and only the combined flow through the NO and NC valves are kept in the calculations, because the flow through the checkball valve is assumed to be very small. dp 4 d t = β V 4 Q 4 s P 4 (s) = β V 4 Q 4 (s) P 4 Q 4 = β s V 4 (4.147) Internal feedback implementation At this point the linearization falls short. The pressure inside the chamber P 4 will keep rising as the flow keeps running in, and eventually P 4 will be greater than the pilot pressure. To regulate the flow as a result of the change in P 4, a feedback will be implemented. Since the duty cycle controls the flow through each valve, the duty cycle needs to be adjusted as a result of a higher pressure. Ultimately, when P 4 is equal to the pilot pressure the flow should stop from flowing through the NC valve, by adjusting the duty cycle to zero. Equation states the transfer function for the feedback. DT p d t dp 4 = 1 P pi l ot P r eser voi r = K PDT (4.148) The implementations made to compensate for changes in flow as a result of changed pressure are illustrated in figure 4.58.

90 82 CHAPTER 4. PROBLEM SOLVING Q NC P B x error DT B x error + - Q NC DT B P B - Q P B Q NC DT B + + Q NO P B DT B P b Figure 4.58: Block diagram from the error in position to a pressure drop between main spool The block from an error in position to the pressures in the chambers in each port, will be calculated in the following equations for the pressure in the equivalent chamber. First new constant are introduced to ease the calculations: K LX P1 = K NC τ NOC s + 1 K LX P2 = K NC QP K LX P3 = β V 4 s K LX P4 = K NO QP K LX P5 = K NO τ NOC s + 1 Using the new constans, the closed loop feedback can be reduced for the pressure as a function of the error in position given in Equation with a new constant introduced to reduce the term for succeeding calculations. P 4 = (x er r or K LX P1 K LX P2 P 4 ) K LX P3 (x er r or K LX P5 + K LX P4 P 4 ) K LX P3 x er r or K LX P3 (K LX P1 K LX P5 ) = P 4 (1 + K LX P2 K LX P3 K LX P4 K LX P3 ) P 4 K LX P3 K LX P1 K LX P3 K LX P5 = x er r or 1 + K LX P2 K LX P3 K LX P4 K LX P3 P 4 = K Peq (4.149) x er r or Multiplying the pressure drop over the main spool with the area of the main spool A m yields a force which are used to move the main spool. In order to compensate for other forces that has an impact on the main spool dynamic behaviour, these forces needs to be examined.

91 4.8. LINEAR MODEL Dynamics of main spool The model of the main spool was derived in a previous section and is stated in equation The input parameters to this model has been linearized in the previous sections. A laplace transform will be applied to derive a linear model of the main spool, with initial conditions assumed to be zero: ) x m (s) (s 2 M m + k m + s c vm s 2 x m (s) M m = s c vm x m (s) k m x m (s) + K Peq x er r (s) A m s 2 x m (s) M m = x m (s) ( k m s c vm ) + x er r r (s) K Peq A m = x er r (s) K Peq A m 1 x m (s) G spool = x er r or (s) G pve The superior transfer function is shortened into smaller transfer function in parts that can be related to the psysical system; The spool, the pve module and the LVDT sensor which are derived in Equation These smaller transfer functions are listed in Equation G spool = 1 M m + k m + s c vm (4.15) G pve = K Peq A m (4.151) G l vd t = K l vd t τ l vd t s + 1 (4.152) Inserting equation 4.138, and in the final expression for the transfer function for the pve module and the main spool yields a closed loop given in equation x m (s) x set (s) = G pve G spool 1 +G pve G spool (4.153) Closed loop control system The closed loop transfer function of the system is now ready to be derived. The input signal is derived in equation and the spool position is measured by the LVDT sensor which was derived in equation The error in spool position is calculated as the difference between the desired and the measured position. Figure 4.59 illustrates the closed loop system with a plant G pl ant to be controlled, which symbolises the linear model of the PVE module and the main spool. The controller is named G c and will be adjusted according to the response and the defined requirements of the controller. X set x set + - x error V rel G c G plant x m' x m LVDT sensor Figure 4.59: The model of the PVE module and the main spool in a closed loop control system named the plant G pl ant

92 84 CHAPTER 4. PROBLEM SOLVING The coefficients for the linear model are calculated based on the values introduced through the sections describing the linerization and the measured constants for the PVG 32. Using these coefficients, the transfer function for each individual part of the linear model can be derived as stated in Table 4.15 G spool = G pve = = 1 M m s 2 +c vm s+k m s s s s+3695 = K Peq A m = K LX P3 (K LX P1 K LX P5 ) 1+K LX ( P3 (K LX P2 K LX P4 ) β K NC V 4 s τ NOC s+1 K NO = = = τ NOC s+1 1+ β V 4 s (K NC QP K NO QP) ( ) s.3 s s s ( ) (s )(s ) G i nput = K i nput =.28 G l vd t K l vd t τ l vd t s+1 = s+1 Table 4.15: How to calculate the coefficients in the linear model of the main spool ) Using these coefficients gives an unstable step response. Figure 4.6 shows a plot of the closed loop transfer functions poles and zeros Pole Zero Map Imaginary Axis Real Axis Figure 4.6: A plot of the poles and zeros in the closed loop transfer function G cl One conjugated pole pair in the right hand plane, is the reason why the transfer function is unstable. The first parameter to analyse is the damping ratio ζ. For a second order linear system the general transfer function is given by: C (s) R(s) = K ω 2 n s 2 + 2ζω n s + ω 2 (4.154) n Basically there are two systems that has an effect on the response of the transfer function, respectively

93 4.8. LINEAR MODEL 85 G pve and G spool. Expanding these functions for s, the parameters determining the damping will be revealed. The damping ratio depends on the coefficients multiplied with the s term. G pve = = β(k NC + K NO ) (τ NOC s + 1) ( V 4 s + K NC QP β K NO QP β ) (4.155) β (K NC + K NO ) s 2 τ NOC V 4 + s (τ NOC K NC QP β τ NOC K NO QP β +V 4 ) + KNC QP β K NO QP β (4.156) For the PVE module the expanded equation shows that the damping ratio depends on the bulk modulus β, the volume V 4 and the transfer functions regulating the flow as a function of the pressure build up (K NC QP and K NO QP ). Adjusting all the parameters in an reasonable range in order to make the damping ratio larger, does seems to move the poles in the right half plane closer to the origine, but no further. Equation 4.15 is allready in an expanded form, and the equation reveals that only the viscous friction coefficient has an effect on the damping ratio. Enlarging this coefficient from c vm = 25 to c vm = 25 is the crucial factor for damping the transfer function as seen in Figure Pole Zero Map Imaginary Axis e+3 1.2e+3 1 1e Real Axis Figure 4.61: A plot of the poles and zeros for the transfer function G cl after the damping ratio is adjusted for the spool To compare the step response of the linear model and the unlinear model, a plot of both models are made using a step input of x set = 3.5mm. The step resonses are shown in figure 4.62

94 86 CHAPTER 4. PROBLEM SOLVING x 1 3 Unlinear model Linear model 3.5 Step response Time [s] Figure 4.62: The figure shows the steprespons of the linear and the unlinear model The dynamics of the linear model seems to match quite well, and the steady state error are not far from the unlinear. One remarkable fact, is the difference in the rise time. This is because of saturations in the unlinear model, especially in the duty cycle, the pressure inside the ports are limited to a range between the pilot pressure P pi l ot and the tank pressure P t ank. In addition, a more accurate implementation of the flow in and out of Port A and Port B chambers in the unlinear model, does also play an important role in deviation between the linear and unlinear model. Even though the linear model contains a simple implementation of the change in flow, the pressure in the chambers still exceeds the limits, resulting in faster step respons time. As for the unlinear model the model can also be optimized and tuned further, but as a tool for understanding the valve, it is suffficient as it is. 4.9 Controlling the valve This section will show a small analysis of the system using the linear model and frequency analysis. It is meant as a small demonstration of how the models can be used in controller designs. The step response for the linear system in Figure 4.62 shows some overshoot O, a steady state error e ss and a rise time t r. These quantities are measured to be: RiseTime: t r =.79s SettlingTime: t s =.619s Overshoot: O = % Steady state error: e ss = lim s s G ol (s) = 4% The rise of the system is very small, causing a somehow large overshoot. The purpose of the controller could be to reduce the overshoot by increasing the damping and increasing the rise time. The tools used to optimize a step response are the bodeplot and the rootlocus given in respectively Figure 4.63 and Figure 4.65.

95 4.9. CONTROLLING THE VALVE 87 5 Bode Diagram Gm = 9.91 db (at 214 rad/sec), Pm = 49.7 deg (at 87.5 rad/sec) Magnitude (db) Phase (deg) Frequency (rad/sec) Figure 4.63: Bodeplot of the linear system The bodeplot shows the following gain and phase margins: At ω g = 214r ad/s the gain margin is G m = 9.91dB At ω p = 87.5r ad/s the phase margin is P m = 49.7 The PVE module already contains a closed loop controller, that controls the position of the main spool through a LVDT sensor. The data from the bodeplot shows that the controller are very well tuned and only a small margin is left for optimization. A reduction in proportional gain will cause both the phase and gain margin to increase and result in a more stable system with more damping on behalf of how fast the system behaves as seen in figure Figure 4.64.

96 88 CHAPTER 4. PROBLEM SOLVING 1.4 Step Response 1.2 G cl 1 Amplitude.8.6 G cl-damped Time (sec) Figure 4.64: Step response with a proportional controller G p =.3 (green) and step respons of the linear system As seen in the Figure 4.65, a larger gain will cause the complex conjugated poles to move to the right half plane and make the system unstable. Because of this, the system is called conditional stable. Using a P controller will change the amplitude characteristics also changing the bandwidth BW of the system. As seen in equation the rise time t r will increase when the bandwith is reduced. Consequently, this will also have an impact on the settling time. t r = K BW (4.157)

97 4.9. CONTROLLING THE VALVE Root Locus 1 5 Imaginary Axis Real Axis Figure 4.65: Root locus of the linear system After analysing the system with the two methods, a controller could be designed. To validate the controller the unlinear model could be used, as this takes saturation and other unlinearities into account. This small section shows that it is possible to use the models as a tool in controller design, utilizing bodeplot and root locus design methods.

98

99 Chapter 5 Conclusion How to model and control a Sauer Danfoss PVG 32 valve? The purpose of this project has been to thoroughly understand the principles and dynamics of the Sauer-Danfoss PVG 32 proportional valve. To accommodate this, the following items were listed as goals in the problem statement: Make a dynamical mathematical description of the valve. When developing the mathematical model, a deep understanding of the equations used to describe hydraulic systems where achieved. In order to determine the quantities used in the equations, it was necessary to dismantle the PVG 32 valve. Several lab tests have been performed to find the precise physical properties of these parts. Create a simulink model which simulates the real nature of a PVG valve. The mathematical model was modelled in Matlab Simulink, in order to solve the equations in a clear and structured manner. The system was recognized as a stiff system with different states, therefore a suitable solver utilizing variable time step was found. Perform tests to use for validation of the model. A test arrangement was established in the laboratory. Several measurements of different load situations were measured. It was posible to log the pressures in the test rig with Labview and the flows were manually read out. The process of making the test, gave a better relationsip to how the valve actually acts, and what physical apperance the different modules of the hydraulic diagram looked like in reality. Validation of the dynamical model. The tests were analysed and because of a good understanding of the characteristics describing the two flowtypes, the load orifices were recognized as laminar. Also, the supply voltage of the test seemed lower than informed, which also shows good knowledge of how the flow is affected by the supply voltage. Upon simulating the different tests in the model, the uncertain parameters were tuned so the dynamics of the simulation were describing the reality more accurate. In order to figure out the right parameters to tune, and how to do it, a methodical trial and error approach was used, along with derivations of linear models of interesting subsystems. At last a comparison of the model to all the 91

100 92 CHAPTER 5. CONCLUSION tests were caried out, and the relative error was all less than 3% as required, and the relative error of two of the tests showed particularry good results, which where under 1% in all of the scope, and less than 5 % in most of it. This was considered a tolerable error due to measurement errors, and unmeasureable quantities. Simplify the advanced dynamical model to a linear model. The part of the dynamic unlinear model, concerning control of the main spool position was simplified to a linear model which made it easier to determine the various parameters affecting the dynamics of the valve. To be able to linearize the model, a point of operation was determined, and certain forces and dynamics were neglected, based upon information recieved from simulations using the unlinear validated model. Understand how to use the developed tools to optimize the valve. The last section of the report showed a small demonstration of how the bodeplot and root locus design methods could be applied to the linear model, after a brief discription of step response design criterias.

101 Chapter 6 Future work When modelling the PVG 32, some parts of the valve was neglected or considered constant. This was done to simplify and make the model faster, but if a more detailed simulation is needed, functions could be derived for; oil density ρ, Bulk modulus β or the flow angles φ. A better controller could be developed to control the main valve, and an external system, like a crane. This could be done using the two validated models. Also if more tests where executed it would be possible to better determine the parameters of the model, and to test whether the model of the turbulent load is also valid. 93

102

103 Bibliography Andersen, Torben Ole, & Hansen, Michael Rygaard. 27a (June). Fluid Power Circuits - System Design and Analysis. Tech. rept. 3.ed. Institute of Energy Technology and Institute of Mechanical Engineering at Aalborg University. Andersen, Torben Ole, & Hansen, Michael Rygaard. 27b (June). Fluid Power Circuits - System Design and Analysis (section The flow force equation). Tech. rept. 3.ed. Institute of Energy Technology and Institute of Mechanical Engineering at Aalborg University. Danfoss. Simulink model of the PVE module. CD/model/PVE29/PVE_CM_19119.mdl. Danfoss. Tryktransmitter til industriel anvendelse Type MBS 32 og MBS 33. Danfoss, Sauer. 2. Technical Information - PVE Series 4 for PVG 32, PVG 1 and PVG 12. Danfoss, Sauer. 26. PVG32 product brochure. june. Parker. 2. Measurement technology for flow volume in hydraulics. Pedersen, Henrik Clemmensen. 29. Conversations during the project period. Power, Danfoss Fluid AMOC (Adaptive Orifice Configuration) Version 1. Beta. Wikipedia. Control Theory. Wikipedia, the free encyclopedia. Linear variable differential transformer. Wikipedia, the free encyclopedia. Transformer. B1

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105 Part I Appendix B3

106

107 Appendix A Govering Equations A.1 Orifice equation In hydraulics, orifices are used to control the flow, This is equivalent to using resistors in electrical circuits. In electronics, the relationship between voltage and currrent is given by Ohm s law. In hydraulics, the relationship between pressure and flow is given by the orifice equation (Equation A.1). Almost every valve is designed to follow this equation. However, this is not always the main design criteria. P A 1 Q P 2 Figure A.1: Turbulent flow through an orifice. The discharge coeffient is a function of the Reynolds number, which again is a function of the velocity, vicosity, density and hydraulic diameter of the orifice. The Reynolds number is proportional to the velocity and the hydraulic diameter of the pipe, hence flows with low velocity and a small orifice opening will have a small Reynolds number. The viscosity changes highly with temperature and pressure. A plot of the viscosity for commonly used mineral oils are shown in Figure A.2. B5

108 B6 ν kinematic viscosity, [cst] C ν,m ν constants C for the specific fluid va 2 2 t a Q absolute temperature, (p [K] 1 p 2 ) 2 1 (A / A ) ρ 2 1 This dependency is normally shown in specially designed APPENDIX charts, A. where GOVERING the kinematic EQUATIONS viscosity shown as function of the temperature becomes a straight line, see Figure A4. C d Defining the discharge coefficient ν [cst] orifice flow by the orifice area. 5 = (1.26) in Equation 1.26 it is possible to express the C C = (1.27) 2 1 v c Cd Cc (A / A1) Q CdA (p1 p 2 ) ρ 12 ISO VG 1 1 ISO VG 68 8 A 7 ISO VG 46 A ISO VG 32 5 ISO VG ISO VG t [ C] Now, combining Equation 1.22, 1.26, and 1.27 the orifice equation (in Danish blændeformlen) can be written = (1.28) Normally is much smaller than and since C v, the discharge coefficient is approximately equal to the contraction coefficient. Different theoretical and experimental investigations has shown that a discharge coefficient of C d.6 is often assumed for all spool orifices. At low temperatures, Fig. A4 Uddebuhle-chart: low orifice The temperature pressure dependency drop, for and/or some of small the most orifice commonly openings, used the Figure A.2: Uddebuhle-chart: Viscosity of commonly used mineral oils. The ISO VG standard refers ν Reynolds number may become mineral sufficiently oils. The ISO VG low standard to refers permit ν at 4 C laminar flow. Although the at 4 C. Andersen & Hansen [27a] analysis leading to Equation 1.28 is not valid at low Reynolds numbers, it is often used The vertical axis of an Uddebuhle chart is a mapping of log log(ν+.8), i.e., anyway by letting the discharge coefficient be a function of Reynolds number. For Re < approximately a double logarithmic axis (especially at higher values of ν). The 1 experimental The orifice horizontal equation results axis only is show a theoretically mapping that of the logt, valid discharge for i.e., non-laminar a logarithmic coefficient flows axis. (flows A is hydraulic with directly high fluid Reynolds proportional is, in number), but general, when the referred discharge to by coefficient its kinematic is viscosity made dependant at 4 C. on the Reynolds number, it can be em- to the square root of Reynolds number; that C ployed on any kind of flow, provided that the characteristic d = δ Re. A typical plot of such a result of the discharge coefficient is available. A is shown typical in plot Figure of C d 1.2. versus Re is given in Figure A.3..8 Hydraulic Fluids Page 5 of 11 Cd Re Figure 1.2 Plot of a discharge coefficient versus Reynolds number for an orifice Figure A.3: Discharge coefficient as a function of the squareroot of the Reynolds number. Andersen & Hansen [27a] Experiments have shown that when the reynolds number is below Re = 1 a linear relationship exist Hydraulic Principles Page 19 of 21

109 A.2. THE ORIFICE EQUATION IN MODELS B7 between the reynoldsnumber and the discharge factor C d = δ Re. Q = C d A 2 ρ (P i n P out ) (A.1) Q A Flow out of orifice Area of the orifice [ m 3 s ] [ m 2 ] C d Flow factor (often assumed to be.6) [ ] [ ] kg ρ Density of the fluid m 3 P i n Up stream pressure. The pressure before the orifice. [Pa] P out Down stream pressure. The pressure after the orifice [Pa] The orifice equation states that the flow through an orifice is proportional to the square root of the pressure drop across it, hence the pressure drop across an orifice is proportional to the flow squared. The relationchip is k eq which is a equivalent constant defined in Equation A.3. p = k eq Q 2 (A.2) k eq = ρ 2 1 Cd A (A.3) Q k eq Flow out of orifice Equivalent discharge factor [ m 3 s ] [ kg m 3 ] p Pressuredrop across the orifice [Pa] Area of the orifice [ m 2 ] A C d Flow factor (often assumed to be.6) [ ] [ ] kg ρ Density of the fluid m 3 If the manufacturer of a valve provides a pressure vs. flow curve, it is possible to obtain k eq. A.2 The orifice equation in models For fluid flow, an orifice works in both directions as long as the direction of the fluid follow the pressure drop. A positive flow value illustrates a correct modelled flow direction, while a negative flow value should illustrate an opposite flow direction. The original equation doesn t take this basic feature into account. Therefore it is assumed, that all the orifice equations are working in both directions if nothing else is stated. This means that the orifce equation will be changed to the following Equation A.4, so the correct flow direction is given by a positive value and a negative value means opposite flow direction.

110 B8 APPENDIX A. GOVERING EQUATIONS 2 Q = C d A ρ P 1 P 2 si g n(p 1 P 2 ) (A.4) The absolute value of the pressure difference will allways give a positive number, while the si g n function take care of the direction by the following method: si g n(p 1 P 2 ) = 1 if P 1 P 2 if P 1 = P 2 1 if P 1 < P 2 A.3 Translatoric friction Consider the free body diagram of Figure A.4. This model contains a mass m, which is moving with the speed v, and a friction force F acting in the opposit direction of the motion. The model isn t considering gravity directly hence no normal force. Indirectly the gravity is included in the friction force. The friction force F acting on the body is composed by 3 types of friction forces: "Striction, Coulomb and Viscous". The different frictions is best described with an illustration. Consider Figure A.5. The figure illustrates the three friction forces, and the resulting friction force which is the sum of them. M V F Figure A.4: A free body diagram illustrating the friction force, the velocity and the mass of an arbitrary mass. Static friction Coulomb friction F F

111 M V A.4. FLOW FORCE F B9 Striction F Coulomb friction F V V Viscous friction F Striction + Coulomb + Viscous F V V Figure A.5: An illustration of the different friction forces composing a resulting friction, which is also illustrated The equation describing the resulting friction as a function of velocity is given in Equation A.5. The equation is quite intuitive when the graphs are considered, hence no further explanation is given. [ ( F f (ẋ) = σẋ + si g n(ẋ) F co + F so exp ẋ )] C s (A.5) ẋ σ Velocity of the spool Viscous friction factor F co Constant Coulomb friction [N ] F so Stribeck friction [N ] C s A factor that determines the decay of the stribeck effect as velocity increases. [ ] [ m s ] [ ] kg s A.4 Flow force When fluid flows across an orifice, it will try to close it. The acting force is called a flow force. If the orifice is fixed, this force is of little interest, because it is often small compared to the forces due to the pressure. When the orifice is variable, this force can lead to errors, because the force required to give a specific discharge area is underestimated. The principle of conservation of momentum is used to describe this phenomenon. F = v (m v) = ṁ v + m t t

112 B1 APPENDIX A. GOVERING EQUATIONS If the flow is set to pass through a control volume ( ) and assumed to be in steady state, the velocity is constant, and the equation can be rewritten. This way only the stationary flow forces are described. This assumption is valid because, in this equation, the change in flow is the dominating force. Q F = ṁ v = ρ Q v = ρ Q ˆ v C d A The discharge coefficient (C d ) is assumed equal to the contraction coeffiecient (C c ), so C d is used in the expression. The terms Q, A and ρ are, the flow out of the control volume, the smallest cross sectional area and the density of the fluid, respectively. The term ˆ v is a unit vector pointing in the direction of the flow. The force of interest is the force acting on the moving part. If the angle between the flow and the line of movement for the moving part is denoted θ, the flow force can be determined by the following eguation. Q 2 F F S = ρ cos(θ) C d A The orifice equation A.1 on page B7 is utilized, so that the equation can be written in terms of pressure instead of flow. F F S = ρ ( ) 2 C d A 2 ρ (P i n P out ) C d A cos(θ) C 2 d A2 2 ρ (P i n P out ) F F S = ρ C d A cos(θ) F F S = 2C d A (P i n P out )cos(θ) (A.6) F F S Flow force. Will always point in the direction closing the valve. [N ] C d Discharge coefficient [ ] The smallest cross sectional area of the opening [ m 2 ] A P i n Upstream pressure. Pressure before the opening. [Pa] P out Downstream pressure. Pressure after the opening. [Pa] θ Angle of the flow [r ad] A.5 Continuity equation The so called continuity equation is an equation used to describe the change of pressure in a cavity of a hydraulic system.

113 A.5. CONTINUITY EQUATION B11 P Q in Q out A Figure A.6: An arbitrary control volume ( ) with flows and pressure. In the Figure A.6 it is seen that if the two flows aren t exactly the same there must be a pressure change in the control volume, unless the volume is changed. This is described in the continuity equation A.7. Q i n Q out = d d t + d p β d t (A.7) Q i n Q out Flow into the cavity Flow out of the cavity Volume of the cavity [ m 3 s [ m 3 s ] ] [ m 3 ] β Bulk modulus of the fluid (Stiffnes of the oil) [Pa] p Pressure in the cavity [Pa] The stiffnes of the oil β isn t a constant. It is highly temperature and pressure dependent. In [Andersen & Hansen, 27a], is a figure (see figure A.7) describing the change in stiffness as the pressure changes, at different temperatures. At an operating point of 2 bar, it is seen that a temperature change from 2 to 4 results in a change of about 1 bar in stiffnes (185bar 175bar ). Which is a change of about 5.5%. In some applications this is a small change, and therefore the bulk modulus can be assumed temperature independent. If the temperature of the oil graphed in figure A.7 is 2 the change in pressure from bar to 2 bar result in a change of stiffnes of 175 bar (185bar 1675bar ), which is a change of about 1%. In some applications this is also a small change, and therefore the bulk modulus can be assumed to be pressure independant, however, this is not allways the case, especially when there is air dissolved into the fluid, which is almost always the case in real systems.

114 BBβ a temperature dependant coefficient, [bar -2 ] B12 Where the temperature dependant coefficients can be determined from Equation (A.4) and Equation (A.5). It should be noted that Equation (A.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 APPENDIX stiffness with A. temperature GOVERINGand EQUATIONS pressure is displayed graphically in Figure A5. β F [bar] p [bar] Fig. A5 The variation of the fluid stiffness with temperature and pressure Figure A.7: The variation of the fluid stiffness with temperature and pressure. Andersen & Hansen [27a] In real systems air will be present in the fluid. The volume percentage at atmospheric pressure will go as high as 2 %. 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. C 2 C 4 C 6 C 8 C 1 C Hydraulic Fluids Page 8 of 11 When air is dissolved into the fluid the stiffness of the oil is dramatically lowered, especially at low pressures. The Figure A.8 illustrates how the effective stiffness of an oil changes with pressure for different volumetric ratios of air entrapped in the fluid. The definition of the ratio is given in Equation A.8. V A ɛ A = V F +V A (A.8) ɛ A The reference volumetric ratio of free air in the fluid at atmospheric pressure [ ] V A Volume of air at atmospheric pressure [ m 3 ] Volume of the fluid at atmospheric pressure and at the reference temperature [ m 3 ] V F

115 A.5. CONTINUITY EQUATION B13 β eff 2 [ bar ] ε A = ε A =. 1 ε A =. 5 ε A =. 2 ε A =. 1 ε.5 A = p [bar] 1 Fig. A6 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 4 C and the compression of the free air is assumed adiabatic Figure A.8: Plots of how the effective stiffness of an oil changes with pressure for different volumetric ratios of air entrapped in the fluid. [Andersen & Hansen, 27a] -----oooo----- Experiments have shown that in most hydraulic systems the bulk modulus is allmost never more than 1. bar Andersen & Hansen [27a]. Hydraulic Fluids Page 11 of 11

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117 Appendix B Test Reports B.1 Compensator spool in PVP module spring Purpose of the experiment The purpose of this experiment is to determine the spring constant of the spring, attached to the pressure sompensator spool in the PVP module, and the preload of this spring. The spring constant was determined by applying a force to the spring, and measure the displacement. Materials used for the experiment Avery (machine for determination of spring constant) Theory The theory behind springs is approximated by Hooke s law of elasticity, and states that the spring force is directly proportional Force kg Travel mm with the displacement. The constant that defines this proportionality is called the force constant or spring constant. This coherence is described by the following equation: F = k p x 9 6 In order to determine the spring constant it is necessary to record a wide range of force-deflection measurements. By inserting these values in a (displacement x, force F)-graph and making a linear approximation, it will be possible to determine the spring constant, from the slope of the graph In theory, one measurement should be sufficient to calculate the constant of the spring. But to increase accuracy, numerous measurements is performed and then linearized to an equation containing information about the spring constant. Figure B.1: Measurements from the experiment Numerous approaches can be used to do the experi- ment, and to record the displacement of the spring under a load force. One approach is, simply, to measure the displacement with a ruler and use a Newtonmeter to determine the load force. By varying the load-force it is possible to get a wide range of measurements. This practice is used by a test machine, which can give a variable load force and simultaneously measure the deflection with a precise ruler that worked as a vernier caliper. The arrangement of the experiment can be seen in figure B.2a and a table with the measuments from B15

118 B16 APPENDIX B. TEST REPORTS the experiment can be seen in Figure B.1. Matlab was used to make a graph of the data from the experiment and the Basic fitting tool to derive an expression. The graph of both the derived expression and the data set can be seen in B.2b. Spring from PVP module, deflection under different forces 35 Measurement points y1= x Force F [N] (a) Arrangement of the experiment Displacement x [m].2.25 (b) Graph from the experiment. Blue line: linearization. Red dots: actual data from the experiment. Figure B.2 Determination of preload The spring, attached to the pressure compensator spool is preloaded, and the force of this will now be determined. To determine the preload, the deformation of the spring when the spool is pushed as far away from the spring as it can be, has to be determined. The equation to determine the preload is expressed in Equation B.1 F pr el oad = (L L 1 ) k p (B.1) Where k p is the spring constant, L is the lenght of the spring before the deformation and L 1 is the lenght when being mounted in the PVP module. The length of the spring, when being applied no force, is measured to be:x unl oad ed = 45.9mm and when it is mounted in the PVP module the lenght is measured to be x l oad ed = 28.9mm. the deformation can then be calculated in Equation B.3 F pr el oad p = (x unl oad ed x l oad ed ) k p (B.2) N = (45.9mm 25.9mm) = 3N m (B.3) Sources of error The most significant source of error is from the reading of the displacement. Other sources of error is from the scale built in in the machine.

119 B.2. COMPENSATOR SPOOL IN PVB MODULE SPRING B17 Conclusion The almost linear behavior of the data set indicates precise and useable measurements and the spring constant was determined to be 15 N m. B.2 Compensator spool in PVB Module spring Purpose of the experiment The purpose of this experiment is to determine the spring constant of the spring attached to the compensator spool in the PVB module. Determination of the spring constant will make it possible to calculate the spring force as a function of spool movement. Because two springs is attached to the compensator spool the graph of the experiment will show an unlinear behavior. Materials used for the experiment Avery (machine for determination of spring constant) Theory The theory behind springs is approximated by Hooke s law of elasticity, and states that the spring force is directly proportional with the displacement. The constant that defines this proportionality is called the force constant or spring constant. This coherence is described by the following equation: F = k c x In order to determine the spring constant it is necessary to record a wide range of force-deflection measurements. By inserting these values in a (displacement x, force F)-graph and making a linear approximation, it will be possible to determine the spring constant. In theory, one measurement should be sufficient to calculate the constant of the spring. But to increase accuracy, numerous measurements is performed and then linearized to an equation containing information about the spring constant. Numerous approaches can be used to do the experiment, and to record the displacement of the spring under a load force. One approach is, simply, to measure the displacement with a ruler and use a Newtonmeter to determine the load force. By varying the Force kg Travel m Force kg Travel m Measurements from the experi- Figure B.3: ment load-force it is possible to get a wide range of measurements. This practice is used by a test machine, which can give a variable load force and simultaneously measure the deflection with a precise ruler that worked as a vernier caliper. The arrangement of the experiment can be seen in figure B.4a and a table with the measuments from the experiment can be seen in Figure B.3.

120 B18 APPENDIX B. TEST REPORTS Matlab was used to make a graph of the data from the experiment and the Basic fitting tool to derive an expression. The graph of both the derived expression and the data set can be seen in B.4b. Pressure compensator springs deflection under different forces 4 35 Measurement points y1= 9942,9 x y2= x Force F [N] (a) Arrangement of experiment Displacement x [m] (b) Graph from the experiment. Blue lines: linearization. Red dots: actual data from the experiment. Figure B.4 Determination of preload The spring in the compensator spool is preloaded and the force of the preload will now be determined. To determine the preload, the deformation of the spring when the spool is pushed as far away from the spring as it can be, has to be determined. The equation to determine the preload is expressed in Equation B.4 F pr el oad = (L 4 L 5 ) k c (B.4) Where k c is the spring constant, L 4 is the lenght of the spring before the deformation and L 5 is the lenght when being mounted in the PVB module. The length of the spring when it is untense is measured to be:x unl oad ed = 36.74mm and when it is mounted in the PVB module the lenght is measured to be x l oad ed = 26.17mm. the deformation can then be calculated inequation B.5 F pr el oad c = (x unl oad ed x l oad ed ) k m = (36.74mm 26.17mm) N = 15.12N m (B.5) Sources of error The most significant source of error is from the analogue read out of the displacement. Other sources of error is from the scale built in in the machine.

121 Figure B.5: Measurements from the experiment B.3. LVDT SPRING B19 Conclusion The almost linear behavior of the data set indicates precise and useable measurements and the spring constant was determined to be 1 N m for the first spring with a force of gravity set to 9.82N. And 35 N m when both springs is compressed. B.3 LVDT Spring Purpose of the experiment The purpose of this experiment is to determine the spring constant of the small spring attached to the LVDT located in the PVE module. Because of the low tolerances of the machine available to determine the constant, it is decided to make a more precise machine. A precise scale is used to measure the force exerted to the spring and a bench drill with a mm gauge to measure the travel of the spring. Materials used for the experiment Scale. Bench drill with mm gauge. Drill. Theory First, the spring is placed on the weight and then the tip of the bench drill is lowered so that it touches the spring. The weight is then reset. For the measurements to be precise, it is necessary to read out the force exerted- and the travel of the spring in small steps. It is decided to measure in steps of 5mm. The arrangement of the experiment can be seen in figure B.6a and a table with the measurements from the experiment can be seen in Figure B.5. Matlab was used to make a graph of the data from the experiment and the Basic fitting tool to derive an expression. The graph of both the derived expression and the data set can be seen in B.6b. Force kg Travel m Force kg Travel m

122 B2 APPENDIX B. TEST REPORTS Spring from PVE module, deflection under different forces Measurement points y1= 19 x Force F [N] (a) Arrangement of the experiment Displacement x [m] (b) Graph from the experiment. Blue line: linearization. Red dots: actual data from the experiment. Figure B.6 Sources of error The most significant source of error is from the analogue read out of the displacement. Other sources of error is from the scale built in in the machine. Conclusion The almost linear behavior of the data set indicates precise N and useable measurements and the spring constant was determined to be 19 m. B.4 Main Spool Spring Purpose of the experiment The purpose of this experiment is to determine the spring constant of the spring located on the main spool. Determination of the spring constant will make it possible to calculate the spring force as a function of spool movement. After having determined the springforce, the preload of the spring will be determined. Materials used for the experiment Machine for determination of spring constant