Modeling of a Hydraulic Braking System

CHRISTOPHER LUNDIN

Master’s Degree Project Stockholm, Sweden 2015

IR-EE-SB 2015:000 Abstract

The objective of this thesis is to derive an analytical model representing a reduced form of a mine hoist hydraulic braking system. Based primarily on fluid mechanical and mechanical physical modeling, along with a number of simplifying assumptions, the analytical model will be derived and expressed in the form of a system of differential equations including a set of static functions. The obtained model will be suitable for basic simulation and analysis of system dynamics, with the aim to capture the fundamentals regarding feedback control of the brake system pressure. The thesis will mainly cover hydraulic servo valve and brake caliper modeling including static modeling of brake lining stress-strain and disc spring deflection- force characteristics. Nonlinearities such as servo valve hysteresis, saturation, effects of under- or overlapping spool geometry, flow forces, velocity limitations and brake caliper frictional forces have intentionally been excluded in order not to make the model overly complex. The hydraulic braking system will be described in detail and basic theory that is needed regarding fluid properties and fluid mechanics will also be covered so as to facilitate the reader in his understanding of the material presented in this work. Overall, the scope of this thesis is broad and more work remains in order to complement the model of the system both qualitatively and quantitatively. Although not complete in its simplified form and with known nonlinearities aside, the validity of the model in the lower frequency domain is confirmed by results given in form of measurements and dynamic simulation. Static analysis of the brake caliper model is also verified to be essentially correct when comparing calculated characteristics against actual measurements, as is also the case for the static models of the brake lining and disc-spring characteristics. Contents

1 Introduction 2 1.1 Mine Hoists ...... 2 1.2 Hoist Brake System ...... 3 1.2.1 Brake Control System ...... 4 1.2.2 Brake Hydraulic System ...... 5 1.2.3 Brake Mechanical System ...... 5 1.3 Project ...... 7 1.3.1 Lab Environment ...... 7

2 The Hydraulic Braking System 10 2.1 The Hydraulic Power Unit ...... 10 2.2 The Hydraulic Control Unit ...... 11 2.3 Transmission Lines ...... 13 2.4 Disc Brake Calipers ...... 15 2.5 Reduced System ...... 16

3 Modeling of the Hydraulic Braking System 18 3.1 Fluid Properties ...... 18 3.1.1 Fluid Mass Density ...... 18 3.1.2 Fluid Bulk Modulus ...... 20 3.1.3 Eﬀective Fluid Bulk Modulus ...... 25 3.1.4 Fluid Viscosity ...... 29 3.2 Fluid Mechanics ...... 32 3.2.1 Navier-Stokes Equations ...... 32 3.2.2 Bernoulli Equation ...... 35 3.2.3 Oriﬁce Flow ...... 35 3.2.4 Pressure Dynamics in a Hydraulic Volume ...... 39 3.3 Hydraulic Components ...... 41 3.3.1 Servo Valve ...... 41 3.4 Mechanical Components ...... 54 3.4.1 Brake Caliper ...... 54 3.4.2 Disc Springs ...... 61 3.4.3 Brake Lining ...... 64 CONTENTS ii

3.4.4 Simpliﬁed Brake Caliper Model ...... 67 3.5 Brake Stand ...... 73 3.6 Dynamic Model ...... 74

4 Model Parameterisation, Simulation and Validation 77 4.1 Model Parameters ...... 77 4.1.1 Fluid Properties ...... 77 4.1.2 Brake Calipers ...... 78 4.1.3 Servo Valve ...... 80 4.2 Static Characteristics ...... 85 4.3 Disc Spring Force ...... 85 4.4 Pressure Dynamics ...... 88 4.4.1 Simulation ...... 88

5 Discussion and Conclusions 91

Bibliography 93 List of Figures

1.1 Double-skip friction hoist...... 3 1.2 Brake control system...... 4 1.3 Brake hydraulic system...... 5 1.4 Brake caliper...... 5 1.5 Brake stand...... 6 1.6 Hydraulic power unit...... 7 1.7 Hardware-In-the-Loop Simulation (HILS)...... 9

2.1 Hydraulic diagram...... 11 2.2 Hydraulic diagram...... 12 2.3 Valve manifold...... 12 2.4 Piping...... 14 2.5 Transmission line...... 14 2.6 Modal approximation...... 15 2.7 Hydraulic cylinder...... 16 2.8 Reduced system...... 17

3.1 Fluid stress and strain...... 22 3.2 Fluid volume and pressure...... 23 3.3 Fluid bulk modulus...... 24 3.4 Container ...... 25 3.5 Effective fluid bulk modulus...... 29 3.6 Shear stress and strain...... 29 3.7 Shear flow...... 30 3.8 Kinematic viscosity as a function of temperature...... 32 3.9 Flow through an orifice...... 36 3.10 Approximated discharge coefficient...... 39 3.11 Hydraulic volume...... 39 3.12 Servo valve symbol...... 42 3.13 Servo valve...... 42 3.14 Torque motor...... 43 3.15 Spool...... 43 3.16 Spool displacement...... 44 LIST OF FIGURES iv

3.17 Valve flow...... 45 3.18 Wheatstone bridge...... 46 3.19 Flow as a function of current...... 49 3.20 Saturation...... 49 3.21 Blocking both actuator ports...... 50 3.22 Pressure Sensitivity...... 51 3.23 Load attachment...... 52 3.24 Internal valve leakage...... 53 3.25 Flow forces...... 53 3.26 Caliper halve...... 55 3.27 Model of the caliper yoke...... 55 3.28 Hydraulic unit...... 56 3.29 Model of the hydraulic unit...... 57 3.30 Brake shoe...... 57 3.31 Brake release...... 58 3.32 Gap adjustment...... 58 3.33 Model of the brake shoe...... 59 3.34 Brake caliper model...... 59 3.35 Spring stack...... 61 3.36 Disc spring...... 61 3.37 Stack deflection and force...... 62 3.38 Stack deflection and spring rate...... 63 3.39 Friction during compression...... 64 3.40 Stress-strain characteristic...... 65 3.41 Stress-strain characteristic and Young’s Modulus...... 66 3.42 Simplified caliper model...... 68 3.43 Brake caliper static characteristics...... 72 3.44 Brake calipers...... 73

4.1 Measurements and pressure sensitivity...... 84 4.2 Measured characteristics...... 86 4.3 Pressure response with diﬀerent signal inputs...... 89 4.4 Simulation...... 90 List of Tables

4.1 Fluid Properties ...... 78 4.2 Brake Caliper Parameters ...... 78 4.3 Brake Lining Parameters ...... 79 4.4 Disc Spring Parameters ...... 79 4.5 Valve Rating ...... 80 4.6 Servo Valve Parameters ...... 83 4.7 Clamping Force and Air Gap Adjustment ...... 85 Chapter 1

Introduction

1.1 Mine Hoists

Mine hoists are used for the purpose of vertical transportation of ore, personnel and equipment in underground mine shafts. With the use of a mine hoist in an underground mine, production may increase significantly as it is the most efficient way of both elevating ore to the surface and transporting personnel to and from the deep levels of a mine. Hoisting distances typically lie in the range between 500 and 1500 meters with payloads up to about 60 tons. Mechanically, a mine hoist may be configured differently depending on a number of factors, but the two main categories of hoists are either of a winding or frictional type. Friction hoists (or Koepe hoists), see the example depicted in figure 1.1, are based on friction for displacement of two conveyances placed on each side of a mine shaft. Through frictional forces between a number of hoisting cables called head ropes (5) and friction inlays mounted on a hoist pulley (1), which is driven by an electrical motor (2) either directly or through a gear box, the conveyances are allowed to ascend and descend in the mine shaft. The motor in turn is mechanically connected to a shaft onto which the rotor and pulley is mounted, which is supported by two bearings (3) on both sides. The conveyances can be either in form of ore containing skips (8) or personnel carrying cages, which may be combined with a counterweight if hoisting with a single skip or cage. The head ropes are connected to the upper end of the conveyances through rope attachments (6) and a number of balancing tail-ropes (10) are fixed to the lower end of both conveyances with the purpose of maintaining constant rope tensional forces on both sides of the hoist pulley. The skips are typically filled through ore-bins (9) or from conveyor belts placed on one or both sides of the shaft depending on if the hoist is used in a single- or double-skip configuration. The diameter of the pulley is decided by rope selection criteria and a number of head sheaves (7) may be installed in order to decrease the size of the shaft if needed. Drum hoists (or winders) are a type of hoist where the hoisting cable is wound around the drum and may be configured in 1.2 Hoist Brake System 3 a single- or double-drum configuration with either one or two rope compartments on each drum and a single or double clutch mechanism for double-drum hoisting in order to be able to adjust the relative distance between the conveyances (this to either adapt for rope elongation or adjustment to different hoisting distances). Regardless of the type of hoist, a disc brake system1 is installed for safety reasons and a number of brake calipers are mounted on brake stands (4) that are placed on each side of the pulley or drum. The mine hoist brake system will be the topic of this work and is presented in more detail in the next section.

Figure 1.1: Double-skip friction hoist.

1.2 Hoist Brake System

From a safety point of view, the braking system is the most critical part of a mine hoist. Failure to bring the hoist to a standstill during operational distur- bances such as a power outage or an emergency situation will lead to a so called overwinding, possibly with serious consequences, which in a worst-case scenario can result in fatal outcome. When also taking into consideration the economic aspects associated with such malfunction of the system, potentially causing halts in production for long periods of time and loss of investments, the conclusion

1Shoe brakes for mine hoists also exist. 1.2 Hoist Brake System 4 must be that safety issues concerning mine hoists are of great importance. Past accidents and incidents have historically lead to more strict regulations world- wide concerning the use of mine hoist systems, with current trends pointing in the direction of an increasing demand for high safety integrity in the mining in- dustry. The hoist brake system consists of three subsystems which spans across three technical domains containing electrical, hydraulic and mechanical components. The diﬀerent subsystems of the brake system are brieﬂy presented in the following.

1.2.1 Brake Control System The control system mainly comprises a main control unit (MCU), a HMI2 oper- ator interface and four channel control units (CCU), see ﬁgure 1.2. The control system is built on a multi-channel architecture with identical, in parallel execut- ing and independent so called brake channels. Each channel of the brake system is controlled by a CCU in form of a PLC3, which includes control logic, moni- toring and closed-loop control functions. The main advantage of a multi-channel

Figure 1.2: Brake control system. system is the ability to both electrically and hydraulically divide the total braking capacity into two or more independent subsystems, thus avoiding the dependency on a single system. In a single channel system (or mono-channel system) one is solely dependent on a single chain of both electrical and hydraulic components. A single component failure will thus aﬀect the entire brake capacity, with the two extremes being complete loss of brake force and full application of the brakes with the hoist in motion. With a multi-channel system, a degree of redundancy is achieved and only a fraction of the brake capacity is aﬀected in case of failure

2HMI - Human Machine Interface 3PLC - Programmable Logic Controller 1.2 Hoist Brake System 5 of the equipment in one of the brake channels (for a four-channel system a fourth and for a two-channel system one half). The disadvantages of a multi-channel system are of course the greater complexity, increased cost of the equipment and reduced availability since the probability of component failure will be higher.

1.2.2 Brake Hydraulic System The hydraulic parts of the brake system consist of a power unit and two control units, see ﬁgure 1.3. Each such control unit in turn contains two so called valve manifolds each, which are the controlling hydraulic elements of the brake channels. As the term indicates, the power unit maintains the system pressure and supplies the control units with hydraulic ﬂuid. The main task of the brake system

Figure 1.3: Brake hydraulic system. is to safely decelerate the mine hoist by friction-based braking in emergency or safety stop situations. During normal operation, hoist deceleration is however performed by the electric motor of the drive system and the brake system only provides a static holding force to securely keep the position of the mine hoist as the drive is disconnected at standstill. The power and control units together hydraulically maneuver the brake calipers that mechanically produce the braking force that is needed during emergency braking and at hoist standstill.

1.2.3 Brake Mechanical System The braking system is of a disc brake type with discs mounted on the drum or pulley of the mine hoist where the clamping force from a number of brake calipers is applied, see ﬁgure 1.4. The calipers are mounted on brake stands (see

Figure 1.4: Brake caliper. 1.2 Hoist Brake System 6

ﬁgure 1.5) which are positioned around the drum or pulley. The number of brake stands and calipers varies and are sized diﬀerently depending on load conditions and regulatory requirements. The number of brake discs are however at least two because of the risk of contamination of the disc surfaces (loss of brake force can amount to half of the total brake capacity in the case of a single disc). The brake calipers acts by applying a clamping force directed normally against both surfaces of the brake discs, which through friction between the brake linings and the brake disc will induce a braking force and in turn also a braking torque. The calipers are of a fail-safe design and do therefore not need any hydraulic pressure applied in order to apply the brake linings against the brake disc surfaces. A restoring spring force instead strives towards a full application of the brakes against the discs and is counteracted by increasing the hydraulic pressure. Maximum working pressure in the system therefore implies that the brake calipers are released from the brake disc with a fully developed gap between the linings and brake disc surface and that the brakes are fully applied at atmospheric pressure (zero gauge pressure) with maximum clamping force. It must however be stressed that the pulley (in

Figure 1.5: Brake stand. the case where transport of the conveyances takes place due to friction between the pulley and head ropes) or drum (when the rope or ropes are wound onto or unwound from the drum) must not be exposed to excessive braking force when the hoist is set in motion as this may give rise to dangerously high deceleration, which in turn can lead to injuries, rope slippage or slack rope conditions. The emergence of this can lead to serious accidents which may involve the risk of endangering the lives of mine personnel. The number of brake calipers is oversized for the case of braking during dynamic conditions, this because the brake system must meet the requirements of static holding force as speciﬁed in applicable mining regulations. For example, the braking system must be able to keep the mine hoist at standstill during overload conditions, which may mean that the mine hoist is subjected to unbalanced forces that exceed twice or possibly even three times that of the rated maximum load. Overcapacity of available braking force is dealt with by dividing the system into several brake channels and limitation of 1.3 Project 7 applied clamping force during dynamic conditions, this because of the risk that could arise if the system would fail with the mine hoist in motion and provide full application of the brakes. For this reason, the braking system is always provided with at least two brake channels.

1.3 Project

The objective of this thesis is to obtain an analytical model of a reduced form of the full-scale hydraulic brake system that has been brieﬂy presented in this introductory chapter. The analytical model will be expressed explicitly in the form of a system of diﬀerential equations along with a number of static functions and is primarily suitable for basic simulation and analysis of system dynamics. Results will be presented in form of dynamic simulation and static analysis of the system, which to some extent will verify the validity of the obtained model. The components and functionality of the system will be described in more detail in the next chapter along with a number of modeling assumptions and the scope of model inclusion.

1.3.1 Lab Environment The hydraulic system is installed in a development and test environment along with the brake control system. A hydraulic power unit (see ﬁgure 1.6) and a single valve manifold mounted therein including a brake stand with a total of three brake calipers, two single and one double unit, are installed in the lab and available for tests. The brake stand is mounted close to the power unit

Figure 1.6: Hydraulic power unit. and hydraulic transmission lines in form of steel pipes connect between them. Additional to what was presented in section 1.2.1, the conﬁguration of the lab control system also consists of both PC- and PLC-based real-time simulation hardware for Hardware-In-the-Loop-Simulation (HILS) of the brake system hydraulics (HySim) and the mine hoist (HoistSim) respectively. Referring to ﬁgure 1.3 Project 8

1.7a, one out of the four channel control units is connected to the valve manifold installed in the power unit while the other control units are connected to HySim. The valves in turn hydraulically control the brake caliper clamping force, which is measured with a force transducer mounted between one of the calipers on the brake stand. The measured clamping force is then fed via an analog input to HoistSim along with the force of three simulated brake channels where a simple summation is made in order to obtain the total applied clamping force for all four channels. HoistSim then simulates the hoist dynamics in order to produce an analog output corresponding to the angular velocity of the hoist pulley or drum, which is then fed back to each one of the channel control units. As shown in figure 1.7b, the deceleration of the hoist is controlled through cascaded closed-loop control of the brake pressure and deceleration calculated from the hoist angular velocity. In an actual hoist installation rotational speed connects back to the control units through incremental pulse encoders, which in the lab environment is emulated through a frequency transducer as illustrated in figure 1.7b. A near full-scale model of the brake hydraulics is implemented through component-based physical modeling and is simulated in HySim. The advantages of component-based modeling is the ease at which large complex models may be configured and changes to existing models are more or less trivial. The disadvan- tage is however the lack of explicit analytical insight into a model of a complete system model, which essentially motivates the objective of this work. 1.3 Project 9

(a) Hardware conﬁguration.

(b) Signal ﬂow and closed-loop control.

Figure 1.7: Hardware-In-the-Loop Simulation (HILS). Chapter 2

The Hydraulic Braking System

A more detailed description of the brake system with its individual subsystems and components will be given in the sections of this chapter that follows. The scope of the system to be modeled only includes a subset of the complete hydraulic brake system and will be presented as a reduced system in the last section of the chapter. This will then form the basis and outline of the analytical model.

2.1 The Hydraulic Power Unit

The main function of the hydraulic power unit is to supply the control units with hydraulic fluid and maintain a constant system pressure. The hydraulic circuitry of the power unit is found in figure 2.1 in form of a hydraulic diagram. The power unit mainly includes a hydraulic pump (4) driven by an electric motor (5), a tank (9) which supplies the braking system with hydraulic fluid and an accumulator (21) with the task of maintaining the system pressure at high loads. Additional to these main items the power unit also includes other components such as control valves (6, 20), an oil heater with thermostat (11, 17), an oil cooler including a circulation pump, electric motor and fan (22-25), shut-off valves with and without an indicator (1-3, 8), oil level indicators (10, 15), oil filters with indicators (13, 26) and (14, 27), a breather filter (18), a thermometer (19), a temperature transducer (16), a check valve and pressure relief valve (29, 7), measurement points (12, 30), a pressure gauge (32) and pressure transducers (28, 31). A vast majority of these components are relatively uninteresting for the modeling of the system dynamics, at least regarding control of the hydraulic brake pressure which will be the main focus of this work. These components will therefore be omitted in the model of the system. Components that are of potential interest in the modeling of the power unit are components (4), (5), (26) and (29). The accumulator (21) has a great impact on the system pressure, but will however be fully loaded with valve (20) de-energized and valve (6) energized (i.e. both valves closed). It can therefore be disregarded when modeling the power unit as it will not have 2.2 The Hydraulic Control Unit 11

Figure 2.1: Hydraulic diagram. any effect on the system pressure. This exclusion is justified by the fact that the brake calipers in the lab environment are few and that the accumulator for that reason is not needed. The hydraulic pump (4) is of a piston type with variable displacement driven by an induction motor (5) and the system pressure is controlled via a built-in mechanical regulator in the pump, which through a swash-plate mechanism controls the flow of oil (the pump can thus be seen as a flow source). When the set pressure is reached, no oil will flow into the system. However, a small leakage flow will still exist but is diverted through a drain line connected to the tank (see the dashed line in the hydraulic diagram in figure 2.1). The filter (26) ensures that the hydraulic fluid entering the system through the pump has a certain degree of cleanliness. This is important as contaminated oil otherwise could damage the control valves both in the hydraulic power unit and the hydraulic control units. A drop in pressure occurs across the filter, but is however of such small magnitude that it may be neglected. A check valve (29) is also installed to prevent the pump supply line from acting as an oil return path.

2.2 The Hydraulic Control Unit

The hydraulic power unit is connected to the two hydraulic control units of the brake system through a pump and tank connection, see P and T in the hydraulic diagram of the power unit in ﬁgure 2.1. The control units are in turn connected to the brake calipers on the brake stands, which are controlled by the brake pressure through a number of valves installed in the control units. The hydraulic diagram for one out of the two channels in each control unit is found in ﬁgure 2.2. The valves used for control of the brake pressure are mounted in so 2.2 The Hydraulic Control Unit 12

Figure 2.2: Hydraulic diagram. called valve manifolds (see ﬁgure 2.3) that physically implements the hydraulic circuitry. Four valve manifolds, one for each hydraulic channel, are placed in the two control units which thus are housing two manifolds each. However, in the lab environment only a single manifold mounted inside the power unit is available for tests. A servo valve (7) is the main component used for pressure control in each hydraulic channel and represents the actuator in the closed-loop control of the brake pressure. The feedback of pressure to the controller in each channel control unit PLC is carried out by the use of one of two redundant pressure transducers (20) and (21), which are mounted on the return line of each hydraulic control unit. In addition to these components, the manifold also contains a number of other items. Among these the relief valve (1), the shut-oﬀ valve (3), measuring points (4), (12) and (13), the return line accumulator (5), the pressure gauge (19) and the pressure transducers (11), (20) and (21) may be excluded as they are of no interest when modeling the system. The pressure relief valve (9) is used to control the pressure at unregulated braking of the hoist (constant brake force)

Figure 2.3: Valve manifold. 2.3 Transmission Lines 13 and is mechanically set to a certain opening pressure. At a pressure below this setting, the valve is closed and no flow will pass through the valve. Above the set pressure level the valve will open and let oil out of the system until the pressure has decreased to the point where the valve will close again. Valve (10) is an electrically controlled pressure relief valve with a large flow capacity which is used at initial application of the brake linings against the brake disc during a safety stop of the mine hoist. In order for the brakes to apply in as short time possible, a large return flow of oil is needed during a time span of about 200 milliseconds. The functionality of the valve is basically the same as for a purely mechanical pressure relief valve, apart from having a pressure level of relief that is set electrically. The check valves (2) and (18) function similar to a relief valve, however the opening pressure of a check valve is much smaller and its main purpose is rectification of the oil flow rather than a limitation of the hydraulic pressure. The directional two-way valves (14), (15), (16) and (17) act as either open or closed elements in the hydraulic circuitry and are controlled through digital signals from the control system. Hence, when energized (or activated) valve (16) will block the path for flow of oil through the valve and allow the brake pressure to be increased above the pressure relief setting of valve (9). When energizing valve (17), oil may flow through the prime application valve (10) and the brake pressure will be constrained to be equal to or below the electrically set level of pressure relief. The normally closed valve (14) implements a similar functionality and blocks the actuator port of the servo valve if needed. The accumulator (6) stores potential energy by compression of a volume of nitrogen in the same way as in the power unit and will during short periods of time allow a much higher rate of flow into the system through the servo valve compared to the flow rate of the pump. It effectively also acts as a shock absorber, this since the mechanical regulator in the pump will not be able to respond to the fast dynamics of the servo valve. If suddenly placing the servo valve in its centered position with the pump delivering its maximum flow rate without an accumulator attached, the system pressure would increase rapidly until reaching the safety pressure relief level which should be avoided. Valve (8) is a pressure compensated flow control valve and is used together with valve (15) to discharge the accumulator during unregulated braking. Through a mechanical function, which regulates the opening of the valve to maintain a constant pressure drop across its ports, a constant flow is achieved.

2.3 Transmission Lines

The different parts of the brake system are interconnected through hydraulic transmission lines in form of steel pipes. The power unit connects to both control units via the pump and tank connectors P and T and the control units are in turn linked to the brake calipers on the brake stands through the L and M 2.3 Transmission Lines 14 connections, corresponding to the supply and return side of the control units respectively (see figure 2.2). With a single manifold mounted inside the power unit in the lab environment, the connections between the pump, tank and the valve manifold may, due to the short distance, be either neglected or considered to be of a purely resistive character from a modeling perspective. The piping between the manifold and brake stand will however form a quite extensive transmission line network as is illustrated in figure 2.4. A main line connects the two ports together and branched connections are made from the main line to each caliper halve mounted on the brake stand. When increasing the brake pressure, and eventually releasing the brake linings from the brake disc, valve (16) and (17) in the control unit are energized and de-energized respectively. Thus, both valves are closed which effectively renders the return connection completely blocked to any throughput of oil. In order to capture the high frequency and resonant behav-

Figure 2.4: Piping. ior of the system, modeling of the transmission line dynamics is necessary. The dynamics of a single hydraulic transmission line may in the frequency domain be described by the following transfer function [16] " # " #" # Pu cosh Γ(s) Z(s) sinh Γ(s) Pd = 1 , (2.1) Qu Z(s) sinh Γ(s) cosh Γ(s) Qd where Pu, Pd, Qu and Qd correspond to the pressure and ﬂow at the upstream and downstream connection of the transmission line as shown in the ﬁgure below.

Figure 2.5: Transmission line.

The functions Γ(s) and Z(s) are the Laplace transforms of the propagation op- erator and characteristic impedance of the transmission line. Now, time-domain 2.4 Disc Brake Calipers 15 expressions of (2.1) may not be obtained without approximation due to the cyclic hyperbolic functions in the frequency domain and several ways of approximating transmission line dynamics exist [8]. The method of modal approximation as presented by Yang and Tobler [16] will yield truncated transfer functions from causal forms of (2.1) with n modes of the transmission line dynamics preserved. An example of modal approximation is shown in figure 2.6. At steady laminar flow, the relation between pressure and flow across and through a pipe with a circular cross-section will be linear and obey the Hagen-Poiseuille law [2] 128µL ∆P = Q, (2.2) πD4 where ∆P is the pressure drop across the ports of the pipe, µ is the dynamic viscosity, L is the length of the pipe, D is the inner diameter of the pipe and Q the mean volumetric rate of flow through the pipe. With sinusoidal input variations of pressure and flow, the dynamic response initially show low-pass characteristic behavior in the lower regions of the spectrum and cyclic anti-resonant and resonant response in both pressure and flow in the higher regions.

Figure 2.6: Modal approximation.

2.4 Disc Brake Calipers

The hydraulic pressure connects to the brake calipers that are mounted on the brake stands through the pipe connections L and M (see ﬁgure 2.4). The calipers are single acting hydraulic cylinders of a spring-return type and may schematically be represented by the symbol shown in ﬁgure 2.7. Functionality of the brake calipers is based on a fail-safe spring-applied and pressure-release principle with a fully developed clamping force at atmospheric pressure. This 2.5 Reduced System 16

Figure 2.7: Hydraulic cylinder. principle therefore requires pressure to be applied to the caliper pressure port in order to decrease the clamping force. At the so called release pressure, the clamping force is balanced by the hydraulic pressure and increasing the hydraulic pressure further will release the brake linings from the brake disc. A number of disc springs in the brake calipers generates the clamping force FC applied against the brake disc of the drum or pulley through the brake lining mounted on the brake shoe of each caliper halve. Disc springs are used due to the large force that can be generated for small deﬂections and also their compactness. The clamping force acts normal to both surfaces of the brake disc and will yield a brake force FB with a magnitude which depends on the coeﬃcient of friction µ between the linings and disc surface FB = 2 × µ FC . (2.3)

The braking force FB that arise from the friction between both surfaces transfers into a braking torque which will stop the rotation of the pulley or drum of the hoist during a safety stop and prevent it from rotating when stopped. At normal service, the brakes simply provide a static balancing holding torque to maintain the hoist at standstill. The magnitude of the brake torque depends on the diameter of the brake disc, the applied brake force and the width of the brake lining DB − wl TB = FB × , (2.4) 2 where TB is the braking torque, DB is the brake disc diameter and wl is the width of the brake lining. Hence, the point of application of the brake torque will eﬀectively be at a distance from the center of the pulley or drum corresponding to the radius of the brake disc corrected for by halve of the width of the brake lining.

2.5 Reduced System

The analytical model of the hydraulic brake system will in this work correspond to a reduced form of the system that has been presented. With focus on closed-loop control of the brake pressure, the system may be reduced to the simplified single channel system shown in figure 2.8. Hence, the pump, motor and check valve need to be included from the hydraulic power unit and only the servo valve and accumulator in the hydraulic control unit. However, the supply pressure will be assumed constant in the modeling of the system, leaving the servo valve and brake 2.5 Reduced System 17 calipers as the remaining components. The directional two-way valve (14) may either be considered as being a hydraulic restriction in form of an orifice or else neglected if the magnitude of the flow through the valve is relatively low, which is the case for system that is modeled. For a complex network of transmission lines, the process of modeling the system dynamics based on modal approximation will yield accurate models but is on the other hand quite involved, see e.g. [9]. Transmission line dynamics will not be taken into account in this work and the validity of the model is therefore limited to the low region of frequency where the transmission lines may be treated as pure hydraulic volumes. Now, as it might seem the presented outline of the model is a highly reduced form of the actual system. However, the essential components of the system are still included and the scope of dynamic and static modeling of the different parts of the system will still be of a considerable extent, as will be presented in the following chapters.

Figure 2.8: Reduced system. Chapter 3

Modeling of the Hydraulic Braking System

In order to model a dynamic system based on physical modeling, insight into and understanding of the underlying fundamentals that play a role in the behavior of the system is of great importance. The first part of this chapter is therefore devoted to the basic concepts and theory of fluid properties and fluid mechanics that is needed to model the system. In the latter parts of the chapter, the different hydraulic and mechanical components that are included in the model of the system are presented and described in detail along with both static and dynamic models.

3.1 Fluid Properties

Fluid is the medium in which energy is transmitted through in a hydraulic system and the type of fluid used will have an impact on a number of different system properties. The fluid itself might be synthetic or non-synthetic and important properties and issues that need to be considered when selecting a hydraulic fluid are lubricity, viscosity, protection against corrosion, tendency to foam, fire resistance and environmental impact. The properties of a fluid medium are apart from the type of fluid used also dependent on temperature, pressure and entrained air. Understanding how these various parameters affect the behavior and control of a hydraulic system constitutes the fundamentals of hydraulic system theory and is therefore presented in the following.

3.1.1 Fluid Mass Density The density of a homogeneous fluid is defined as its mass per unit volume, denoted with the symbol ρ ∆m m ρ = lim = , (3.1) ∆V →0 ∆V V 3.1 Fluid Properties 19 where m, V , P and T correspond to fluid mass, volume, pressure and temperature respectively. The definition above assumes an isothermal and isobaric condition, where the Latin prefix iso used in these terms translates into equal or similar, i.e. a condition where the fluid pressure and temperature are held constant such that P = P0 and T = T0. The mass density of a fluid is dependent on temperature and pressure, both causing volumetric change to a fluid under varying conditions.

The relation between fluid density, temperature and pressure may be formulated through the equation of state that describes the state of a matter under a given set of physical conditions. An example of an equation of state is the ideal gas law PV = nRT, (3.2) where P is the absolute pressure, V is the gas volume, n is the amount of sub- stance (number of moles), R is the universal gas constant and T is the absolute m temperature of the gas. By noting that n = M , where m is the mass and M is the molar mass, equation (3.2) may be reformulated to explicitly give an expression for the density as a function of the state variables m m M P PV = nRT = RT ⇒ ρ = = · . (3.3) M V R T For a liquid, no such simple relation between fluid density, temperature and pressure exists. However, a linearized equation of state in form of a first-order Taylor expansion may be used to describe the variations of density with small changes in pressure and temperature [10]

∂ρ ∂ρ ρ = ρ0 + (P − P0) + (T − T0) (3.4) ∂P ∂T P =P0 T =T0 where ρ0, T0 and P0 represent the density, temperature and pressure of the ﬂuid respectively at a known reference point.

Equation (3.4) can be expressed on the form    1  ρ = ρ0 1 + (P − P0) −α0 (T − T0) , (3.5) β0 | {z } | {z } ∆P ∆T where β is the isothermal bulk modulus and the reciprocal 1/β is defined as being the compressibility of a fluid 1 1 ∂ρ = . (3.6) β ρ0 ∂P The isobaric coefficient of fluid thermal expansion α in (3.5) is defined such that 1 ∂ρ α = − , (3.7) ρ0 ∂T 3.1 Fluid Properties 20 which quantifies the volumetric expansion of a fluid due to varying thermal conditions under constant pressure. With a fluid bulk modulus β that is numerically large in magnitude (approximately 1.7 × 109 Pa for a petroleum based mineral oil at 40° C) and a coefficient of thermal expansion α that is numerically small in magnitude (approximately 7 × 10−4° C−1 for a petroleum based mineral oil), fluid density is more or less unaffected by changes in temperature and pressure and may therefore be regarded as constant ρ ≈ ρ0 throughout a range of normal variation in temperature and pressure about a known working point. Effects of temperature on the coefficient of fluid thermal expansion are moderate throughout a relatively wide range of normal working temperatures, which is also the case for the fluid bulk modulus. The density of a hydraulic fluid typically lies between 850 kg/m3 and 900 kg/m3. For an incompressible fluid, considered to be an ideal condition, the density ρ is assumed to be constant.

3.1.2 Fluid Bulk Modulus The fluid bulk modulus of elasticity, or more specifically the isothermal fluid bulk modulus of elasticity, is a measure of the tendency of a fluid volume to deform elastically when under uniform compression. Denoted with the letter β, it is defined as [10]

∆σ ∂σ β = lim = , (3.8) ∆ε→0 ∆ε ∂ε T =T0 where σ and ε are the fluid stress and strain respectively. Stress is defined as the force causing a deformation of the fluid divided by the area of the surface on which the force is applied and fluid strain may be defined as the negative relative volumetric change (commonly known as the engineering strain) caused by an increase or decrease of fluid pressure

V0 − V ∆V ε = = − . (3.9) V0 V0 With the definition of fluid strain according to above, strain will increase with a decreased volume relative to the volume of the fluid at a reference pressure, typically under atmospheric conditions. Since stress is identical to fluid pressure

σ = P, (3.10) the bulk modulus may be expressed with fluid pressure explicitly ∂P β = . (3.11) ∂ε Further to the above, by differentiating strain as defined by (3.9) one will obtain 1 dε = − dV, (3.12) V0 3.1 Fluid Properties 21 as well as an equivalent definition of the tangential fluid bulk modulus ∂P ∂P β = = −V0 . (3.13) ∂ε ∂V From the definition of density and the assumption on constant mass, it is concluded that 1 1 dm = d(ρV ) = ρ dV + V dρ = 0 ⇒ − dV = dρ, (3.14) V ρ and additionally the bulk modulus may thus also be given by ∂P β = ρ0 , (3.15) ∂ρ which is the form of the bulk modulus in equation (3.6).

The secant isothermal bulk modulus is defined as the slope between two points on the stress-strain curve ∆P K = . (3.16) ∆ε The secant bulk modulus of liquids varies linearly with pressure over a limited range [10] according to K = K0 + mP, (3.17) where K0 represents the fluid bulk modulus at a pressure P = 0 relative to the atmospheric pressure (gauge pressure). For a mineral oil, which is a fluid typically used in a hydraulic system, equation (3.17) is valid up to a pressure of about 800 bar1. At a normal working temperature of the oil in the range of 30 − 70 °C the fluid bulk modulus K0 lies in the range of about 1.8 - 1.5 GPa and the coefficient of proportionality m is considered constant throughout the whole normal working range of fluid temperature, and for a mineral oil m ≈ 5.6 [10].

Calculating the strain with the reference point taken at atmospheric pressure yields V0 − V V0 − V V0 − V ∆ε = − ε0 = − 0 = . (3.18) V0 V0 V0 With the atmospheric pressure deﬁned as the reference pressure the diﬀerence in stress, or pressure, will simply equal

∆P = P − P0 = P − 0 = P. (3.19) Calculation of the secant bulk modulus now gives

∆P P V0 P K . = = V0−V = (3.20) ∆ε V0 − V V0 1 1 bar = 1 × 105 Pa 3.1 Fluid Properties 22

An expression relating the ﬂuid strain and stress may be obtained by equating (3.20) with (3.17)

∆P V0 P P K = = = K0 + mP ⇒ ε = . (3.21) ∆ε V0 − V K0 + mP This function is depicted in ﬁgure 3.1, where the origin corresponds to atmospheric conditions and the bulk modulus and the coeﬃcient of proportionality are given the values of K0 = 1.8 GPa and m = 5.6.

30 [MPa] σ 20 Stress 10

0 0 0.5 1 1.5 2 Strain ε [%]

Figure 3.1: Fluid stress and strain.

Rearranging the above also allows for identification of the relation between fluid volume and pressure P V = V0 (1 − ε) = V0 1 − . (3.22) K0 + mP The relative fluid volume 1 − ε as a function of fluid pressure is shown in figure 3.2. With a relation between stress and strain at hand, the tangential isothermal bulk modulus defined as in (3.11) may be calculated through differentiation of (3.21) with respect to P [10]

2 1 ∂ε K0 (K0 + mP ) = = 2 ⇒ β = . (3.23) β ∂P (K0 + mP ) K0 Equivalently, one may arrive at the same expression for the tangential bulk modulus through (3.13) and (3.22).

2 ∂V V0 K0 ∂P (K0 + mP ) = − 2 ⇒ β = −V0 = . (3.24) ∂P (K0 + mP ) ∂V K0 3.1 Fluid Properties 23

100 [%] V 99 Volume

98 0 10 20 30 Pressure P [MPa]

Figure 3.2: Fluid volume and pressure.

If instead a logarithmic deﬁnition of the ﬂuid strain (commonly known as the true strain) is used V ε = − ln , (3.25) V0 the tangential isothermal bulk modulus is calculated according to 1 ∂P dε = − dV ⇒ β = −V . (3.26) V ∂V

For values close to V/V0 = 1, a Taylor expansion may be used to show that

" 2 # V V 1 V V0 − V − ln = − − 1 − − 1 + ... ≈ . (3.27) V0 V0 2 V0 V0

Now, using equation (3.22) together with the derivative ∂V/∂P in (3.24) yields [10]

2 P (K0 + mP ) β = 1 − K0 + mP K0

(m − 1) P ! mP = K0 1 + 1 + . (3.28) K0 K0

A comparison between the different moduli is plotted in figure 3.3, where K0 = 1.8 GPa and m = 5.6. All show a linear characteristic in the given pressure range and as expected, the secant bulk modulus is lower than the tangential forms. The difference between the tangential bulk modulus based on the engineering strain or true strain is more or less insignificant in the lower range of fluid pressure as 3.1 Fluid Properties 24

2.5

1.5

1 K

Bulk Modulus [GPa] 0.5 β, ε = −∆V/V0 β, ε = − ln (V/V0) 0 0 10 20 30 40 Pressure P [MPa]

Figure 3.3: Fluid bulk modulus. was indicated by equation (3.27).

Linear forms of (3.23) and (3.28) may be obtained by noting that

2 2 (K0 + mP ) m 2 = K0 + 2 mP + P ≈ K0 + 2 mP (3.29) K0 K0 and (m − 1) P ! mP K0 1 + 1 + K0 K0

m(m − 1) 2 = K0 + (2m − 1)P + P K0 ≈ K0 + (2m − 1)P. (3.30) 3.1 Fluid Properties 25

3.1.3 Effective Fluid Bulk Modulus The fluid bulk modulus quantifies the elasticity of a fluid as it undergoes volumetric deformation. However, apart from the fluid bulk modulus, one must also take into consideration the influence of additional elasticity due to entrained air and mechanical compliance in a hydraulic system, which together with the compressibility of the fluid forms an effective fluid bulk modulus. Elasticity is a source of resonance within a hydraulic system, which from a control point of view is not a problematic issue as long as the effective bulk modulus is of a sufficient magnitude. However, the presence of a relatively large fraction of unsolved air within the hydraulic fluid and the use of flexible hydraulic hoses instead of fixed metal pipelines will however undoubtedly affect the hydraulic system in a soften- ing manner. This will effectively lower the frequency of resonance and thus can potentially become a concern if feedback control is employed.

Consider an uncompressed mixture of ﬂuid and air in a container as depicted in ﬁgure 3.4a.

(a) Uncompressed (b) Compressed

Figure 3.4: Container

The total volume of the system is assumed to equal the combined volume of liquid and entrained air [10] V = Vl + Vg. (3.31) A pressurization of the system through compression of both the fluid and air content inside the container as in figure 3.4b will give rise to a decrease in volume. Pressurizing the combined volume in the container will also lead to mechanical deformation of the container itself. The total volume of the system can therefore also be expressed as V = V0 − Vε + Vδ = Vl + Vg, (3.32) where V0 represents the initial volume of the system, Vε a pressurizing decrease of fluid volume and Vδ an increase in system volume due to mechanical compliance 3.1 Fluid Properties 26 of the container.

An eﬀective volume may be deﬁned such that

Ve = V0 − Vε = Vl + Vg − Vδ. (3.33) Diﬀerentiating the above yields the diﬀerential change in volume

dVe = dVl + dVg − dVδ. (3.34) Referring to equation (3.26), the following expression for the tangential eﬀective bulk modulus may be obtained from (3.34) [10]

1 1 ∂Ve 1 ∂Vl 1 ∂Vg 1 ∂Vδ = − = − − + . (3.35) βe Ve ∂P Ve ∂P Ve ∂P Ve ∂P By complementing each term in (3.35), a relation between the eﬀective bulk modulus, the bulk modulus of the liquid, the entrained air and that of the container may be obtained as ! ! 1 Vl 1 ∂Vl Vg 1 ∂Vg 1 ∂Vδ = − + − + βe Ve Vl ∂P Ve Vg ∂P Ve ∂P Vl 1 Vg 1 1 = + + . (3.36) Ve βl Ve βg βδ In the above, the bulk modulus of the container has been deﬁned such that ∂P βδ = Ve . (3.37) ∂Vδ

By noting that Ve = Vl + Vg − Vδ, equation (3.36) can be arranged as follows

1 Vδ 1 Vg 1 1 = 1 + + + . (3.38) βe Ve βl Ve βg βδ It can be shown that the above is analogous to a series connection of three mechanical springs representing the hydraulic ﬂuid, the entrained air and the container respectively [10].

The bulk modulus of the container is assumed much greater in magnitude than both the bulk modulus of the hydraulic fluid and entrained air and therefore has no significant impact on the effective bulk modulus. In the absence of any hose connections or alike, which could have the potential effect of lowering the effective bulk modulus, no attention will thus be paid to mechanical compliance in the model of the system. Based on this assumption, (3.38) is reduced to the following form 1 1 Vg 1 = + , (3.39) βe βl V βg 3.1 Fluid Properties 27

where the volume of the system is Ve = V since Vδ = 0.

The tangential bulk modulus for liquid βl in (3.39) was stated in section 3.1.2. It may however be justified to simplify the effective bulk modulus by assuming a constant fluid bulk modulus βl if the region of interest lies in a range relatively close to the atmospheric pressure, which is the case for the system being modeled.

To obtain an explicit expression for the bulk modulus of air βg needed in equation (3.39), one may assume that the pressurization and compression of the entrained air occurs under adiabatic conditions (i.e. implying that no heat transfer occurs to or from the surroundings of the gas). For an adiabatic process, the following holds true γ γ (P + P0) V = P0 V0 , (3.40) where P is the pressure of the gas relative to the atmospheric pressure P0, V is the volume of the gas, V0 the gas volume at atmospheric pressure and γ the adiabatic index2.

Now, diﬀerentiating (3.40) will yield

γ γ−1 γ d((P + P0) V ) = (P + P0) γ V dV + V dP = 0. (3.41)

The tangential adiabatic bulk modulus for air is found by arranging the above such that

γ γ−1 dV dP V dP = −(P + P0) γ V dV ⇒ − = , (3.42) V γ(P + P0) which gives [10] βg = γ (P + P0). (3.43) The bulk modulus of air linearly increases with pressure and the presence of entrained air in the hydraulic fluid has a significant impact on the effective bulk modulus. As a comparison, the adiabatic bulk modulus of air at atmospheric 5 3 pressure is βg = γP0 ≈ 1.4 × 10 Pa , which is to be compared with the bulk modulus of oil which typically lies in the range of 1.5×109 Pa to 1.8×109 Pa. Also, from (3.39) note that the volume of the entrained air in the fluid is an important factor which determines the characteristics of the effective bulk modulus in the lower region of the pressure range, and for this reason thorough deaeration of the hydraulic fluid is often needed in hydraulic systems. When the pressure is increased, the bulk modulus of the entrained air in the fluid also increases until the effective bulk modulus asymptotically approaches that of the fluid bulk modulus. What has not been mentioned is also the fact that the entrained air dissolves into 2 γ = 1.4 for air 3 1 atm = 1.01325 bar = 1.01325 × 105 Pa 3.1 Fluid Properties 28 the fluid as pressure is increased. Dissolved air has no significant effect on the effective bulk modulus and will effectively decrease the volume of entrained air in the liquid [6]. The effect of air dissolving into the fluid and its effect on the bulk modulus will however not be taken into account in the following.

In equation (3.39), the gas volume Vg may be calculated through the relation between gas pressure and volume as given in (3.40) and thus

− 1 P0 γ Vg = Vg0 . (3.44) P0 + P With the bulk modulus of the liquid assumed constant, the volume of the liquid Vl can be obtained through integration of (3.26) as

∂P βl = − ⇒ P = −βl ln Vl + C. (3.45) ∂Vl Vl Exponentiation of both both sides and solving for the liquid volume gives

− P βl Vl = Vl0 e . (3.46)

An explicit expression for the eﬀective bulk modulus is found by solving for βe in equation (3.39)

(Vg + Vl) βl βg Vg + Vl βe = = βl . (3.47) βl (Vg + Vl) βg + Vg βl V V V g + l + βg g

Replacing Vg, Vl and βg in (3.47) with (3.44), (3.46) and (3.43) respectively yields

 1 P  P γ − R + 1 + e βl β β  P0  , e = l  1 P  (3.48)  β P γ −  R l + 1 + e βl γ(P +P0) P0

V where R = g0 . Vl0 P − β If βl P , which normally is the case, then e l ≈ 1 and the above may be simpliﬁed so that  1  R + 1 + P γ  P0  βe = βl   , (3.49)  1  R βl + 1 + P γ γ(P +P0) P0 which corresponds to the model of the eﬀective bulk modulus derived by Cho et al. [4].

The eﬀective bulk modulus as given in (3.49) is plotted in ﬁgure 3.5 for R ∈ {0.1, 0.5, 1.0} % and βl = 1.8 GPa. 3.1 Fluid Properties 29

2 [GPa] e β R = 0.1 % 1.5 = 0.5 % R = 1.0 % 1 R

0.5

Eﬀective Bulk Modulus 0 0 1 2 3 4 5 Pressure P [MPa]

Figure 3.5: Eﬀective ﬂuid bulk modulus.

3.1.4 Fluid Viscosity In solid mechanics, shear strain is directly proportional to the shear stress considering a linear elastic deformation of a solid material. This constant of proportionality that relates shear strain to the shear stress is called the shear modulus of elasticity. If shear stress is applied to a solid, as illustrated in ﬁgure 3.6, the material deforms and remains in this state until the shear stress is removed whereby the solid elastically regains its original state. This is not replicated by a

Figure 3.6: Shear stress and strain.

fluid however as a constantly applied shear stress will deform a fluid continuously. Consider the situation in figure 3.7 where two plates, both having an area A, are closely spaced apart at a certain distance h and separated by a homogeneous fluid. For a Newtonian fluid with a parallel flow between the plates (referred to as shear flow or Coutte flow), shear stress τ is proportional to the rate of shear strainγ ˙ in the direction of the fluid flow dγ τ = µ . (3.50) dt 3.1 Fluid Properties 30

With a shear stress τ applied on the upper plate at a time instant t = 0 two points p1 = (x1, y1) and p2 = (x2, y2) are situated in two different layers y1 =6 y2 of the fluid between the plates. The shear strain γ equals zero since x1 = x2. Now, at a second time instant t = ∆t, the different layers of fluid have traveled a distance ∆x relative to each other such that the points now have moved in 0 0 0 0 0 0 the x-direction and are equal to p1 = (x1, y1) and p2 = (x2, y2). Note that the velocity u of each fluid layer differs and varies from u = 0 at the bottom layer to u = U at the top layer beneath the top plate. Assuming that the traveled

Figure 3.7: Shear flow. distance ∆x is small, the change in shear strain may be expressed as ∆x! ∆x ∆γ = tan ≈ , (3.51) ∆y ∆y since for small angles θ ≈ 0 sin θ tan θ = ≈ θ. (3.52) cos θ The change in the horizontal direction ∆x equals 0 0 ∆x = x2 − x1 = u2∆t − u1∆t = (u2 − u1) ∆t = ∆u∆t, (3.53) where u1 and u2 are the velocities of the two different fluid layers. Relating this to the change in shear strain yields ∆x ∆u ∆t ∆γ ∆u = ≈ ∆γ ⇒ ≈ . (3.54) ∆y ∆y ∆t ∆y Taking the limit now gives dγ du = . (3.55) dt dy Hence, the rate of shear strainγ ˙ is equal to the gradient of the fluid velocity.

From (3.50) the shear stress may now be calculated as du τ = µ . (3.56) dy 3.1 Fluid Properties 31

The velocity varies linearly from 0 to U between the plates and the velocity gradient therefore equals du U = . (3.57) dy h Replacing the velocity gradient in (3.56) with (3.57) now yields U τ = µ . (3.58) h The force applied on the upper plate causing shear strain to the fluid is proportional to the area A of the plate U F = τA = µ A . (3.59) h The coefficient of proportionality µ is termed absolute viscosity, or the dynamic viscosity of the fluid. Viscosity describes the internal resistance to flow of a fluid and may be thought of as a measure of fluid friction. A fluid with a low viscosity will flow more smoothly compared to a fluid with a high viscosity, which is true when comparing for example water with a typical hydraulic oil, where water having a lower viscosity than oil at a certain temperature exhibits a more fluent characteristic. This is seen analytically when rearranging equation (3.56) such that du 1 = τ. (3.60) dy µ An increased viscosity will decrease the velocity gradient and thus yield a flow where the velocities of the different fluid layers deviate less than with a lower viscosity. Dividing the dynamic viscosity with the density of the fluid gives the kinematic viscosity ν µ ν = . (3.61) ρ The properties of a fluid are dependent on conditions such as pressure and temperature, which is also the case for fluid viscosity. A varying fluid pressure has a modest impact on viscosity, whereas changes due temperature variations are significant. Pressure and temperature affect the viscosity exponentially according to [8] α(P −P0) µ = µ0 e (3.62) and − λ(T −T0) µ = µ0 e (3.63) where µ0 is the viscosity given at a certain temperature T0 and pressure P0. The viscosity of the fluid is highly dependent on temperature, which is clearly seen 4 in figure 3.8 for the case where ν0 = 46 cSt at T0 = 40 °C and ν = 6.8 cSt at T = 100 °C with fluid pressure held at the level of atmospheric pressure. All real 4 1 cSt (centiStoke) = 1 × 10−6 m2/s 3.2 Fluid Mechanics 32

150 [cSt] ν

100

50 Kinematic Viscosity 0 0 20 40 60 80 100 120 Temperature T [◦C]

Figure 3.8: Kinematic viscosity as a function of temperature.

fluids exhibit resistance to shear stress and an ideal fluid, which lacks any viscous forces, is known as an inviscid fluid. In the modeling of the system, isothermal conditions will be assumed and fluid viscosity µ will thus be considered as being constant.

3.2 Fluid Mechanics

The flow of a fluid is elementary to hydraulic systems and basic knowledge of the fundamental laws and equations which govern fluid flow is therefore necessary in order to understand and be able to analyze and model fluid mechanical systems. This section will briefly present the essential theory and what is needed for the modeling of orifice flow and pressure dynamics within hydraulic volumes, which will be needed in later sections of this chapter.

3.2.1 Navier-Stokes Equations The governing laws of fluid mechanics are expressed in form of momentum, mass continuity and energy equations for fluid flow. The Navier-Stokes equations are fundamental partial differential equations that describe the conservation of momentum within a fluid. In differential form, the equations are

∂u ! ρ + (u · ∇) u − f = −∇p + µ∇2u, (3.64) ∂t where ρ is the fluid density, u is the velocity vector of the fluid flow, t is time, f is a vector containing all the body forces acting on the fluid, p is the fluid pressure 3.2 Fluid Mechanics 33 and µ is the fluid viscosity. If considering the fluid as being incompressible and assuming isothermal conditions, both the fluid density ρ and the dynamic viscosity µ are constant. A solution of the Navier-Stokes equations will be in form of a flow field describing the velocity of the fluid at a given point in space and time.

Reynolds Number The Navier-Stokes equations may under certain conditions be simplified such that more simple analytical solutions can be obtained. Equation (3.64) can be written in a non-dimensional form by introducing the following non-dimensional variables ˆ ˆ ˆ 1 2 ˆ 2 1 u = û U, f = f F , t = t T , p =p ˆ P , ∇ = ∇ L and ∇ = ∇ L2 , where U, F , T , P and L represent a characteristic reference of fluid velocity, force, time, pressure and length respectively. Implemented into (3.64), the non-dimensional form of the Navier-Stokes equations is given by [10]

ρ U ∂û ρ U 2 P µ U + (û · ∇ˆ ) û − ρF ˆf = − ∇ˆ pˆ + ∇ˆ 2û. (3.65) T ∂tˆ L L L2 Fluid flow may be characterized as either being laminar, with well-defined streamlines of smooth, constant and orderly fluid motion without disruptions, or turbulent with a more erratic, swirling and fluctuating type of description. Laminar flow occurs when viscous (frictional) forces are dominant while turbulent flow is dominated by inertial forces. With a non-dimensional form of the Navier-Stokes equations at hand, the relation between the inertial and viscous terms may be quantified by relating the terms of (3.65) to the viscous term (containing the factor µ) on the right side of the equation. Dividing both sides of the equation with µ U/L2 yields

ρL2 ∂û ρF L2 PL + Re (û · ∇ˆ ) û − ˆf = − ∇ˆ pˆ + ∇ˆ 2û, (3.66) T µ ∂tˆ µ U µ U where ρ UL UL Re = = . (3.67) µ ν In fluid mechanics the Reynolds number, denoted Re, is a dimensionless number that describe whether flow conditions lead to laminar or turbulent flow. A large Reynolds number indicate that the inertial terms are dominant, whereas a small number indicate dominance of viscous terms. Laminar flow tends to occur at low fluid velocities, below a threshold of the Reynolds number at which it becomes turbulent. This corresponds with what is indicated by (3.67). Hence, above a critical Reynolds number a transition from laminar to turbulent flow will occur. However, in practice a transition may take place across a wider range of values. 3.2 Fluid Mechanics 34

In the case of a steady ﬂow, where ∂u/∂t = 0, the Navier-Stokes equations are reduced to ρ(u · ∇) u − µ∇2u − ρf = −∇p. (3.68) For low Reynolds numbers Re 1, the inertial terms are much smaller than the viscous term and can therefore be ignored. Equation (3.68) then becomes

µ∇2u − ∇p + ρf = 0, (3.69) which is known as Stokes ﬂow. If body forces are also neglected, (3.69) simplify to ∇p = µ∇2u. (3.70) If instead Re 1, the viscous term is much smaller compared to the inertial terms and (3.68) then becomes

ρ(u · ∇) u + ∇p − ρf = 0, (3.71) leaving Euler’s equation of inviscid flow (µ = 0). Neglecting body forces in the above yields ∇p = −ρ(u · ∇) u. (3.72) Apart from the fluid properties of density ρ and viscosity µ, the definition of the Reynolds number also includes the fluid velocity U and a characteristic length L. This dimension is defined differently depending on the type of flow and geometry. As a common example, for a cylindrical pipeline with a circular cross-section the characteristic dimension is chosen to be the diameter of the pipe by convention. Hence, for a pipeline the Reynolds number is defined as

ρ UDh UDh Re = = . (3.73) µ ν

The hydraulic diameter Dh is used for calculation of the Reynolds number in circular as well as non-circular ducts or pipes and is deﬁned such that 4A Dh = , (3.74) P where A is the cross sectional area and P is the wetted perimeter (encircling the ﬂow) of a duct or pipe. In the case of a circular cross-section the hydraulic diameter is equal to the geometrical diameter since πD2 1 Dh = 4 = D. (3.75) 4 πD For a rectangular duct with dimensions a and b the hydraulic diameter will equal ab 2ab Dh = 4 = . (3.76) 2(a + b) (a + b) 3.2 Fluid Mechanics 35

3.2.2 Bernoulli Equation By assuming an incompressible, inviscid and steady ﬂow of ﬂuid (3.64) may be reduced to the following form of the Euler’s equation 1 (u · ∇) u − f = − ∇p. (3.77) ρ

The vector algebraic identity

∇(A · B) = (A · ∇)B + (B · ∇)A + A × (∇ × B) + B × (∇ × A) (3.78) may be used to rewrite equation (3.77) such that 1 (u · ∇)u = ∇(u · u) − u × (∇ × u). (3.79) 2 The above may be evaluated further to give 1 1 ∇u2 + ∇p − ∇(−gz) = u × (∇ × u), (3.80) 2 ρ where the body forces are gravitational and assumed equal to the gradient of the potential of gravity. With an irrotational velocity ﬁeld the curl of the velocity is zero and (3.80) becomes

1 1 ! 1 ∇ u2 + p + gz = ∇ ρu2 + p + ρgz = 0 (3.81) 2 ρ 2 which implies that 1 ρu2 + p + ρgz = C, (3.82) 2 where C is a constant. The equation mathematically formulates the Bernoulli principle, which states that an increase in ﬂuid velocity simultaneously implies a decrease in ﬂuid pressure and/or potential energy, which is to be considered as a statement of the conservation of energy.

3.2.3 Orifice Flow An orifice is a form of restriction with the purpose of controlling and restricting flow and pressure within a fluid mechanical system. In a hydraulic system, flow through valves is often characterized as orifice flow and the analytical relation between flow and pressure through and across an orifice therefore constitute one of the most important equations in hydraulic system theory. Flow of fluid through a sharp-edged orifice is depicted by the streamlines as seen in figure 3.9. Due to 3.2 Fluid Mechanics 36

Figure 3.9: Flow through an orifice. the requirement on conservation of mass (with the mass flow rate calculated as ρQ) the volumetric flow rate through the orifice must equal

Q = u1A1 = u2A2, (3.83) where Q is the flow and u1, A1, u2 and A2 are the fluid velocity and stream area on the upstream and downstream side of the orifice respectively. If an incompressible, inviscid and steady flow through the orifice is assumed, the Bernoulli equation (3.82) holds on both the upstream and downstream side of the orifice. Applying (3.82) to both sides of the orifice at A1 and A2 in the same plane and equating the two equations gives

1 2 1 2 p1 + ρu + ρgz1 = p2 + ρu + ρgz2 = C 2 1 2 2 1 2 2 ⇒ p1 − p2 = ρ u − u . (3.84) 2 2 1 The Bernoulli equation as stated in (3.82) is not applicable at the orifice because of fluid acceleration, instead u2 and p2 refers to the point of contraction on the downstream side of the orifice called the vena contracta, where the stream area A2 < A0. Also note the indicated fluid turbulence on the downstream side of the orifice. Because of the turbulence, this type of flow is termed turbulent orifice flow (as opposed to non-turbulent laminar orifice flow). The fluid velocity u1 and u2 in (3.84) may be expressed in terms of the volumetric flow and stream area through the use of (3.83) 1 Q 2 Q 2! p1 − p2 = ∆p = ρ − . (3.85) 2 A2 A1 Solving for the flow Q yields s s 1 2 A2 2 Q = ∆p = ∆p. (3.86) r 2 2 ρ r 2 ρ 1 − 1 1 − A2 A2 A1 A1 3.2 Fluid Mechanics 37

Now, the jet area at the vena contracta may instead be expressed in terms of the orifice area A2 = Cc A0, (3.87) where Cc is the contraction coefficient. As was mentioned earlier in the text, the area at the vena contracta A2 < A0, and thus Cc < 1. Moreover, in order to account for discrepancies between the theoretically predicted and the actual jet velocity u2 a velocity coefficient Cv < 1 is needed for correction. Since Q = u2 A2 = u2 Cc A0, complementing (3.86) with the coefficients Cc and Cv gives [11] s Cv Cc A0 2 Q = ∆p. (3.88) r 2 ρ 1 − C2 A0 c A1

The orifice equation may now be written on the form s 2 Q = Cd A ∆p, (3.89) ρ where Cd is the discharge coefficient of the orifice and

Cv Cc Cd = . (3.90) r 2 1 − C2 A0 c A1

In practice, Cv ≈ 1 and A1 A0 and thus consequently Cd ≈ Cc. Alternatively, the oriﬁce equation may also be expressed on the form q Q = K ∆p, (3.91)

q 2 where K = CdA ρ .

Laminar - Turbulent Orifice Equation Equation (3.89) describes the relation between pressure drop across and turbulent flow through an orifice. For a sufficiently low Reynolds number however, the flow will become laminar with a linear relation between orifice flow and pressure drop and equation (3.89) will in this case therefore not describe the flow through the orifice correctly. Moreover, when the fluid flow Q and pressure drop ∆p approach zero one is likely to encounter numerical difficulties if (3.89) is implemented in its current form since a discontinuity of the derivative exist at the origin

dQ C A d 0 = √ = ∞. (3.92) d∆p ∆p=0 2ρ∆p ∆p=0 It is possible to combine laminar and turbulent orifice flow into a single continuous relation by allowing the coefficient of discharge to vary with the Reynolds number 3.2 Fluid Mechanics 38

[8]. Solving for the pressure drop ∆p as a function of flow Q in (3.89) and replacing 2 the constant 1/Cd with the variable ξ gives 1 ∆p = ξ ρ Q2. (3.93) 2A2 For high Reynolds numbers, equation (3.93) will be equal to (3.89) and thus 1 lim ξ = . (3.94) Re→∞ 2 Cd At low Reynolds numbers it has empirically been found that the discharge coefficient is directly proportional to the square root of the Reynolds number [11] which implies that 1 √ 1 √ = δ Re ⇒ ξ = , (3.95) ξ δ2 Re where δ is the laminar flow coefficient. With the Reynolds number defined as per (3.73) and the fluid velocity equal to U = Q/A

ρ UDh ρ QDh Re = = , (3.96) µ µA the orifice flow at low Reynolds numbers will be linear and laminar since using (3.95) in (3.93) will yield 2 2 3 µ 2δ DhA πδ Dh ∆p = 2 Q ⇒ Q = ∆p = ∆p. (3.97) 2δ DhA µ 2µ The laminar flow coefficient δ depends on both the type of orifice restriction and its geometry.

Now, by deﬁning the coeﬃcient ξ such that [8]

k1 ξ = + k2, (3.98) Re 2 2 a laminar-turbulent orifice equation is obtained√ where k1 = 1/δ and k2 = 1/Cd . The approximated discharge coefficient 1/ ξ is plotted in figure 3.10 for the case q of a sharp edged orifice for which Cd = π/(π + 2) ≈ 0.611 and δ = Cd/ (20) ≈ 0.137 [11].

Solving for the oriﬁce ﬂow Q in equation (3.93) gives q πDh 2 2 2 Q = −k1µ + µ k1 + 8k2ρDh∆p . (3.99) 8ρk2 The derivative of (3.99) is continuous at ∆p = 0 since

dQ πD3 πδ2D3 h h = = . (3.100) d∆p ∆p=0 2µk1 2µ Note that the tangent of (3.99) at the origin is equal to the coeﬃcient of proportionality in (3.97). 3.2 Fluid Mechanics 39

0.6

. √1 0 4 ξ

0.2

0 0 2 4 6 8 10 12 14 16 √ Re

Figure 3.10: Approximated discharge coeﬃcient.

3.2.4 Pressure Dynamics in a Hydraulic Volume The fundamental assumption on conservation of mass within a fluid system leads to another important equation in hydraulic system theory, namely the description of the dynamics of pressure in hydraulic volumes. Consider the flow of fluid into and out of a volume V as depicted in figure 3.11. By the principle of conservation

Figure 3.11: Hydraulic volume. of mass, the rate at which mass is accumulated inside the volume must equal the net difference between the rate of mass flow of incoming flow and outgoing. This is mathematically described by the mass conservation continuity equation in integral form d ρ dV + ρ u · dS = 0 (3.101) dt ˆV ˆS d ⇒ ρ dV = − ρ u · dS, (3.102) dt ˆV ˆS where V is the hydraulic volume, S is the boundary surface of the hydraulic volume, ρ is the density of the fluid and u is the vector field of fluid velocity across the boundary surface of the volume.

The left hand side of (3.102) equals the rate of change of the fluid mass stored 3.2 Fluid Mechanics 40 in the hydraulic volume. With a fluid that is homogeneous and a total volume of the system known and equal to V , this may be rewritten such that d d ρ dV = (ρV ). (3.103) dt ˆV dt The right hand side of equation (3.102) equals the total net rate of mass flow across the boundary of the volume V . With a single ended entry of fluid into the hydraulic volume (Qout = 0) and a mean flow Qin = Q, the rate of mass flow into the volume is calculated as Q ρ u · dS = −ρ A = −ρ Q, (3.104) ˆS A where the mean fluid velocity is equal to Q/A.

Equating (3.103) and (3.104) according to (3.102) and applying the chain rule to (3.103) gives d dV dρ (ρV ) = ρ + V = ρ Q. (3.105) dt dt dt Dividing with ρ and complementing (3.105) such that the bulk modulus and the time rate of change of pressure may be identiﬁed yields dV 1 dρ dp dV V dp + V = + dt ρ dp dt dt β dt dV V dp ⇒ + = Q. (3.106) dt β dt

Finally, the pressure dynamics of the volume is given by rearranging the above such that dp β dV ! = Q − . (3.107) dt V dt Now, with the volume V constant the hydraulic volume is eﬀectively identical to a ﬁxed capacitance Ch = V/β in the hydraulic domain since

dp β 1 = Q = Q dt V Ch dp ⇒ Q = Ch , (3.108) dt which is analogous with an electrical capacitance du i = C . (3.109) dt 3.3 Hydraulic Components 41

The transfer function between the volumetric flow rate Q and the pressure P of the hydraulic volume V is found by applying the Laplace transform to (3.108) 1 1 sP (s) = Q(s) ⇒ G(s) = , (3.110) Ch Ch s which shows that the pressure of the hydraulic volume is identical to a scaled integration of the flow input. With a constant hydraulic capacitance Ch, a pure proportional feedback control of the pressure in the hydraulic volume gives the following transfer function of the closed-loop system Gc K 1 F (s)G(s) Chs Gc(s) = = 1 + F (s)G(s) 1 + K 1 Chs K ⇒ Gc(s) = . (3.111) Ch s + K Applying the final value theorem (which holds) with a step input to the above shows that control of pressure in the volume through control of the input flow proves especially simple if a pure proportional control principle is used since the steady-state error becomes zero 1 lim Y (t) = lim s Y (s) = lim s Gc(s) U = U. (3.112) t→∞ s→0 s→0 s

The assumption on a constant hydraulic capacitance Ch and fixed volume V is valid in the mid-region of fluid pressure in the hydraulic system where the brake linings are still applied against the brake disc and the value of the effective bulk modulus is close to reaching that of the liquid.

In the lower pressure region the volume is constant but the hydraulic capacitance will vary with the pressure due to its dependency on the bulk modulus, which is identical to what is given in (3.49). Hence, the pole of the closed-loop system will directly depend on the eﬀective bulk modulus and be equal to K K s = − = −βe . (3.113) Ch V As seen, the bulk modulus has a great impact on system bandwidth and the fraction of entrained air within the ﬂuid should therefore be kept at a minimum.

3.3 Hydraulic Components

3.3.1 Servo Valve The ability to achieve continuous and stepless control of ﬂow is made possible through the use of a directional valve known as the hydraulic servo valve. This 3.3 Hydraulic Components 42 valve forms the basis of pressure control in the hydraulic braking system and is controlled by closed-loop feedback of the brake pressure through a pressure transducer connected to the control system. A hydraulic symbol representing the servo valve is shown below. The servo valve has four ports to which connections

Figure 3.12: Servo valve symbol. to the hydraulic pump, reservoir tank and the actuator are made. The ports P and T are the pump and tank connectors of the valve, where the pump connection serves as a pressure source and fluid supply to the brake system and the tank connection as a return exhaust path for flow of fluid out of the system. The ports A and B are actuator connectors of the valve and are connected to an actuator of some kind, such as a hydraulic cylinder or a hydraulic motor. In this application the A port is connected to the brake calipers and the B connector is blocked, hence only the A connection of the valve is used and this due to the fact that the brake calipers are of a spring-return type and thus have the ability to self-retract without any need of flow from the pump.

A cross-sectional view of the servo valve and its internals is found in ﬁgure 3.13. Electrically, the control system connects to the valve through a connector (A) with

Figure 3.13: Servo valve. 3.3 Hydraulic Components 43 a signal in form of a current that feeds a so called torque motor (B) that is placed internally. An illustration of the torque motor is seen in ﬁgure 3.14. The torque motor basically consists of a permanent magnet with a north pole piece (a) and a south pole piece (b), an armature (c) with two coils (d) attached and a lever (e). When a current is driven through the coils, an electromagnetic force is induced on

Figure 3.14: Torque motor. the armature which will then pivot about its equilibrium position either clockwise or counterclockwise depending on the direction of the current driven through the coils. Hydraulically, the servo valve essentially consists of two spool valve stages which indirectly control the direction and magnitude of fluid flow through the servo valve. The connecting ports of the valve are located under the valve body, of which only the two actuator ports (G) and (H) are seen in the figure. The internal connections of the supply P and return T are referred to as (I) and (J) respectively. Through the lever (C) the torque motor connects to the pilot spool stage (D), which is placed inside a hollow main spool stage (E). The main spool in turn is placed inside the spool sleeve (F). By displacement of the pilot stage from its centered equilibrium position by the torque motor, an induced hydraulic force will move the main stage axially until the forces on the main spool stage again are balanced. The main stage is indirectly and hydraulically controlled by the pilot spool (hence the term pilot) which is indicated in figure 3.12 by the triangular shaped symbol seen on the left together with an arrow indicating electro-hydraulic control of the valve. An illustration of the main stage spool is shown in figure 3.15. The spool consists of a number of so called control lands that block oil flow through the valve sleeve, which internally connects to each

Figure 3.15: Spool. 3.3 Hydraulic Components 44 of the four ports P , T , A and B as seen in the figure. In its centered position, flow through the valve is blocked with all the connections to the valve ports in the spool sleeve completely covered, but not overlapped, by the spool lands. However, with displacement of the main spool from its center position, flow of fluid is allowed through the valve and can either be in a direction from P to A and B to T or P to B and A to T . These different spool positions are reflected in the symbol of the component and are illustrated more clearly in figure 3.16 together with actual spool position in the valve. Initially the valve prevents any

Figure 3.16: Spool displacement.

flow in its centered position, which is indicated by the symbol A to the left where all paths are blocked by the spool lands. When a current is driven through the coils of the torque motor the pilot spool, and thus also the main spool, will be shifted axially and allow fluid to flow pass the control lands. Depending on the direction of the current through the coils, displacement of the main spool will be in either direction B or C as shown in the figure and symbols on the left reflect the allowed path of flow through the valve. With no current present in the coils of the torque motor the pilot and the main spool will return to the center position, which symbolically is indicated by springs on both sides of the valve symbol in figure 3.12. The term critical center is used to describe the type of spool that is illustrated above. The lands of the spool may however in some cases also overlap or underlap the ports of the sleeve in which case the spool is 3.3 Hydraulic Components 45 referred to as closed center and open center respectively.

A model of the servo valve may in a simplified form be divided into a static and dynamic part, where the static part relates spool displacement and fluid pressure with flow of fluid through the valve and the dynamic part corresponds to the dynamic relation between input current and displacement of the actuating main spool of the valve. Models of both the static and dynamic part of the servo valve are presented in the following.

Static Model It is important to understand that the servo valve itself does not directly control pressure or flow, it merely prevents or allows flow in either of the directions indicated in figure 3.16. Flow through the valve is determined by the pressure differential that exists across the valve ports and the opening area between the spool and sleeve of the valve, which is controlled by the supplied electrical current through the torque motor. Flow within the valve is depicted in figure 3.17 with a displacement x of the main spool according to B in figure 3.16. Note that flow of the fluid through the valve is distributed over a number of different paths located along the perimeter of the sleeve and that the flow as designated therefore represents the total flow to simplify the analysis. Now, as expected fluid will flow

Figure 3.17: Valve flow. from P to A and from B to T . However, due to radial clearance between the spool and the sleeve a leakage flow inevitably also exists between the path P to B and A to T . Hence, in practice no path is completely blocked regardless of spool position. Although small in magnitude, the leakage flow is dominant at small displacements of the main spool (typically in the case of feedback control of pressure when the valve is balancing about a set point at steady state) and may not be neglected in the null region of the valve (i.e. about the centered position of the main spool) and must be accounted for in the analysis of the system. The restriction will be in form of an annulus that form between the 3.3 Hydraulic Components 46 spool and sleeve and the internal leakage flow of the valve may be described by the orifice equation. Through fundamental circuit analysis, the following equations are obtained describing the relations between the supply, return and actuator port flow  QS = QSA + QSB (3.114a)    QR = QRA + QRB (3.114b)

 QA = QS − QR (3.114c)  A A  QB = QRB − QSB . (3.114d) Equating the ﬂow through the actuator ports gives

QA = QB ⇒ QSA + QSB = QRA + QRB ⇒ QS = QR. (3.115) With a displacement of the main spool according to C in ﬁgure 3.16, the actuator ﬂow QA and QB change directions such that ( QA = QRA − QSA (3.116a)

QB = QSB − QRB . (3.116b) Further to the above, the relation between the pressure drop across the valve chamber and the actuator ports may be formulated as

PS − PR = PSA + PRA = PSB + PRB . (3.117) The orifices that effectively form between the spool and sleeve of the valve together with what has been stated above may analogously be described as a Wheat- stone bridge, see figure 3.18, to facilitate the understanding and analysis of the servo valve.

Figure 3.18: Wheatstone bridge.

The ﬂow through and pressure drop across the actuator load may be deﬁned as ( PL = PA − PB (3.118a)

QL = QA = QB, (3.118b) 3.3 Hydraulic Components 47

where PA and PB is the pressure at the A and B actuator ports of the valve respectively.

Flow through the valve will be in form of orifice flow and with an assumption on valve symmetry, modeling of both the actuator and leakage flow may be combined to yield the following [5]  q x0 + x x ≥ 0 Q K P − P SA = S A 2 −1 (3.119a) x0 ( x0 − k x ) x < 0  2 −1 q x0 ( x0 + k x ) x ≥ 0 QRA = K PA − PR (3.119b) x0 − x x < 0

 2 −1 q x0 ( x0 + k x ) x ≥ 0 QSB = K PS − PB (3.119c) x0 − x x < 0  q x0 + x x ≥ 0 Q K P − P RB = B R 2 −1 (3.119d) x0 ( x0 − k x ) x < 0.

q 2 The coefficient K is defined such that K = Cd ρ w, where w is equal to the area gradient of the valve (for a fixed orifice K is defined as given in section 3.2.3). Multiplying the valve area gradient with the valve spool position yields the opening area A of the orifice. With positive displacement of the valve spool as in B of figure 3.16 and 3.17, direction of flow through the valve and attached load will be according to what is depicted in figure 3.18. The internal leakage area of the valve, due to radial clearance between the spool and sleeve, may be interpreted as a minimum opening x0 of the spool and for an ideal valve x0 = 0 (i.e. free from internal leakage). With a displacement of the spool from its centered position, the internal leakage will decrease which is reflected in the model by the inversely proportional terms found in (3.119) that will effectively decrease the area of leakage in the valve. The four piecewise defined functions may be rewritten to explicitly include the valve area  1 q A0 + (Amax − A0)x ˆ xˆ ≥ 0 Q K P − P SA = S A 2 −1 (3.120a) w A0 (A0 − k (Amax − A0)x ˆ ) xˆ < 0  2 −1 1 q A0 (A0 + k (Amax − A0)x ˆ ) xˆ ≥ 0 QRA = K PA − PR (3.120b) w A0 − (Amax − A0)x ˆ xˆ < 0

 2 −1 1 q A0 (A0 + k (Amax − A0)x ˆ ) xˆ ≥ 0 QSB = K PS − PB (3.120c) w A0 − (Amax − A0)x ˆ xˆ < 0  1 q A0 + (Amax − A0)x ˆ xˆ ≥ 0 Q K P − P RB = B R 2 −1 (3.120d) w A0 (A0 − k (Amax − A0)x ˆ ) xˆ < 0, 3.3 Hydraulic Components 48

where the area has been calculated using the area gradient w and the spool position has been normalized. With the area gradient w factored out, the valve 1 q 2 coeﬃcient will be w K = Cd ρ . By showing the area gradient equal to

A0 w (x0 + xmax) = w x0 + w xmax = x0 + w xmax = Amax x0 A0 Amax − A0 ⇒ w = = , (3.121) x0 xmax equation (3.119a) is modiﬁed such that

Amax − A0 w (x0 + x) = A0 + x = xmax x A0 + (Amax − A0) = A0 + (Amax − A0)x ˆ (3.122) xmax and x2 w2 x2 A2 w 0 = 0 = 0 , (3.123) x0 − kx w x0 − kx A0 − k (Amax − A0)x ˆ which gives (3.120a). Equations (3.119b) - (3.119d) are modified in the same way to yield (3.120b) - (3.120d). The position of the spool may thus be defined as either relative or absolute and may alternatively also be indirectly given as the valve coil control current. As was discussed in section 3.2.3, the orifice equation should for numerical reasons be implemented in its laminar-turbulent form as given in equation (3.99). In order to obtain the flow Q through each orifice, the hydraulic diameter Dh is needed and may be calculated from the orifice area as given in (3.120). Due to the lack of knowledge regarding the exact valve geometry, the hydraulic diameter will be calculated based on circular orifice geometry.

If the pressure ∆p across the valve supply and return port is held constant and neglecting leakage, flow will linearly depend on the spool position since s 2 q q Q = Cd w ∆p x = K ∆p w x, (3.124) ρ | {z } constant where a no-load (short-circuit) connection across the actuator ports of the valve is assumed. For an ideal valve, one would thus expect a linear relation between current and flow with pressure across the valve held constant. Due to saturation, the servo valve characteristic however deviate from the ideal as is shown in figure 3.19. The internals of the valve will constitute a restriction in the higher regions of flow and effectively saturate the flow. If simply adding two orifices on the supply and return connection of the valve in series with the bridge in figure 3.18, 3.3 Hydraulic Components 49

Figure 3.19: Flow as a function of current. the graph in figure 3.20 may be plotted to illustrate the deviation of the actual valve current-flow characteristic from the ideal. Modeling of the servo valve will despite the nonlinearity of the valve assume a linear current-flow characteristic with a constant area gradient proportional to the slope of the tangent w ∝ ∂Q/∂i through the origin. This is motivated by the fact that the control signal will be kept close to the null-region in the simulation of the system. Another important

Flow Q [%] 100

Current I [%]

−100 100

Linear −100 Saturated

Figure 3.20: Saturation. characteristic of the servo valve may be obtained by blocking the actuator ports as illustrated in ﬁgure 3.21. With the ports blocked, no load is attached to the valve and QSA = QRA as well as QSB = QRB . Hence, two identical and parallel legs are now connected between the supply and return port of the valve. By equating (3.119a) with (3.119b) and (3.119c) with (3.119d) one may solve for the 3.3 Hydraulic Components 50

Figure 3.21: Blocking both actuator ports.

actuator port pressure PA and PB respectively such that [5]  fPS +PR  1+f x ≥ 0 PA = (3.125a) PS +fPR  1+f x < 0  PS +fPR  1+f x ≥ 0 PB = (3.125b) fPS +PR  1+f x < 0

where f is a calculation coeﬃcient and is equal to

|x|!2 |x|!2 f(x) = 1 + 1 + k . (3.126) x0 x0

Plotting actuator port pressure against input current for both port A and B will give the pressure sensitivity of the servo valve, see figure 3.22a. As expected, the pressure is divided to half of the supply pressure at the centered position x = 0 for both ports. When displacing the spool from the centered position the pressure will either increase or decrease depending on if the displacement is positive or negative. If the spool is displaced such that x > 0, pressure will increase at port A and decrease at port B and vice versa if x < 0. Notice that in a range of displacement less than about 1 % of the rated current, the pressure at the actuator ports shifts from half of the supply pressure to about maximum. The actual pressure sensitivity curve of the valve is found in figure 3.22b. Also note the hysteretic behavior of the valve which will form a loop shaped current-pressure curve. The characteristics of the valve might also be affected by a slight overlap or underlap at the centered position. The model of the system will however not implement any of these nonlinearities.

When connecting a load to the A port of the valve, in this case the brake calipers, the valve and load will be connected according to the conﬁguration illustrated 3.3 Hydraulic Components 51

100 [%] P 50 Pressure

0 −2 −1 0 1 2 Current I [%] (a) Modeled.

(b) Actual.

Figure 3.22: Pressure Sensitivity. 3.3 Hydraulic Components 52 in figure 3.23. As was discussed in section 3.2.4, the brake calipers may, when they are applied against the brake disc, be interpreted as being identical to a hydraulic capacitance. Note well that what is shown in 3.23 is used to illustrate an analogy intended to simplify the understanding of the hydraulic circuit. When controlling the hydraulic pressure in the brake system one may with another analogy of an electrical system interpret the servo valve as a being a variable resistance in which flow enters or leaves the hydraulic volume. If the actuator port pressure PA initially is zero and x = 0, flow to the load QA will be nonzero and at its maximum when t = 0. As the pressure transiently increase in the hydraulic volume, flow to the load will decrease until an equilibrium is reached where PA = (PS − PR)/2 and QA = 0. At this point the flow through the valve must be QSA = QRA , and the system will be in a steady state. If displacing the spool from the centered position, the hydraulic pressure will increase or decrease until an equilibrium anew is reached with pressure at the actuator port as given in figure 3.22a.

With the actuator ports blocked as shown in 3.21, the valve internal leakage will be equal to the supply ﬂow QS. By substituting the expressions of (3.119) into (3.114a) - (3.114d) and calculating the supply ﬂow as QS = QSA + QSB the following is obtained [5]

q x0 + |x| QS = 2 K PS − PR q . (3.127) 1 + f(x) The valve internal leakage is plotted against the valve displacement x and shown in ﬁgure 3.24. Leakage will be at its maximum in the spool centered position where x = 0 and decrease when the spool is displaced such that x =6 0 in either direction.

Flow through the valve will create forces on both spool stages, see ﬁgure 3.25.

Figure 3.23: Load attachment. 3.3 Hydraulic Components 53

0.4

0.3 min] / [l

Q 0.2 Flow 0.1

0 −4 −2 0 2 4 Current I [%]

Figure 3.24: Internal valve leakage.

The reactive force due to the acceleration of fluid through the orifice between the main spool and sleeve will exert a force F on the spool in both the radial and axial direction at steady-state flow

F = Fx i + Fy j. (3.128)

The magnitude of the components of the force depends on the jet angle θ, which will vary with displacement of the spool. The jet angle is θ = 21° for small displacements and will asymptotically approach θ = 69° for large displacements [11]. Due to the symmetry of the valve, the lateral component is compensated and

Figure 3.25: Flow forces. will not affect the spool significantly. The axial forces on the spool will however tend to close the valve and effectively function as a return spring on the spool. Displacement of the main spool stage is carried out indirectly and hydraulically through the pilot stage, which is controlled by the torque motor. The magnitude of the flow forces on the pilot stage is small and will not be included in the model of the valve. 3.4 Mechanical Components 54

Dynamic Model In order to accurately model the dynamic characteristics of the valve, modeling of the torque motor and spool dynamics is needed (see e.g. [11]). Detailed modeling will require knowledge of parameters not readily available, and because of the difficulties a simplified and general black-box model approach is preferred. A second-order model approximation of the servo-vale dynamics of the following form is suggested [8] 1 ζ 2 x¨ + 2 x˙ + x = Ku, (3.129) ω0 ω0 where x is the spool position, ω0 is the valve natural frequency, ζ is the valve damping coefficient and K the valve gain. The model can be modified to also include valve hysteresis and spool velocity limitation due to viscous friction, this will however be left out. The current input u will control the position x of the spool and, as was mentioned earlier in the text, the static part of the model then relates the spool position to the flow of hydraulic fluid through the valve.

3.4 Mechanical Components

3.4.1 Brake Caliper The disc brake calipers consist of a number of different mechanical parts that from a modeling point of view together form a quite complex object. Simplification is therefore needed in order to obtain a model with a reasonable amount of dynamic states that captures the essential characteristics of the brake calipers. However, before a more simple model may be formulated, insight into the detailed functionality of the brake caliper and its individual parts is nevertheless needed in order to first gain fundamental understanding. A cross-sectional top view and an exploded isometric view of a caliper half is shown in figure 3.26a and 3.26b respectively. Each caliper half essentially consists of five different main parts and these are the (A) yoke, (B) hydraulic unit, (C) disc springs, (D) brake shoe and (E) brake lining. Each individual part is described in more detail in the following.

Yoke The yoke of the caliper is split into two halves and is fitted onto the brake stand with one half on each side of the stand. The two pieces are then bolted together to form a complete brake caliper. The yoke serves as a caliper housing and also need to withstand axial and radial forces of considerable magnitude. Clamping force applied on both surfaces of the brake disc will yield an axial deflection of the yoke. Deflection will decrease the maximum clamping force, and thus also the resultant brake force, and must therefore be limited. Apart from axial clamping 3.4 Mechanical Components 55

(a) Cross-sectional view.

(b) Exploded view.

Figure 3.26: Caliper halve. forces on both sides of the brake disc, the yoke is also subjected to radial forces from the brake disc and gravitational forces due to the own weight of the caliper. From a modeling point a view, where clamping force of the brake caliper is of interest first and foremost, only the yoke deflection due to axial forces needs to be taken into account. In a model of the complete caliper, which will be presented later in the section, the yoke axial deflection is modeled with a mass and spring element as shown in figure 3.27.

Figure 3.27: Model of the caliper yoke. 3.4 Mechanical Components 56

Hydraulic Unit Hydraulic pressure is connected to the brake caliper through ports on the connector plate (A) of the hydraulic unit, see ﬁgure 3.28a. Three out of the four

(a) Front view. (b) Cross-sectional view.

Figure 3.28: Hydraulic unit. connector ports P , A, and B function as pressure ports (B) and all connect hydraulically to the piston (E) through the cylinder chamber (D) inside the hydraulic unit. The fourth connector port R is a drain port (C) which removes hydraulic fluid from the caliper in case of any internal leakage between the cylinder chamber and the piston seals (F) and (G) (internal paths between the cylinder chamber and drain port not shown). The connector plate is mounted onto the disc spring support piece (H) through a set of screws (I). The hydraulic unit is then fixed to the yoke of the caliper through an adjustment screw (J) which is placed between the connector plate and the support piece. The adjustment screw enables an axial positioning of the hydraulic unit relative to the caliper yoke through manual adjustment to both set the gap between the brake lining and brake disc initially and to compensate for wear of the brake lining when needed. In order to accurately model the hydraulic unit a total of four masses is needed, see figure 3.29. First, the piston is allowed to move freely inside the hydraulic unit and is therefore represented by a single isolated mass. However, as can be seen in figure 3.28b, the connector plate acts as a rear stop for the piston and a gap and spring must therefore be added in the model to account for the end of travel of the piston inside the hydraulic unit. The connector plate, the support piece and the adjustment screw are each represented by individual masses in the model. A flexible interconnection is added between the connector plate and the support piece in form of a spring element, this due to the forces that are exerted on the connector plate by the piston at the rear end of the hydraulic unit at full release of the brake lining. Backlash between the adjustment screw, the support 3.4 Mechanical Components 57

Figure 3.29: Model of the hydraulic unit. piece and connector plate exists and needs to be incorporated into the model. This is added through a gap and spring on both sides of the mass representing the adjustment screw. Finally, two forces are indicated to act on both the piston and the support piece (with diﬀering area of ﬂuid contact) due to the applied hydraulic pressure.

Brake Shoe The brake shoe (A) (see ﬁgure 3.30) holds the brake lining (D) in place, which is clamped and secured by a set of screws (C) on both sides. A connection of the piston (E) to the center rod (B) transfers the force generated by the hydraulic pressure to the brake shoe. This force in turn acts on the disc springs (also

Figure 3.30: Brake shoe. called cup springs or Belleville springs), which are placed between the brake shoe and the hydraulic unit (see figure 3.26). The disc springs are compressed to an initial deflection by the nut (F), which essentially holds the brake shoe, 3.4 Mechanical Components 58 center rod, disc spring and piston assembly together. The force generated by the hydraulic pressure counteracts the clamping force generated by the disc springs and when the force on the piston equals the force on the brake shoe the pressure is said to balance the clamping force, in which case no force is applied on the brake disc. If the hydraulic pressure is increased beyond the release pressure the brakes will release and a gap between the brake disc and the brake lining will form. This is illustrated in figure 3.31. Maximum displacement of the brake

Figure 3.31: Brake release. shoe is achieved when the piston reaches its end of travel at the rear stop in the hydraulic unit, i.e. when the piston has reached the connector plate. The end of travel of the piston occurs below the maximum working pressure, but increasing the hydraulic pressure further will only displace the piston insignificantly. With the brake linings fully released, gap adjustment is performed if needed during commissioning or maintenance. This is needed due to the fact that wear of the brake linings will eventually affect the clamping force. Since brake force is indirectly controlled by the hydraulic pressure it is important that the gap lies in a close range of what is nominally intended. An inductive proximity sensor is therefore installed in each caliper half on the back cover of the connector plate (see figure 3.26b) to monitor the gap by measuring the distance between the back cover and the center rod. By manually turning the adjustment screw the gap between the brake lining and brake disc is adjusted to normally lie in the range of 1-3 mm, see figure 3.32. Now, although it may not be obvious, the gap

Figure 3.32: Gap adjustment. adjustment effectively controls the maximum clamping force achieved with full application of the brakes onto the brake disc at atmospheric pressure. In figure 3.32, the maximum clamping force for each of the three cases of gap adjustment will be increasing from the left example of adjustment of the gap to the example 3.4 Mechanical Components 59 on the right. This since the spring deflection at full application, and thus the maximum clamping force, depends on the adjustment of the gap with the brakes fully released. From a modeling point of view, the brake shoe and piston assembly may be seen as a single mass and thus identical to the rightmost mass seen in the model depiction of the hydraulic unit in figure 3.29. The model of the brake shoe is shown in the figure below. The brake lining attached to the brake shoe is

Figure 3.33: Model of the brake shoe. compressible and therefore need to be modeled with a spring element. Hydraulic pressure applied to the surface of the piston in the hydraulic unit generates a force on the piston and brake shoe assembly and therefore acts on the mass representing the whole assembly in the model.

Brake Caliper Model From what has been discussed and presented earlier in the section, a complete model of the brake caliper may now be formed. Interconnecting the model of the caliper yoke, brake shoe and the hydraulic unit will give the following mechanical model

Figure 3.34: Brake caliper model.

where the five different masses m1,..,5 correspond to the brake shoe, hydraulic unit, connector plate, adjustment screw and yoke respectively. The springs k1 and k2 represent the brake lining and the disc springs and k3,..,9 flexible intercon- nections and hard stops of high stiffness. Two rigid points of reference (a) and 3.4 Mechanical Components 60

(h) are placed at the brake disc and brake stand. The gap (b) is the gap between the brake lining and brake disc that form when the brake lining is released from the brake disc and (c) corresponds to the distance between the piston and the connector plate. Backlash between the adjustment screw and hydraulic unit and between the threads of the yoke and the adjustment screw is represented by the gaps (d), (e), (f) and (g). Two forces F1 and F2 are applied on the piston and the hydraulic unit by the hydraulic pressure that acts in both directions. No dissipative restrictions in form of frictional elements (e.g. between the piston and the seals in the hydraulic unit) or dampers have been added.

The brake caliper operates in one of three different discrete states which depend on the applied hydraulic pressure. Without any force applied on the piston the brake lining is applied against the brake disc with a fully developed clamping force and the gap (b) is closed. With the brake in its applied state, the gap (c) is open and the piston is positioned at a maximum distance from the connector plate. The disc springs are pushing the hydraulic unit against the adjustment screw such that the gap (e) and (f) are closed and the gap (d) and (g) are open. The force exerted on the adjustment screw is transferred to the caliper yoke which will be deflected due to the torque that is created by the shear force from the disc springs. When hydraulic pressure is applied on the piston and the support piece, the disc springs will compress and the brake lining will decompress. Since the force applied on the piston will counteract the force that is applied on the brake shoe by the disc springs, the clamping force on the brake disc will decrease. Increasing the hydraulic pressure will also decrease the deflection of the caliper yoke due to the supporting hydraulic force on the hydraulic unit. If the hydraulic pressure is increased to the caliper release pressure, the force of the disc springs will be balanced by the force on the piston and brake shoe assembly. At this point no clamping force is applied on the brake disc by the brake lining. If the hydraulic pressure is increased beyond the release pressure, the brake lining will be released from the brake disc and the gap (b) between the brake disc and the brake lining will be nonzero. Due to the backlash, the hydraulic unit will be pushed inward by the hydraulic pressure whereby the gap (e) and (f) will open and the gap (d) and (g) will close. The force on the hydraulic unit will thus transfer to the connector plate which is pressed onto the adjustment screw and the caliper yoke. If the pressure is increased until the piston reaches the connector plate, the gap (c) will close and the gap (b) will obtain its maximum. At this point the brake is fully released from the brake disc. At the rear stop of piston travel the force on the piston will counteract the force applied on the hydraulic unit, adjustment screw, connector plate and caliper yoke.

The current form of the brake caliper model is overly complex for dynamic modeling and therefore needs to be simpliﬁed. However, the model is suitable for detailed static modeling which will oﬀer great insight into the fundamental func- 3.4 Mechanical Components 61 tionality of the caliper. The static analysis of the complete brake caliper model will not be presented here.

3.4.2 Disc Springs Disc springs are stacked back-to-back in series as shown in ﬁgure 3.35 to achieve the caliper clamping force that is needed. Through the use of disc springs large

Figure 3.35: Spring stack. forces are obtained for small deflections. Force is exerted on both the brake shoe and the hydraulic unit as shown in the figure. The relation between spring deflection and disc spring force is nonlinear and is described by the following equation for a single disc [1]

4 ! ! ! 4 E t s h0 s h0 1 s F = 2 2 − − + 1 , (3.130) (1 − µ ) α Do t t t t 2 t where s is the spring deflection, E is the Young’s modulus and µ the Poisson’s ratio for the disc spring material used, Di and Do the inner and outer diameter of the disc, t is the material thickness and L0 and h0 is the unloaded inner and outer height of the disc respectively. Definitions are given in figure 3.36.

Figure 3.36: Disc spring.

The calculation coeﬃcient α in equation (3.130) is deﬁned in the following way

2 −1 1 δ − 1! δ + 1 2 ! α = − , (3.131) π δ δ − 1 ln δ 3.4 Mechanical Components 62 where δ is the ratio between the inner and outer diameter of the disc

Do δ = . (3.132) Di Equation (3.130) gives the force generated by a single disc. To obtain the force generated by a number of identical discs stacked in series as shown in 3.35, the total deflection of the stack is divided by the number of discs N to give the deflection of each individual disc N 1 X s = si. (3.133) N i=1 Hence, calculation of the total disc spring force may be carried out with the deflection of each individual disc spring. The total deflection of the stack of springs will then simply be equal to the individual disc spring deflection multiplied with the number of disc springs in the spring stack. As indicated in (3.130), the relation between disc spring force F and deflection s is nonlinear and therefore may generally not be linearly approximated. A plot of the characteristic for the disc springs used in the calipers however reveals a close to linear relation between spring force and deflection as seen in figure 3.37. One may linearize about a

100

[kN] 60 F

40 Force

0 0 5 10 15 20 Deﬂection s [mm]

Figure 3.37: Stack deﬂection and force. working point by calculating the slope of a tangent to (3.130) or use a secant between two points on the curve of 3.37.

The derivative of (3.130) with respect to spring deﬂection is

4 dF 4 E t 2 2 2 k = = 3 2 2 3s − 6h0s + 2(t + h0) . (3.134) ds 2 t (1 − µ ) α Do 3.4 Mechanical Components 63

The spring rate of a spring stack with a deﬂection-force characteristic as per 3.37 is calculated with (3.134) to yield the spring rate shown in ﬁgure 3.38. Despite

5.4 m] / 5.2 [MN

k 5

4.8

Spring Rate 4.6

0 5 10 15 20 Deﬂection s [mm]

Figure 3.38: Stack deﬂection and spring rate. the possible linearization of the force characteristic, (3.130) will be used to model the force generated by the disc springs.

Friction Frictional forces arise during compression and decompression of the spring stack because of internal friction due to elastic deformation of the material, contact between the discs and the guide rod and at the edges of the spring stack [12], see ﬁgure 3.39. Thus, when compressing the spring stack additional hydraulic force is needed in order to also overcome frictional forces. At decompression of the disc springs the frictional forces will act in the opposite direction, and the force on the spring stack must be lowered to a level below that of the spring force including the added resisting frictional forces in order to decompress the discs. Friction will thus have a hysteretic eﬀect on the control of the brake calipers. With friction acting against the applied hydraulic pressure, this will imply that release of the brakes will occur at a pressure higher than expected and that the level of pressure when applying the brakes will be lower than without the contribution of frictional forces. Calculation of the total spring force, including edge and surface frictional forces between individual discs when loading or unloading the springs, may be carried out as per DIN5 2092, where a percentile addition of frictional force to the calculated spring force is made n F ± Ff = F, (3.135) 1 ∓ wM (n − 1) ∓ wR 5DIN - Deutsches Institut für Normung 3.4 Mechanical Components 64

Figure 3.39: Friction during compression.

where n is the number of discs in parallel, wR is the coeﬃcient of edge friction and wM is the coeﬃcient of disc surface friction. For series stacking of the discs n = 1 and the friction force may therefore be calculated as

1 wR ±Ff = − 1 F = ± F. (3.136) 1 ∓ wR 1 ∓ wR A friction factor µ can be deﬁned such that

wR µ = , (3.137) 1 ∓ wR with the disc spring force F as a normal force acting perpendicular to the surface of contact between the springs and the abutment structure. The standard DIN 2093 divides disc springs into three diﬀerent categories depending on the ratio between the outer diameter Do and the thickness t of the disc. For a series C disc, which is the category of springs used in the brake calipers, the coeﬃcient of edge friction according to the standard is stated to lie in the range wR ∈ [0.03, 0.05]. Frictional forces due to edge friction is thus estimated to equal about 3-5 % of the spring force and is the dominant of the resistant frictional forces in the brake caliper.

3.4.3 Brake Lining When clamping force is applied on the brake disc by the calipers, the brake linings are compressed. The resulting deformation, due to the applied compressive force, is not negligible and thus must be accounted for in the modeling of the mechanical parts of the system. A characteristic is available in form of a stress-strain diagram for the linings in use [3] and the relationship between stress and strain is shown in ﬁgure 3.40. The stress-strain diagram is initially nonlinear and progressive in the elastic region before reaching a linear relation between stress and strain. For 3.4 Mechanical Components 65

Figure 3.40: Stress-strain characteristic. a sufficiently large stress the material will reach its yield strength, which implies a transition from the elastic region to the region of plastic deformation (not modeled). From the data available, the Young’s modulus E may be determined as the slope of the tangent to the stress-strain curve of the brake linings ∆σ dσ E = lim = . (3.138) ∆ε→0 ∆ε dε In the model of the system, the brake lining is represented by a spring element with stiffness k (see figure 3.34). Now, the stress-strain characteristic is material specific for the brake linings used whereas the stiffness depends on the dimensions of the linings. However, by assuming that strain is defined as the negative relative change in lining height L (engineering strain) and noting that

dσ d(F/A) dF L0 L0 E = = = − · = k , (3.139) dε d(−∆L/L0) d∆L A A where A and L0 are the area and the uncompressed height of the brake linings respectively, the spring stiffness k can be calculated through the following relation A k = E . (3.140) L0 This may intuitively be more well understood if one adopts the view of the brake lining as consisting of distributed infinitesimal spring elements connected both in series and in parallel. From the empirical data available, spline curve fitting may be employed to obtain numerical expressions of the stress-strain characteristic in figure 3.40. A plot of the sampled data points from the stress-strain diagram together with the fitted spline curve is shown in figure 3.41a. The Young’s modulus 3.4 Mechanical Components 66

5000 Empirical Data Curve Fit 4000

[kPa] 3000 σ

2000 Stress 1000

0 0.00 1.00 2.00 3.00 4.00 Strain ε [%] (a) Empirical stress-strain and curve ﬁt.

Approximation 0.20 Model [GPa]

E 0.15

0.10

0.05 Young’s Modulus 0.00 0.00 1.00 2.00 3.00 4.00 Strain ε [%] (b) Young’s Modulus.

5000 Empirical Data Curve Fit Model 4000

[kPa] 3000 σ

2000 Stress 1000

0 0.00 1.00 2.00 3.00 4.00 Strain ε [%] (c) Empirical, curve ﬁt and modeled stress-strain.

Figure 3.41: Stress-strain characteristic and Young’s Modulus. 3.4 Mechanical Components 67

E of the brake lining may be approximated through numerical derivation of the fitted characteristic and is shown in figure 3.41b. The Young’s modulus shows exponential increase in the lower region of strain before asymptotically reaching a maximum and constant value in the elastic region. An empirical model of the Young’s modulus may be represented by a piecewise defined function  0 ε ≤ 0 E = −ε/τ (3.141) E0 + (Emax − E0) 1 − e ε > 0, where E0 is the Young’s modulus at zero strain, Emax is the maximum value of the Young’s modulus in the elastic region and τ is a factor of exponential decay. Since no tensile forces are present, the Young’s modulus is non-zero only for positive strain. A plot of the model is found in figure 3.41b along with the estimated Young’s modulus. Through integration of (3.141) the relation between stress and strain is obtained  0 ε ≤ 0 σ = −ε/τ (3.142) E0 ε + (Emax − E0) ε − τ(1 − e ) ε > 0.

A plot of the available empirical data together with the spline fitted curve and modeled data is shown in figure 3.41c. Using (3.140), the spring stiffness k may finally be expressed as  A 0 ε ≤ 0 k = · −ε/τ (3.143) L0 E0 + (Emax − E0) 1 − e ε > 0.

3.4.4 Simplified Brake Caliper Model A simplification of the caliper model is needed in order to minimize the number of dynamic states and reduce the overall complexity. By neglecting backlash and assuming rigidity of the hydraulic unit, the adjustment screw, the connector plate and the caliper yoke assembly the model of the caliper will reduce to the single mass mechanical model shown in figure 3.42. The three distinct states of operation of the caliper described in section 3.4.1 are clearly illustrated in the figure. A frictional element has been added because of the significant friction that arises due to the use of disc springs in the caliper as was discussed in section 3.4.2. Applying Newton’s second law of motion yields the following second order differential equation

mx¨ = −F ± Ff − F1 + F2 + F3, (3.144) where m is the mass of the brake shoe and piston assembly, F is the force applied on the piston by the hydraulic pressure p, F1 is the force from the brake shoe and brake lining (see section 3.4.3), F2 is the force generated by the disc springs 3.4 Mechanical Components 68

(a) Brake applied.

(b) Brake released.

Figure 3.42: Simpliﬁed caliper model.

(see section 3.4.2), F3 the force applied on the piston at the rear stop by the connector plate and Ff represents the frictional forces. The force applied on the cylinder piston by the hydraulic pressure is F = pA, where A is the piston area. The forces F1 and F2 are calculated through the nonlinear functions given in section 3.4.2 and 3.4.3 and the force F3 generated by the caliper rear stop may be represented by a linear relation between spring deﬂection and force. Brake caliper frictional force Ff , as discussed in section 3.4.2, is added or subtracted depending on if the brakes are released or applied (i.e. if the brake pressure is increased or decreased).

Linear Analysis To simplify an analysis of the caliper model, a linear model of (3.144) may be obtained by neglecting friction and assuming linear models of the springs

mx¨ = −F − k1(x10 + x) + k2(x20 − x), (3.145) where x = 0 corresponds to the position where the brake lining is fully applied against the brake disc as per 3.42a and x10 and x20 are the initial deﬂections of 3.4 Mechanical Components 69 the springs. A static analysis of the caliper is of interest, therefore mx¨ = 0

−F − k1(x10 + x) + k2(x20 − x) = 0. (3.146)

With the brake applied and x = 0 when p = 0 equilibrium is required and thus

−k1x10 + k2x20 = 0

⇒ k1x10 = k2x20 = F0, (3.147) where F0 corresponds to the maximum applied clamping force produced by the caliper disc springs. With the brake applied onto the brake disc, i.e. when

−x10 < x ≤ 0, equation (3.146) will reduce to

−F − k1x − k2x = 0

⇒ F = −(k1 + k2) x. (3.148)

Replacing the force applied by the hydraulic pressure on the piston as equal to F = pA and solving for the position x yields A x = − p. (3.149) k1 + k2 When releasing the brake as shown in ﬁgure 3.42b, no clamping force is applied by the brake lining on the brake disc. Removing the term which represents the force from the brake lining in (3.146) will give

−F + F0 − k2x = 0

⇒ F = F0 − k2x. (3.150)

Solving for the position −L < x ≤ −x10 gives

F − F0 A x = − = − (p − p0), (3.151) k2 k2 where L − x10 is the length of travel of the caliper piston equal to the distance of the gap between the brake disc and lining at the rear stop in the hydraulic unit and p0 is the hydraulic pressure level corresponding to the maximum clamping force F0. The intersection of (3.149) and (3.151) is obtained by equating and solving for the pressure p ! A A k2 − p = − (p − p0) ⇒ p = 1 + p0, (3.152) k1 + k2 k2 k1 which is referred to as the brake release pressure or balancing pressure since at the release of the brake lining from the brake disc the force applied by the hydraulic pressure is balancing the force generated by the disc springs. For an 3.4 Mechanical Components 70

incompressible brake lining limk1→∞ p = p0 and in practice p ≈ p0 since k1 k2. Substituting the pressure p in both (3.149) and (3.151) with (3.152) gives x =

−F0/k1 = −x10 as deﬁned in (3.147).

At full release of the brake, i.e. where x ≤ −L, a force F3 is added to account for the rear stop of the caliper piston

−F + F0 − k2x + k3(−L − x) = 0. (3.153)

Solving for x in the above now yields ! 1 A k3L x = − (F − F0 + k3L) = − p − p0 + . (3.154) k2 + k3 k2 + k3 A

If k3 k2 and F − F0 k2 + k3 the above may be rewritten to show that

F − F0 k3L k3L x = − − ≈ − ≈ −L, (3.155) k2 + k3 k2 + k3 k2 + k3 which indicates a hard stop at x = −L.

The linear static model of the brake caliper may be summarized in a piecewise deﬁned function

 A − p 0 ≤ p ≤ pr  k1+k2  A x = − (p − p0) pr < p ≤ pf (3.156)  k2  A k3 − (p − p0 + L) p > pf , k2+k3 A where pr = p0(1 + k2/k1) is the release pressure and pf = p0 + k2A/L is the pressure at full release of the brakes. A plot of the function is shown in figure 3.43a for the case where k1 = 30 MN/m, k2 = 5 MN/m, k3 = 10 GN/m and F0 = 80 kN. As seen in the figure, below the release pressure pr at about 10 MPa the brake lining and disc springs will compress and decompress respectively. Once the brake lining is released from the brake disc, the rate of change in position will increase. It is important to understand that potential energy is stored in the caliper disc springs when increasing the pressure in the system and that the power required to open the calipers at a certain rate will increase throughout the whole range above release of the brake lining from the brake disc. This since power is calculated as P = ∆p Q, where ∆p is the pressure drop across the hydraulic pump (which will increase as the caliper pressure is increasing) and Q is the flow of fluid into the calipers.

Nonlinearity A nonlinear static analysis of the calipers can be carried out in much the same way as before, but with the nonlinear relations between disc spring deﬂection and force 3.4 Mechanical Components 71 according to (3.130) as well as between stress and strain given in (3.142) instead of the linear spring models used previously. Leaving out friction in (3.144), the condition at equilibrium is given by

−F − F1 + F2 + F3 = 0. (3.157)

By replacing the terms above the equation may be rewritten x − pA − Al σ1 x10 + + F2(x20 − x) + k3 max(0, −(x + L)) = 0, (3.158) tl where Al and tl is the brake surface area and thickness of the brake lining. The initial strain x10 and deﬂection x20 of the brake lining and disc springs are solved for by letting x = 0 and p = 0 which gives

−Al σ1(x01 ) + F2(x02 ) = 0

⇒ Al σ1(x01 ) = F2(x02 ) = F0. (3.159)

Equation (3.158) is solved numerically to qualitatively compare the linear and nonlinear static characteristic of the caliper. The position as a function of the hydraulic pressure is found in ﬁgure 3.43b. The essential diﬀerence between the linear and nonlinear characteristic in the static model of the caliper lies in the lower region below the release pressure where the brake lining characteristic is nonlinear. Friction may also be added in a static analysis such that x − pA − Al σ1 x10 + + F2(x20 − x) tl

± min( pA, µF2(x20 − x)) + k3 max(0, −(x + L) = 0. (3.160)

When increasing the brake pressure with initial full application of the brake linings, this will imply that the hydraulic pressure must overcome the friction force of the discs springs before any movement of the caliper brake shoe is possible. Plotting the full characteristic for both release and application of the brakes shows the hysteretic eﬀect of disc spring friction and that the actual release pressure pr of the brakes due to friction is increased above what was stated earlier. A pressure level of brake application when applying the brakes from a released state may also be deﬁned and is denoted pa, where pa < pr. 3.4 Mechanical Components 72

[mm] −1 x

−2 Position

−3

0 2 4 6 8 10 12 14 Pressure p [MPa] (a) Linear characteristic.

[mm] −1 x

−2 Position

−3

0 2 4 6 8 10 12 14 Pressure p [MPa] (b) Nonlinear characteristic.

[mm] −1 x

−2

Position Without Friction − Release 3 Application

0 2 4 6 8 10 12 14 Pressure p [MPa] (c) Nonlinear characteristics with friction.

Figure 3.43: Brake caliper static characteristics. 3.5 Brake Stand 73

3.5 Brake Stand

The calipers of the hydraulic brake system are mounted on a brake stand and connected in parallel. Since the hydraulic transmission lines are left out in the model of the system, a simplification is justified as the calipers may be seen as lumped into a single equivalent caliper unit. Referring to figure 3.44, the disc

Figure 3.44: Brake calipers. springs are effectively connected in parallel and the total clamping force Ftotal applied on the brake disc thus equals 2N X Ftotal = FCi = 2N × FC , (3.161) i=1 where N is the number of calipers and FC is the clamping force generated by a single caliper half. The hydraulic pressure is applied on each piston and the total piston area Atotal may thus in the same fashion be calculated as 2N X Atotal = Ai = 2N × A, (3.162) i=1 where A is the piston area in each caliper half. From a modeling point, this will have no significance regarding calculation of caliper disc spring force since there is a constant proportion between clamping force and piston area. However, calculation of the total volume of the hydraulic system must take into account the number of calipers as this will affect the pressure dynamics and release time.

The total volume of fluid in the system will increase when the brake pressure is increased and the change in volume will equal ∆V = −2NA ∆x where N is the number of brake calipers, A is the piston area and x is the position of the piston in each caliper as defined in figure 3.42. Initially the position of the piston will be x = 0 with all brake linings fully applied onto the brake disc and the fluid volume of the system will be equal to V = V0. With change of position, the volume may be calculated as V = V0 − 2NA x. (3.163) The time rate of change in system volume will equal dV dx = −2NA . (3.164) dt dt 3.6 Dynamic Model 74

Substituting (3.163) and (3.164) into (3.107) will yield the pressure dynamics in the system as ! dp βe dx = Q + 2NA . (3.165) dt V0 − 2NA x dt

At a pressure level p below the release pressure pr of the brakes, and with a ﬂow Q into the volume equal to the ﬂuid displacement of the hydraulic pump, the change in position will be small and pressure will rise quite rapidly. Once the brake linings are released from the brake disc, increase in the system volume will however lower the time rate of change of the brake pressure as is seen in equation (3.165).

An equivalent lumped single brake caliper must also take into account the total moving mass mtotal, which simply will equal the sum of the mass m of the moving parts of each caliper half

2N X mtotal = mi = 2N × m. (3.166) i=1 3.6 Dynamic Model

In this section a complete dynamic model is presented along with a number of static functions that has been presented throughout the text. The dynamic model of the system consists of a system of nonlinear differential equations including a number of different static functions, see equations (3.167a) - (3.167e) and equations (3.167f) - (3.167r). The model has five different dynamic states xi where x1 represents the spool position, x2 spool velocity, x3 brake pressure, x4 brake caliper piston position and x5 piston velocity. Equation (3.167a) integrates the spool velocity x2 to obtain the spool position x1 and equation (3.167b) describes the dynamics between the valve input current u and the main spool position x1, as presented in section 3.3.1. The servo valve has both a mechanical and electrical offset which effectively will shift the pressure sensitivity curve with respect to the input current depending on the sign and magnitude of the offset. An offset u0 has therefore been added to the input to account for both mechanical and electrical offset of the spool position at zero input u = 0. Without any offset, the pressure will ideally be equal to PS/2 (which may adjusted mechanically by shift- ing the valve sleeve). The dynamics of the brake pressure is given by equation (3.167c), which was presented in section 3.2.4 and section 3.5. Equation (3.167d) integrates piston velocity to obtain piston displacement and equation (3.167e) calculates piston acceleration as was presented in section 3.4.4. The force from the brake lining, the disc springs, the rear hard stop and the hydraulic pressure are summed to calculate the resultant accelerating force acting on the mass as seen in equation (3.167e). Disc spring friction is not included in the dynamic model 3.6 Dynamic Model 75 of the system. Equation (3.167f) calculates the effective bulk modulus as given in section 3.1.3. Flow through the valve is calculated with equations (3.167g) - (3.167o), which were given in section 3.2.3, 3.3.1 and 3.2.1. Calculation of the force from the brake lining is carried out through the empiric stress-strain function in (3.167p) as given in section 3.4.3 multiplied with the lining width and height, which yields the lining area in contact with the brake disc and thus the resultant force. Equations (3.167q) - (3.167r) calculates the force from the disc springs and were presented in section 3.4.2. Dampers are not included, but are simple to add in equation (3.167e) if needed. 3.6 Dynamic Model 76

 x x  ˙ 1 = 2 (3.167a)   x −ω2 x − ζ ω x ω2 u u  ˙ 2 = 0 1 2 0 2 + 0 ( + 0) (3.167b)   βe(x3)  x Q x , x NA x  ˙ 3 = ( ( 1 3) + 2 5 ) (3.167c)  V0 − 2NA x4   x˙ 4 = x5 (3.167d)    1 1  x˙ 5 = −A x3 − F1 x10 + x4 (3.167e)  Nm t  2 l    + F2(x20 − x4) + max 0, −K (x4 + L)     1   P γ  R + 1 +   P0   βe(P ) = βl   (3.167f)   1   βl P γ  R γ(P +P ) + 1 + P  0 0   Q(x, p) = QS(x, p) − QR(x, p) (3.167g)   πD x q  hS ( ) 2 2 2  QS(x, p) = −k1µ + µ k + 8k2ρD (x)(PS − p) (3.167h)  1 hS  8ρk2   q  πDhR (x) 2 2 2  QR(x, p) = −k1µ + µ k + 8k2ρD (x)(p − PR) (3.167i)  1 hR  8ρk2  2 k1 = 1/δ (3.167j)  2  k2 = 1/C (3.167k)  d  s   4AS(x)  DhS (x) = (3.167l)  π   s  A x  4 R( )  DhR (x) = (3.167m)  π     x0 + x x ≥ 0  A x w  S( ) = 2 −1 (3.167n)  x (x0 − kx) x < 0  0     x0 − x x ≥ 0  AR(x) = w (3.167o)  2 −1  x0 (x0 + kx) x < 0      ε <  0 0  F1(ε) = wl hl −ε/τ (3.167p)  E0 ε + (Emax − E0) ε − τ(1 − e ) ε ≥ 0    4 ! ! !  4 E t s h0 s h0 1 s  F2(s) = − − + 1 (3.167q)  − µ2 α D2 t t t t t  (1 ) o 2   !2 !−1  1 δ − 1 δ + 1 2  α = − (3.167r)  π δ δ − 1 ln δ Chapter 4

Model Parameterisation, Simulation and Validation

In the previous chapter, models of the different components of the system were obtained through physical modeling and a complete model of the system was given in form of a system of differential equations along with a number of static functions. Parameterisation of the model is required in order to be able to simu- late the system and will be presented in this chapter. Parameters may either be gathered directly from component specifications and other known data or indirectly through calculations and measurements. Comparisons between measured data against static and dynamic model computations and simulation will also be made to verify the correctness of the model.

4.1 Model Parameters

4.1.1 Fluid Properties Most of the information needed regarding the properties of the hydraulic fluid is either stated by the supplier of the fluid or given as tabulated values. Both density ρ and kinematic viscosity ν are found in the product data sheet of the oil and are given in table 4.1. However, data regarding the liquid bulk modulus βl is not readily available and will be assumed equal to 1.8 GPa (see section 3.1.2). The total volume of fluid in the system V0 may be estimated by calculation of volume in the pipelines connected between the valve manifold and the brake stand. Pockets of oil also exist in the piston chamber of each caliper, the manifold and ball valves which gives a small discrepancy between the actual oil volume and the calculated. The atmospheric pressure P0 and the adiabatic index γ are both given as known constants and the tank return pressure PR is assumed equal to zero (gauge pressure). Regarding supply pressure PS, this will be taken from available measurements of the accumulator pressure. The fraction of entrained air 4.1 Model Parameters 78

Table 4.1: Fluid Properties

Parameter Symbol Unit Value Fluid Density ρ kg/m3 876 at 15 °C Fluid Bulk Modulus βl GPa 1.8 Kinematic Viscosity ν cSt 46 at 40 °C cSt 6.8 at 100 °C Fraction of Air R % Adiabatic Index γ 1.4 Atmospheric Pressure P0 bar 1.01325 Supply Pressure PS bar Return Pressure PR bar 0.0 Volume V0 liter 4.44

R may be identified in different ways, but will in this case be roughly estimated by curve fitting to measured data.

4.1.2 Brake Calipers Data needed for the overall parameterisation of the brake caliper model is listed in table 4.2 Table 4.2: Brake Caliper Parameters

Parameter Symbol Unit Value Mass m kg 43.0 Stroke L mm 2.0 Piston Area A cm2 74.6 Maximum Clamping Force F0 kN 75.6 Number of Calipers N 4 Rear Stop Spring Rate K GN/m 10.0

Two single unit calipers and a double unit caliper is installed on the brake stand, which is equivalent to a total of N = 4 calipers. Mass of the individual parts in each brake caliper halve is specified in the installation and maintenance manual available from the manufacturer, and the total moving mass m is obtained by summation. Through direct measurements of the inner and outer diameter of the caliper piston, the area A is calculated as π π A = D2 − D2 = 122 − 72 = 74.6 cm2, (4.1) 4 o i 4 which is confirmed by the manufacturer data where piston area is specified to be 75 cm2 in each brake caliper halve. Force transducer measurements of the clamping force F0 gives a measured maximum force of about 76 kN at full application 4.1 Model Parameters 79 of the brakes. The spring rate of the caliper rear stop is set to a large value, in this case 10 GN/m.

Brake Linings

Dimensions of the width wl, height hl and thickness tl of the brake linings are given in table 4.3. As mentioned in the previous section, brake lining force is calculated from the lining area wl × hl applied against the brake disc and strain of the lining is obtained using the thickness of the lining according to what is given in (3.158). Through curve ﬁtting of the static stress-strain function given in section 3.4.3 against the available empirical data, values of the maximum and minimum Young’s modulus Emax and E0 and the exponential decay τ may be estimated. Table 4.3: Brake Lining Parameters

Parameter Symbol Unit Value Young’s Modulus E0 GPa 0.01 Emax GPa 0.15 Decay τ 0.08 Width wl mm 300.0 Height hl mm 202.5 Thickness tl mm 25.0

Disc Springs All information that is needed for model parameterisation of the disc springs are either provided directly by the manufacturer or found available in form of tabulated material specific data. The dimensions of the individual disc springs as defined in figure 3.36 are given in table 4.4 below. The number of disc springs

Table 4.4: Disc Spring Parameters

Parameter Symbol Unit Value Young’s Modulus E GPa 206 Poisson’s Ratio ν 0.3 Outer Diameter Do mm 125 Inner Diameter Di mm 64 Inner Height h0 mm 2.6 Thickness t mm 8.0 Number of Discs N 9

N in each spring stack is specified by the manufacturer and confirmed through 4.1 Model Parameters 80 visual inspection in figure 3.26. The Young’s modulus E and Poisson’s ratio ν are both given through readily available spring material data.

4.1.3 Servo Valve Data needed for parameterisation of the servo valve model can to some extent be derived from what is found available in the valve datasheet. Other values are either indirectly obtained from measurements or approximated. Rated values of valve pressure, flow and current are given in table 4.5. Rated flow is measured under a no-load condition with a short-circuit connection between the actuator ports at maximum spool actuation, which corresponds to rated current input to the valve. Pressure ∆P = PS − PL − PR = PS − PR is held constant across the supply and return line connections (see figure 3.17 and 3.18), thus the pressure drop across each orifice will ideally equal (PS −PR)/2. By convention the pressure across the supply and return line is chosen to be 70 bar (about 1000 psi1) which yields a pressure drop ∆P = 35 bar across the valve orifices. Referring to figure

Table 4.5: Valve Rating

Parameter Symbol Unit Value Rated Pressure Prated bar 210 Rated Flow Qrated l/min 40 Rated Current Irated mA 300

3.19, a linearization of the current-flow graph is desired. By extrapolating the tangent of the curve at the origin to the point on the graph where Q = Qrated, ∗ a linearized rated current Irated is obtained (i.e. the current at which rated flow is obtained without saturation, see the example depicted in figure 3.20). The ∗ linearized rated current Irated < Irated is estimated from the current-flow graph ∗ ∗ and Irated ≈ 249 mA. A linearized rated flow Qrated may also be defined such that ∗ Qrated Qrated = ∗ . (4.2) Irated Irated ∗ Calculation of Qrated now yields 0.300 Q∗ = × 40 ≈ 48.2 l/min. (4.3) rated 0.249 From the above, and by using the orifice equation, the maximum orifice opening area Amax may be calculated as √ 2 × 0.300/0.249 × 40/60000 2 Amax = q √ ≈ 14.3 mm . (4.4) 2 5 0.63 × 876 × 70 · 10

1 1 bar = 14.5037738 pounds/inch2 4.1 Model Parameters 81

Since Amax A0, and thus w ≈ Amax/xmax, the area gradient w may be calculated from the slope of the tangent through the origin of the current-ﬂow characteristics √ 2 Qrated w ≈ √ , (4.5) q 2 ∗ Cd ρ ∆p Irated which yields √ 2 × 40/60000 2 w ≈ q √ ≈ 47.5 mm /A. (4.6) 2 5 0.63 × 876 × 70 · 10 × 0.249

However, a normalized spool position is used in the model which implies that the gradient must be scaled with the rated current to obtain the normalized area gradient such that √ 2 × 40/60000 2 w ≈ q √ ≈ 14.3 mm . (4.7) 2 5 0.63 × 876 × 70 · 10 × 0.249/0.300

Thus, the maximum opening area of each oriﬁce is equal to about 14.3 mm2, as was calculated in (4.4).

The minimum spool opening x0 may be calculated by solving for x0 with x = 0 in equation (3.127) [5]

q x0 QS0 QS0 = 2K PS − PR q ⇒ x0 = √ √ , (4.8) 1 + f(0) 2K PS − PR where the flow gain K is equal to √ 2 Qrated K = √ . (4.9) 5 ∗ 70 · 10 Irated The total maximum valve leakage flow (quiescent flow), including both the flow through the pilot and main spool stage, is stated by the manufacturer to be less than 5 % of the rated flow at rated pressure. The valve leakage flow is thus distributed between the pilot and main spool stage. With lack of data regarding the main stage spool leakage, QS0 represents a degree of freedom in the model of the servo valve. Other ways of obtaining x0 exists however. If x = x0 one may solve for the leakage coefficient k in equation (3.125). Evaluating (3.126) for x = x0 gives

!2 !2 |x0| |x0| 2 f(x0) = 1 + 1 + k = 4 (1 + k) . (4.10) x0 x0 4.1 Model Parameters 82

Solving for k with the above yields v v u u 1 u PA(x0) 1 u PA(x0)/PS k = t − 1 = t − 1, (4.11) 2 PS − PA(x0) 2 1 − PA(x0)/PS where PA(x0)/PS is obtained from the valve pressure sensitivity graph, see ﬁgure 3.22b. Note that in order to obtain a feasible solution k > 0, it is required that

PA(x0)/PS 4 > 4 ⇒ PA(x0)/PS > = 0.80, (4.12) 1 − PA(x0)/PS 5 which numerically will constrain the minimum spool opening x0 and in turn also the main spool leakage flow QS0 . From the pressure sensitivity graph one sees that x0 > 0.006. Calculating the corresponding null main spool leakage flow gives √ 2 Prated Qrated QS = √ x0 0 5 ∗ 70 · 10 Irated √ 2 × 210 · 105 × 40/60000 = √ × 0.006 × 0.300 ≈ 1.00 l/min, (4.13) 70 · 105 × 0.249 which is half of the maximum total quiescent flow. With the actuator ports blocked as illustrated in figure 3.21, the orifice leakage area may be calculated as √ 2 × 0.5 × 1.00/60000 2 A0 ≈ q √ ≈ 0.086 mm . (4.14) 2 5 0.63 × 876 × 210 · 10

Referring to equation (3.121), the minimum oriﬁce area A0 may also be calculated from the area gradient and the minimum spool opening since

A0 w = ⇒ A0 = wx0, (4.15) x0 which gives 2 A0 ≈ 14.3 × 0.006 ≈ 0.086 mm . (4.16)

The maximum attained value of the minimum spool opening x0 may be calculated for the case where the main spool stage alone consumes the whole quiescent ﬂow at rated pressure √ 70 · 105 × 0.249 x0 = √ × 0.05 × 40/60000 ≈ 3.59 mA. (4.17) 2 × 210 · 105 × 40/60000

Calculating the relative spool opening gives 3.59/300 ≈ 0.012. The nominal working pressure of the system is 145 bar and thus well below that of the rated pressure. A lower working pressure will decrease the quiescent leakage ﬂow in the 4.1 Model Parameters 83

main spool stage and calculating the flow QS0 for the case where PS = 145 bar and x0 = 0.006 yields √ 2 × 145 · 105 × 40/60000 QS = √ × 0.006 × 0.300 ≈ 0.83 l/min. (4.18) 0 70 · 105 × 0.249 The calculations presented correspond to parameters valid for a newly manufac- tured valve. The control lands of the valve spool will however wear with time and increase the null leakage and thus also the minimum spool opening x0. Without available measurements of the center flow one may instead estimate x0 through curve fitting of the pressure sensitivity graph against measurements of equilibrium pressure obtained with open-loop current input to the servo valve, see figure 4.1a and 4.1b. By plotting the relative input current and pressure (the rated current and supply pressure is used to normalize the current and brake pressure), adjustments of x0, PA(x0) and u0 are made to fit the pressure sensitivity graph against the measured data as seen in figure 4.1c. By adjustment of the parameters, the minimum spool opening is found to be approximately x0 ≈ 0.0165, which is higher than what was calculated as the expected maximum according to the data sheet. This may however be explained by the inevitable wear of the valve spool which will increase x0 with time. Valve port pressure PA(x0) with the actuator ports blocked or under equilibrium conditions (which determines the leakage coefficient k), is set to PA(x0)/PS = PÂ(x0) ≈ 0.82. The offset u0 simply shifts the curve to adapt for mechanical and electrical offset, hence u0 = u0m +u0e , and is found to be u0 ≈ −0.02. The valve frequency response from which the natural frequency ω0 and damping ζ may be estimated is provided by the valve manufacturer. The valve coefficient of discharge Cd and laminar flow coefficient δ are given typical values. It should be noted that the rated values provided by the manufacturer are rounded and therefore approximate. Nevertheless, the calculated values serve an indicative purpose.

Table 4.6: Servo Valve Parameters

Parameter Symbol Unit Value Natural Frequency ω0 rad/s 160 Damping ζ 0.6 Laminar Flow Coefficient δ 0.10 Discharge Coefficient Cd 0.63 Area Gradient w mm2 14.3 Minimum Spool Opening x0 0.0165 Normalized Pressure Sensitivity PÂ(x0) 0.82 at Minimum Spool Opening Spool offset u0 -0.02 4.1 Model Parameters 84

0 [%] i −5 Current −10

−15 0 20 40 60 80 100 120 Time t [s] (a) Signal input.

100

80 [bar]

p 60

40 Pressure 20

0 0 20 40 60 80 100 120 140 Time t [s] (b) Step response.

100

80 [%] p 60

40 Pressure 20

0 −6 −4 −2 0 2 4 6 8 10 Current i [%] (c) Pressure sensitivity.

Figure 4.1: Measurements and pressure sensitivity. 4.2 Static Characteristics 85

4.2 Static Characteristics

The static characteristic of the brake caliper in figure 3.43b is obtained by simultaneously measuring both the brake pressure and displacement of the center rod as a function of time. By slowly varying the pressure linearly with the brakes applied and increasing the pressure until the brakes are fully released, the characteristics can be plotted in a xy-graph as shown in figure 4.2a. The position of the center rod is measured using an inductive proximity sensor, which is positioned a few millimeters from the rod in the fully applied state of the brake caliper. Hence, when the brake pressure is at its minimum level the measured distance between the sensor and the center rod will be at its maximum (which in the model of the system is used as a position of reference for which x = 0). The brake pressure varies from about 20 bar2 to the maximum working pressure of about 145 bar. When increasing the hydraulic pressure the brake lining will decompress and the distance between the center rod and the sensor will decrease. Note that nonlinearities exist in the range above the release pressure which is not reflected by the linear model of the brake caliper due to the simplification of the model as was discussed in section 3.4.1 and 3.4.4. Hysteretic effects of friction on the static characteristics of the brake caliper may also be measured by slowly decreasing the brake pressure from the fully released state and maximum working pressure to the pressure at the minimum level. This will yield the hysteresis loop of the static characteristics as seen in figure 4.2b.

4.3 Disc Spring Force

The calculated disc spring force is conﬁrmed correct when compared to available manufacturer clamping force data as tabulated in table 4.7

Table 4.7: Clamping Force and Air Gap Adjustment

Air Gap Unit Force Unit 1 mm 80100 N 2 mm 75500 N 3 mm 70900 N

The air gap corresponds to the gap between the brake lining and the brake disc at full release of the brakes (see figure 3.32). Manufacturer data in form of an inspection certificate specifies a force of 80100 N at a stack length of 79.1 mm. With an unloaded spring height of L0 = h0 + t = 2.6 mm + 8.0 mm = 10.6 mm per disc, the total stack length equals L = N × L0 = 9 × 10.6 mm = 95.4 mm.

2 Brake pressure and center rod displacement measured using a diﬀerent system with a lower pressure limitation set to 20 bar. 4.3 Disc Spring Force 86

5.5

5 [mm] x 4.5

Position 4

0 20 40 60 80 100 120 140 Pressure p [bar] (a) Static characteristics.

5.5

5 [mm] x 4.5

Position 4

0 20 40 60 80 100 120 140 Pressure p [bar] (b) Hysteresis.

Figure 4.2: Measured characteristics. 4.3 Disc Spring Force 87

Thus, at a stack deflection of s = 95.4 mm − 79.1 mm = 16.3 mm the calculated clamping force should be equal to about 80100 N. This is confirmed by calculating the disc spring force with equation (3.130), where the deflection of each single disc in the spring stack is equal to 16.3/9 ≈ 1.81111 mm. Calculation of δ and α gives 0.125 δ = = 1.953125 (4.19) 0.064 and 2 δ−1 1 δ α = δ+1 2 ≈ 0.684474. (4.20) π δ−1 − ln δ The coefficient is now calculated as 4 × 2.06 · 1011 × (8 · 10−3)4 ≈ 3.467919 · 105, (4.21) (1 − 0.32) × α × (125 · 10−3)2 which yields a spring force equal to

F ≈ 3.467919 · 105× (4.22)

1.811111 2.6 1.811111 2.6 1 1.811111 − − · + 1 8.0 8.0 8.0 8.0 2 8.0 ≈ 3.467919 · 105 × 0.231117 ≈ 80150 N.

For a gap of 2 mm between the lining and brake disc at full release of the brakes, the deﬂection of the spring stack will be equal to 15.3 mm with maximum caliper clamping force. The spring force is calculated in the same way with a spring deﬂection of each disc equal to 15.3/9 = 1.7 mm to give

F ≈ 3.467919 · 105× (4.23)

1.700000 2.6 1.700000 2.6 1 1.700000 − − · + 1 8.0 8.0 8.0 8.0 2 8.0 ≈ 3.467919 · 105 × 0.217729 ≈ 75506 N.

At a gap of 3 mm, the deﬂection of the spring stack will equal 14.3 mm with maximum clamping force. Spring deﬂection of each spring will now equal 14.3/9 = 1.588889 mm, which gives

F ≈ 3.467919 · 105× (4.24)

1.588889 2.6 1.588889 2.6 1 1.588889 − − · + 1 8.0 8.0 8.0 8.0 2 8.0 ≈ 3.467919 · 105 × 0.204277 ≈ 70842 N. 4.4 Pressure Dynamics 88

4.4 Pressure Dynamics

It was concluded in section 3.2.4 that the calipers, when applied against the brake disc, were identical to a hydraulic capacitance and that the time rate of change of the system pressure therefore should be proportional to the input of flow to the system (i.e. the system is integrating, see equation (3.108)). This is confirmed when applying a sinusoidal input current signal to the servo valve as is shown in figure 4.3a. The working point is placed in the upper region of the pressure range below the release pressure pr (at about 100 bar) where the effective bulk modulus is close to that of the fluid and the amplitude of the current signal is chosen to limit the peak-to-peak pressure response such that the changes in the pressure drop ∆p is moderate. As seen in figure 4.3a, the pressure lags the input signal by 90° (the input current signal is scaled to the same amplitude and centered at the peak-to-peak midpoint of the pressure response for phase comparison). The integrating nature of the system is also confirmed when measuring the response to a triangular and square wave input as in figure 4.3b and 4.3c.

4.4.1 Simulation The cyclic input signals confirms the integrating character of the system in steady state. By input of a step-shaped open-loop signal to the servo valve, the transient behavior of the system is measured, as was shown in figure 4.1b. The dynamic model given in (3.167a) - (3.167e) together with the static functions (3.167f) - (3.167r) is simulated with an input signal identical to the one shown in figure 4.4a, and the simulated pressure response along with actual measured data and supply pressure (dashed) are plotted in figure 4.4b. It must be stressed that this is far from a normal operating mode of the valve and used here only for the purpose of validating the dynamic model. Apart from the hysteresis, the qualitative behavior of the actual system is captured quite well by the simulation. Quantitatively the pressure sensitivity function has not been optimally fitted to the measured data, which also explains deviations between the actual and simulated system. The effects of entrained air is shown close to t = 0 s, which is also matched by the model. 4.4 Pressure Dynamics 89

100 Signal Pressure 90 [bar]

p 80

70 Pressure

0 0.2 0.4 0.6 0.8 Time t [s] (a) Sinusoidal response.

Signal 80 Pressure [bar]

p 70

Pressure 60

50 0 0.2 0.4 0.6 0.8 Time t [s] (b) Triangular response.

60 Signal 50 Pressure

[bar] 40 p 30

Pressure 20

0 0.2 0.4 0.6 0.8 Time t [s] (c) Square response.

Figure 4.3: Pressure response with diﬀerent signal inputs. 4.4 Pressure Dynamics 90

0 [%] i −5 Current −10

−15 0 20 40 60 80 100 120 140 Time t [s] (a) Signal input.

150

100 [bar] p

50 Pressure Measurement Simulation 0 0 20 40 60 80 100 120 140 Time t [s] (b) Actual and simulated pressure response.

Figure 4.4: Simulation. Chapter 5

Discussion and Conclusions

The work presented in this report covers a quite broad range of fluid mechanical and mechanical physical modeling. Although the scope of the modeling of the system was limited to essentially only include the servo valve and brake caliper models, the complete model yet attains an appreciable number of differential equations and static functions. From a perspective of model complexity, this of course speaks in favor of computer aided component based modeling where initial configuration and changes in the model, if needed, are more or less trivial. This will especially facilitate the modeling of large complex systems such as the complete configuration of the hydraulic brake system. One may of course also claim this way of modeling to be less prone to error. However, a trade-off between simplifying the process of system modeling and insight to and understanding of the underlying analytical relations exists. In the case of component based physical modeling, thorough documentation of each individual model of course is necessary. Yet, one lacks analytical completeness of an interconnected system and is therefore dependent on numerical analysis in order to analyze system dynamics. The same argument may be used when comparing so called white-box physical modeling with black-box modeling. Another advantage of a physically based model is of course also the ability to extrapolate and make predictions of the effects of changes to the model parameters (e.g. pipe length, volume, mass etc). This is not be replicated by a black-box model due to its numerical nature. However, even if an accurate, more detailed and complex physically based model is readily available, model parameterisation will prove to be difficult if data is not found a priori. Even if parameters are at hand, the resulting model may not qualitatively or quantitatively outweigh the benefits of a more simple model, as is the case of the model of the servo valve dynamics presented in section 3.3.1 or the complete brake caliper model in section 3.4.1 compared to the simplified model that was presented in section 3.4.4. This is also the case for the calculation of disc spring force given in section 3.4.2, where a secant based simple linear model may be used instead. Hydraulic systems are challenging in the sense of fluid dependency on temperature, pressure and entrained air but also from the perspective 92 of mathematical complexity due to a vast number of nonlinear relations between fluid pressure and flow, fluid pressure and effective bulk modulus etc. A lack of knowledge regarding exact valve geometry may also be problematic. For this reason numerous simplifications and assumptions are needed in order to be able to handle the modeling of a hydraulic system rationally and examples of this has been shown throughout the text. The model of the system presented in this work is by no means complete and much work still remains regarding validation of the dynamic model. The aim has been to capture the fundamentals of the system regarding feedback control of the brake pressure, this in order to obtain an analytical model that can be used for e.g. closed-loop control analysis. Many of the nonlinearities such as valve hysteresis, saturation, effects of under- or overlapping spool geometry, flow forces, velocity limitations and brake caliper frictional forces have been intentionally left out in order not to make the model overly complex. In its current form, the model is limited to the lower region of the frequency domain only. The inclusion of transmission lines is therefore required to complement the model and modeling of the piping network may be carried out through the principles of modal approximation techniques for hydraulic transmission lines. In practice, dynamics will differ and vary depending on actual installations. It is therefore of interest to predict resonance frequencies, bandwidth and also the required duration of initial brake application at emergency stops of a mine hoist to be able to give guidelines and recommendations regarding pipe installation length and pipe dimensions. Assumption on an ideal pressure source in form of a constant supply pressure is not correct, and modeling of the accumulator and variable displacement pump connected to the servo valve is needed to include the effects of a varying system supply pressure. A more complex and near full scale model with inclusion of auxiliary components into the model should prefer- ably be implemented in a component based modeling environment due to the reasons discussed earlier. Static modeling of the mechanical parts of the system in form of disc spring force and brake lining stress-strain characteristics were verified quantitatively correct against available data. More detailed insight into and understanding of the disc spring frictional forces, such as the frictional force between the discs and the guide rod, are however needed. A complete model of the brake caliper was presented in section 3.4.1, and as mentioned before a static analysis of the caliper may be carried out in order to gain detail understanding of the caliper functionality. Static analysis of the simplified caliper model was also verified to be essentially correct against actual measurements of brake caliper center rod displacement and brake pressure. Measurements taken with cyclic inputs in form of sine, triangular and square shaped signals clearly showed the integrating character of the system in the brake applied state. Regarding parameterisation of the model, most fluid data is normally available through supplier data or given as known constants. The fraction of fluid entrained air is however a common unknown which needs to be identified. Other mechanical data is also to a large extent available or easily measured. Bibliography

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