Mobile Hydraulics

Mobile Hydraulics

Workbook TP 800

With CD-ROM

C (X1)

D (X2)

A

B

LS P T

B

A

T

T C

D

P

LS LS1

LS2

P

Festo Didactic 574166 en

Order No.: Edition: Authors: Graphics:

574166 11/2011 (V0.9) Levent Unan Levent Unan

© Festo Didactic GmbH & Co. KG, 73770 Denkendorf, Germany, 2011 Internet: www.festo-didactic.com E-mail: [email protected] The copying, distribution and utilization of this document as well as the communication of its contents to others without expressed authorization is prohibited. Offenders will be held liable for the payment of damages. All rights reserved, in particular the right to carry out patent, utility model or ornamental design registration.

Contents Use for intended purpose _________________________________________________________________ VII Preface ________________________________________________________________________________ VIII Introduction ______________________________________________________________________________X Work and safety instructions ________________________________________________________________ XI Training package for mobile hydraulics (TP 800) _______________________________________________ XIII

Equipment set TP 801 ______________________________________________________________________1 Allocation of components to exercises _________________________________________________________2 Notes for the teacher/trainer_________________________________________________________________3 Exercises and solutions TP 801 ______________________________________________________________5 Exercise 1 Earth driller (uncompensated pressure) _______________________________________________5 Exercise 2 Earth driller (compensated pressure) _______________________________________________ 13 Exercise 3 Earth driller (proportional control) _________________________________________________ 19 Exercise 4 Earth driller (proportional control) _________________________________________________ 23 Exercise 5 Earth driller (proportional control with open-centre load-sensing system) _________________ 27 Exercise 6 Earth driller (6/3-way proportional valve) ____________________________________________ 31 Exercise 7 Boom cylinder of an excavator _____________________________________________________ 35 Exercise 8 Boom cylinder of an excavator (pilot check valve) _____________________________________ 39 Exercise 9 Boom cylinder of an excavator (counterbalance valve) _________________________________ 43 Exercise 10 Boom cylinder of an excavator (piloted counterbalance valve) __________________________ 49 Exercise 11 Two hydro motor of a crane system (various valve configurations) ______________________ 55

Equipment set TP 802 ____________________________________________________________________ 61 Allocation of components to exercises _______________________________________________________ 61 Notes for the teacher/trainer_______________________________________________________________ 62 Exercises and solutions TP 802 ____________________________________________________________ Exercise 1 Steering system of an agricultural machine __________________________________________ Exercise 2 Calculating the turning of the steering wheel _________________________________________ Exercise 3 Steering system of an off-road vehicle (shock valve) ___________________________________ Exercise 4 Steering system of a street sweeping vehicle (priority function) __________________________ Exercise 5 Steering system of an articulated vehicle ____________________________________________

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65 65 71 75 79 83

III

Equipment set TP 803 ____________________________________________________________________ 87 Allocation of components to exercises _______________________________________________________ 88 Notes for the teacher/trainer_______________________________________________________________ 89 Exercises and solutions TP 803 ____________________________________________________________ 91 Exercise 1 Mobile block on an excavator (with load-sensing unit) _________________________________ 91 Exercise 2 Mobile block on an excavator (with load-sensing unit) _________________________________ 97 Exercise 3 Remote control of mobile valves (hydraulic joystick) ___________________________________ 99 Exercise 4 Earth driller (load and speed) ____________________________________________________ 103 Exercise 5 Earth driller (pre-compensation) __________________________________________________ 107 Exercise 6 Earth driller (Pressure compensation with LS system) _________________________________ 111 Exercise 7 Two hydro motors of a crane system (with LS system) ________________________________ 115 Exercise 8 Two hydro motors of a crane system (pre-compensation) ______________________________ 119 Exercise 9 Two hydro motors of a crane system (mobile block with pre-compensation for each actuator) 123 Exercise 10 Two hydro motor of a crane system (post-compensation) _____________________________ 127

Exercises and worksheets TP 801 Exercise 1 Earth driller (uncompensated pressure) _______________________________________________1 Exercise 2 Earth driller (compensated pressure) _________________________________________________9 Exercise 3 Earth driller (proportional control) _________________________________________________ 17 Exercise 4 Earth driller (proportional control) _________________________________________________ 23 Exercise 5 Earth driller (proportional control with open-centre load-sensing system) _________________ 27 Exercise 6 Earth driller (6/3-way proportional valve) ____________________________________________ 31 Exercise 7 Boom cylinder of an excavator _____________________________________________________ 37 Exercise 8 Boom cylinder of an excavator (pilot check valve) _____________________________________ 41 Exercise 9 Boom cylinder of an excavator (counterbalance valve) _________________________________ 45 Exercise 10 Boom cylinder of an excavator (piloted counterbalance valve) __________________________ 51 Exercise 11 Two hydro motor of a crane system (various valve configurations) ______________________ 57

Exercises and worksheets TP 802 Exercise 1 Steering system of an agricultural machine __________________________________________ Exercise 2 Calculating the turning of the steering wheel _________________________________________ Exercise 3 Steering system of an off-road vehicle (shock valve) ___________________________________ Exercise 4 Steering system of a street sweeping vehicle (priority function) __________________________ Exercise 5 Steering system of an articulated vehicle ____________________________________________

IV

63 69 73 77 81

© Festo Didactic GmbH & Co. KG 574166

Exercises and worksheets TP 803 Exercise 1 Mobile block on an excavator (with load-sensing unit) _________________________________ 85 Exercise 2 Mobile block on an excavator (with load-sensing unit) _________________________________ 91 Exercise 3 Remote control of mobile valves (hydraulic joystick) ___________________________________ 93 Exercise 4 Earth driller (load and speed) _____________________________________________________ 97 Exercise 5 Earth driller (pre-compensation) __________________________________________________ 101 Exercise 6 Earth driller (Pressure compensation with LS system) _________________________________ 105 Exercise 7 Two hydro motors of a crane system (with LS system) ________________________________ 109 Exercise 8 Two hydro motors of a crane system (pre-compensation) ______________________________ 113 Exercise 9 Two hydro motors of a crane system (mobile block with pre-compensation for each actuator) 117 Exercise 10 Two hydro motor of a crane system (post-compensation) _____________________________ 121

Fundamentals 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Definition________________________________________________________________________ I-1 Hydraulics fundamentals ___________________________________________________________ I-3 Closed Hydrostatic System ________________________________________________________ I-10 Load-sensing systems ____________________________________________________________ I-19 Variable-displacement pumps ______________________________________________________ I-34 Unloading system ________________________________________________________________ I-39 Flow-dividing valves ______________________________________________________________ I-41 Mobile valve blocks ______________________________________________________________ I-47 6/3-way directional control valves and valve configurations ______________________________ I-63 Pressure compensation ___________________________________________________________ I-70 Load holding and motion control ____________________________________________________ I-79 Hydraulic joystick ________________________________________________________________ I-87 Priority valve ____________________________________________________________________ I-90 Steering ________________________________________________________________________ I-95

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Use for intended purpose The training package for mobile hydraulics may only be used:  For its intended purpose in teaching and training applications  When its safety functions are in flawless condition The components included in the training package are designed in accordance with the latest technology, as well as recognised safety rules. However, life and limb of the user and third parties may be endangered, and the components may be impaired, if they are used improperly. The training system from Festo Didactic has been developed and manufactured exclusively for training and vocational education in the fields of automation and technology. The respective training companies and/or trainers must ensure that all trainees observe the safety precautions which are described in this workbook. Festo Didactic hereby excludes any and all liability for damages suffered by trainees, the training company and/or any third parties, which occur during use of the equipment set in situations which serve any purpose other than training and/or vocational education, unless such damages have been caused by Festo Didactic due to malicious intent or gross negligence.

© Festo Didactic GmbH & Co. KG 574166

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Preface Festo Didactic’s learning system for automation and technology is geared towards various educational backgrounds and vocational requirements. Correspondingly, the training system is broken down as follows:  Technology oriented training packages  Mechatronics and factory automation  Process automation and control technology  Mobile robotics  Hybrid learning factories The training system for automation and technology is continuously updated and expanded in accordance with developments in the field of education, as well as actual professional practice. The technology packages deal with various technologies including pneumatics, electro-pneumatics, hydraulics, electro-hydraulics, proportional hydraulics, mobile hydraulics, programmable logic controllers, sensor technology, electrical engineering, electronics and electric drives.

The modular design of the training system allows for applications which go above and beyond the limitations of the individual training packages. For example, PLC actuation of pneumatic, hydraulic and electric drives is possible.

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© Festo Didactic GmbH & Co. KG 574166

All training packages are comprised of the following elements:  Hardware  Media  Seminars Hardware Hardware included in the training packages consists of industrial components and systems that are specially designed for training purposes. The selection and design of the components encompassed by the training packages are especially well matched to the projects included in the accompanying media. Media The media provided for the individual groups of topics are allocated to the teachware and software categories. The practically oriented teachware includes:  Technical books and textbooks (standard works for imparting basic knowledge)  Workbooks (practical exercises with supplementary instructions and sample solutions)  Lexicons, manuals and technical books (which provide technical information on groups of topics for further exploration)  Transparency sets and videos (for easy-to-follow, dynamic instruction)  Posters (for clear-cut representation of facts) From the software category, programmes are made available for the following applications:  Digital training programmes (didactically and medially prepared learning content)  Simulation software  Visualisation software  Software for measurement data acquisition  Project engineering and design engineering software  Programming software for programmable logic controllers The teaching and learning media are available in several languages. They are intended for use in classroom instruction, but are also suitable for self-study. Seminars Comprehensive seminar offerings covering the contents of the training packages round out the programme for training and vocational education.

Do you have suggestions or criticism regarding this manual? If so, send us an e-mail at [email protected]. The authors and Festo Didactic look forward to your feedback.

© Festo Didactic GmbH & Co. KG 574166

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Introduction This workbook is part of the training system for automation technology from Festo Didactic GmbH & Co. KG. The system provides a solid basis for practice oriented training and vocational education. The TP 501 basic level is suitable for fundamental training in the field of hydraulic control technology. Knowledge regarding the fundamentals of hydraulics, as well as the function and use of hydraulic components, is imparted. Simple hydraulic controllers can be set up with the equipment set. TP 801, 802 and 803 is targeted at vocational training in the field of mobile hydraulics. This workbook imparts knowledge regarding the physical relationships which prevail in the field of hydraulics, and its most important basic circuits. Topics covered by the exercises include:  Recording the characteristic curves of individual components  Comparing usage of various components  Setting up various basic circuits  Applying basic hydraulic equations Technical prerequisites for setting up the controllers include:  A Learnline workstation equipped with a Festo Didactic slotted profile plate. The slotted profile plate has 14 parallel T-slots at 50 mm intervals.  A hydraulic power unit (operating voltage: 230 V, 50 Hz, operating pressure: 6 MPa (60 bar), volumetric flow rate: 4 l/min.)  A power pack with short-circuit protection (input: 230 V, 50 Hz, output: 24 V, max. 5 A) for supplying power to the flow sensor  Laboratory safety cables You will need components included in equipment set TP 801, 802 and 803 in order to complete the exercises. The theoretical fundamentals for understanding these exercises are included in the workbook. Data sheets for the individual components are also available (cylinders, valves etc.).

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Work and safety instructions

General  Trainees should only work with the circuits under the supervision of a trainer.  Operate electrical devices (e.g. power supply units, compressors, hydraulic power units) only in laboratories that are equipped with a Residual Current Device (RCD).  Observe specifications included in the technical data for the individual components, and in particular all safety instructions!  Faults which may impair safety must not be generated in the training environment and must be eliminated immediately.  Wear your personal protective equipment (safety goggles, safety shoes) if you are working on the circuits. Mechanical setup  Only reach into the setup when it is at a complete standstill.  Mount all of the components securely onto the slotted profile plate.  Limit switches may not be actuated frontally.  Danger of injury during troubleshooting! Use a tool to actuate the limit switches, for example a screwdriver.  Set all components up so that activation of switches and disconnectors is not made difficult.  Adhere to the instructions regarding positioning of the components.  Always set up cylinders together with the appropriate cover. Electrical specifications  Only use extra low voltages: max. 24 V DC.  Electrical connections may only be established and interrupted in the absence of voltage!  Only use connecting cables with safety plugs for electrical connections.  Pull the plug only when disconnecting connector cables – never pull the cable. Hydraulics  Limit system pressure to 6 MPa (60 bar). Maximum permissible pressure for all devices included in the training package is 12 MPa (120 bar).  Danger of injury from oil temperatures > 50 ° C! Hydraulic oil with a temperature > 50 ° C can cause burns or scalding.  Danger of injury when switching on the hydraulic power unit! Cylinders may advance and retract automatically.  All valves, devices and hose lines are equipped with self-sealing quick-connect couplings.

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Connecting hose lines – Never connect or disconnect hose lines when the hydraulic power unit is running, or while under pressure! Couplings must be connected in the unpressurised state. – Set the coupling socket squarely onto the coupling nipple! The coupling socket and the coupling nipple must not be fitted askew. Setting up hydraulic circuits – The hydraulic power unit and the electrical power pack must be switched off when assembling the circuit. – Before commissioning, make sure that all tank lines have been connected and that all couplings have been securely fitted. Commissioning – Cylinders may only be commissioned with their covers in place. – Switch on the electrical power pack first, and then the hydraulic power unit. Dismantling hydraulic circuits – Assure that pressure has been relived before dismantling the circuit. – Switch off the hydraulic power unit first, and then the electrical power pack. If connections are decoupled while under pressure, pressure is trapped in the device by the non-return valve in the coupling. This pressure can be vented with the pressure relief unit.

Mounting technology The mounting boards for the components are equipped with mounting variant A, B or C:  Variant A, snap-in system Lightweight components that are not load-bearing (e.g. directional control valves and sensors). Simply clip the components into the slots on the slotted profile plate. Release the components by turning the blue lever.  Variant B, bolt system Components with medium load capacity (e.g. hydraulic or pneumatic cylinder). These components are clamped onto the profile plate using T-head bolts. The blue, knurled nut is used for clamping and loosening.  Variant C, screw system For components with high load capacity and components which are seldom removed from the profile plate (for example on-off valve with filter-regulator). The devices are secured with socket head screws and T-head bolts. Required accessories A digital multimeter is required in order to evaluate exercises which make use of the flow sensor. The output voltage of the flow sensor is measured with the multimeter. You will need a stopwatch in order to measure hydraulic cylinder retracting and advancing times.

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Training package for mobile hydraulics (TP 800) The TP 800 training package consists of a multitude of individual training materials and seminars. The subject matter of this package is strictly mobile hydraulics. Individual components from training package TP 800 may also be included in other packages. Important TP 800 components  Permanent workstation with Festo Didactic slotted profile plate  Equipment sets or individual components (e.g. cylinders, valves and pressure gauges)  Complete laboratory setups Media The workbook includes exercise sheets for each exercise, the solutions to each individual worksheet and a CD-ROM. A set of ready-to-use exercise sheets and worksheets is included in each workbook for all of the exercises. Data sheets for the hardware components are made available along with the equipment set. Available software for use in combination with training package TP 800 includes FluidLab® M. With the help of examples based on actual practice, the learner works through the basic principles of mobile hydraulics and becomes familiar with components used in mobile hydraulic vehicles. The media are offered in several languages. You will find further training materials in our catalogues and on the Internet.

Component designations Hydraulic components are designated in circuit diagrams in accordance with ISO 1219-2.

CD-ROM contents The workbook is included on the CD-ROM as a PDF file.

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Equipment set TP 801 The mobile hydraulic equipment set (TP 801) has been put together for basic training in the field of mobile hydraulic. It includes all of the components which are necessary for mastering the learning objectives, and can be supplemented with TP802 and 803 equipment sets.

Equipment set for mobile hydraulics, basic level (TP 801), order no. 574161 Designation

Order number

Quantity

Counterbalance valve

572149

1

Pressure compensator for open centre load sensing

572123

1

3-way pressure reducing valve

544337

1

Pressure relief valve

544335

2

Flow control valve

152842

1

Non-return valve, 0.6 MPa opening pressure

548618

1

Shuttle valve

572122

1

Double non-return valve, delockable

572151

1

Shut-off valve

152844

1

Proportional 6/3-way hand lever valve

572141

2

Loading unit/cylinder load simulator

572145

1

Diaphragm accumulator with shut-off block

152859

1

Hydraulic motor

152858

2

4-way distributor with pressure gauge

159395

2

T-distributor

152847

3

Pressure switch, electronic

548612

2

Flow sensor

567191

2

© Festo Didactic GmbH & Co. KG 574166

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Equipment set TP 801

Allocation of components to exercises TP 801 equipment set Exercise

1

2

1

1

3

4

5

6

7

8

9

10

1

1

1

1

1

1

11

Component Flow control valve Non-return valve, 0.6 MPa opening pressure Shuttle valve

1

Double non-return valve, delockable Shut-off valve

1 1

1

1

1

1

1

1

1

Counterbalance valve

1

Pressure compensator for open centre load sensing

1

1

3-way pressure reducing valve Pressure relief valve

1 1

1

1

1

2

1

1

1

1

1

1

Diaphragm accumulator with shut-off block

1

1

1

2

1

1

1

1

1

1

1

1

1

1

1

1

2

Pressure switch, electronic

2

2

2

2

2

2

2

2

2

2

2

Flow sensor

1

1

1

1

1

1

1

1

1

1

2

1

1

1

1

1

1

1

1

2

2

2

2

2

2

2

2

2

2

1

1

1

2

3

Proportional 6/3-way hand lever valve 4-way distributor with pressure gauge T-distributor

2

1

1

Loading unit/cylinder load simulator Hydraulic motor

1

2

2 1

2

© Festo Didactic GmbH & Co. KG 574166

Equipment set TP 801

Required accessories Exercise

1

2

3

4

5

6

7

8

9

10

11

4

7

6

6

10

8

10

10

9

10

10

2

4

4

4

4

Component Hose line with quick release couplings, 600 mm Hose line with quick release couplings, 1000 mm

2

Hose line with quick release couplings, 1500 mm

2

2

2

2

2

2

2

2

2

2

2

Set of safety laboratory cables

1

1

1

1

1

1

1

1

1

1

1

Multimeter

1

1

1

1

1

1

1

1

1

1

2

Hydraulic power pack with constant displacement pump

1*

1*

1*

1*

1*

1*

1*

1*

1*

1*

1*

Hydraulic power pack with LS variable and constant displacement pump combination

1*

1*

1*

1*

1*

1*

1*

1*

1*

1*

1*

Hydraulic oil (DIN 51524), HLP22, 20 Litres

2

2

2

2

2

2

2

2

2

2

2

Power supply unit for mounting frame

1

1

1

1

1

1

1

1

1

1

1

(> 3.5 l/min)

* for TP 801 and TP 802 a hydraulic power pack with a constant displacement pump with  3.5 l/min flow is required. Order number 572128 is recommended to cover TP 803 also.

Notes for the teacher/trainer Learning objectives The basic learning goal of this workbook is to become familiar with the fundamentals of mobile hydraulics, as well as the practical setup of circuits on the slotted profile plate. This direct interaction involving both theory and practice ensures faster progress and longer-lasting learning. Concrete, individual learning objectives are assigned to each exercise. Required time The time required for working through the exercises depends on the learner’s previous knowledge of the subject matter. Apprentices in the field of metalworking or electrical installation: approx. 4 days. With training as skilled engineers: approx. 2 days. Device set components The workbook and the equipment set are matched to each other. Only the components included in TP 801 equipment set are required for all exercises. Each exercise can be set up on a slotted profile plate with a width of at least 1100 mm.

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Equipment set TP 801

Standards The following standards are used in this workbook: ISO 1219-1: Fluid power systems and components – Graphic symbols and circuit diagrams, symbols ISO 1219-2: Fluid power systems and components – Graphic symbols and circuit diagrams, circuit diagrams EN 60617-7: Graphical symbols for circuit diagrams EN 81346-2 Industrial systems, installations and equipment and industrial products – Structuring principles and reference designations Identification of the solutions Solutions and supplements in graphics or diagrams appear in red. Designations in the worksheets Texts which require completion are identified with a grid or grey table cells. Graphics which require completion include a grid. Training notes Additional information is provided here regarding the individual components and the completed controllers. These notes are not included in the exercise book. Solutions The solutions given in this workbook result from test measurements. The results of your measurements may deviate from these.

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Exercises and solutions TP 801 Exercise 1 Earth driller (uncompensated pressure) Learning objectives After completing this exercise you will:  Be familiar with the relationship between flow, pressure and energy consumption.  Be able to calculate the energy consumption of a hydraulic system under various conditions  Be familiar with the energy waste of flow sharing

Presentation of the application The motor speed of an earth driller is to be controlled by a flow control valve.

Layout

Earth driller

Application 1. The constant pump has a flow of (qP) 4 l/min. 2. The motor rotates as soon as the pump starts running. 3. The motor rotates in one direction only.

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Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

4. 5. 6. 7.

The speed of the motor is adjusted by turning a flow control valve. The load acting on the motor changes dependent on soil conditions. The pressure acting on the pump is indicated by pressure sensor p1. The pressure of the load is indicated by pressure sensor p2.

Tasks Measure the pressure acting on the pump and calculate the energy consumption of the pump when: 1.1 The motor rotates at full speed without load. (qA = 4 l/min) 2.1 The motor rotates at full speed with a load of 30 bar. (3 MPa). (qA = 4 l/min) 3.1 The motor rotates at half speed without load. (qA = 2 l/min) 4.1 The motor rotates at half speed with a load of 30 bar (3 MPa). (qA = 2 l/min) Read the flow on the flow meter when 5.1 The load is increased to 40 bar (4 MPa) (qA = ?) 6.1 Calculate the wasted energy for each task and show graphically.

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Hints Build the circuit shown on the next page. For the tasks 2 and 4, the relief valve (1V2) for load simulation should be set to 30 bar (3 MPa) which is read on p2. During load simulation setting, the motor should be run at the desired speed (full speed for 1 and 2, half speed for 3 and 4) and the setting adjusted so that 30 bar (3 MPa) of load is obtained at p2 in all cases. In the tasks where no load should be acting on the motor, valve 1V2 should be turned until the minimum pressure is reached. (Note: It is not possible to get 0 bar (0 MPa).) The theory related to this exercise can be found in chapter 2.3, 4.1

Safety note Limit the pump pressure to 60 bar (6 MPa) by adjusting 0Z1.

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© Festo Didactic GmbH & Co. KG 574166

Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

Circuit for Exercise 1

Calculating the energy consumption of the pump: The theoretical energy requirement (P) of the pump can be calculated by the following formula: Ppump =

qP ∆p 600 η

Ppump =

Hydraulic power (kW)

qp

=

Pump flow (l/min)

p

=

Pressure difference between two points where the hydraulic power is converted into mechanical energy (bar (MPa))



=

Efficiency of the pump (will be neglected for all exercises/ = 1)

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Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

1.1 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed without load. qP = 4 l/min qA = 4 l/min p1 = 15 bar (1.5 MPa) p2 = 13 bar (1.3 MPa)

Ppump =

415 = 0.1 kW 600

In this task, the pump flow is 4 l/min and the tank pressure is assumed to be 0 (that’s why p is 15 – 0). So the power required to drive the pump under the given conditions depends on flow of 4 l/min at 15 bar (1.5 MPa) acting on the pump.

1.2 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed with a load of 30 bar (3 MPa). qP = 4 l/min qA = 4 l/min p1 = 33 bar (3.3 MPa) p2 = 30 bar (3 MPa)

Ppump =

433 = 0,22 kW 600

In this task, the power required to drive the pump under the given conditions depends on flow of 4 l/min at 30 bar (3 MPa) acting on the pump. As the pressure acting on the pump is more than the double the value from the previous exercise, the power required to drive the pump has increased proportionally.

1.3 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed without load. qP = 4 l/min qA = 2 l/min p1 = 57 bar (5.7 MPa) p2 = 5 bar (MPa) (5 MPa)

Ppump =

457 = 0,38 kW 600

In this task, to achieve the flow sharing between the motor and the pressure relief valve, the cracking pressure should be reached by turning the flow control valve to raise the system pressure to a high level. Because of the increased pressure acting on the pump, the power required to drive the pump is almost at its maximum. The pressure p2 decreases because the resistance created by the oil flow has also decreased.

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© Festo Didactic GmbH & Co. KG 574166

Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

1.4 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed with a load of 30 bar (3 MPa). qP = 4 l/min qA = 2 l/min p1 = 57 bar (5.7 MPa) p2 = 30 bar (3 MPa)

Ppump =

457 = 0,38 kW 600

In this task, although there is 3 bar (0.3 MPa) of load acting on the motor, the pressure acting on the pump is at its highest. The cracking pressure has to be reached any way to achieve flow sharing.

1.5 Read the flow on the flow meter when the load is increased to 40 bar (4 MPa). qP = 4 l/min qA = 1.5 l/min p1 = 57 bar (5.7 MPa) p2 = 40 bar (4 MPa) When the load is increased from 30 bar (3 MPa) to 40 bar (4 MPa), the pressure drop across the flow control valve decreases (57 to 40 bar (5.7 to 4 MPa)). When the pressure drop decreases, the flow also decreases, which causes the motor flow to decrease to 1.5 l/min. If the pressure drop was constant, the motor would rotate at the same speed although the load has increased.

1.6 Calculate the wasted energy for each task and show graphically. The useful energy of the system is the energy that is converted into mechanical energy by the hydro motor. The pressure to calculate this energy can be read on pressure sensor p2. The wasted energy is the difference between the pump power and the motor power. Or in other words, there is waste whenever there is a pressure difference without conversion to mechanical energy. (In tasks 4 and 5 such energy waste is caused by the pressure relief valve and the flow control valve.) 1.1

413 Pmotor = = 0,086 kW 600 Ppump  Pmotor = 0,1- 0,086 = 0,014 kW

1.2

430 Pmotor = = 0,2 kW 600 Ppump  Pmotor = 0,22 - 0,2 = 0,02 kW

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Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

1.3

25 Pmotor = = 0,016 kW 600 Ppump  Pmotor = 0,38 - 0,016 = 0,364 kW Another way to calculate the wasted energy is:

257 Prelief valve = = 0,19 kW 600 2( 57 -5) Pthrottle alve = = 0,173 kW 600 Prelief valve + Pthrottle valve = 0,19 + 0,173 = 0,364 kW

1.4

230 Pmotor = = 0,1 kW 600 Ppump  Pmotor = 0,38 - 0,1 = 0,28 kW Another way to calculate the wasted energy is:

Prelief valve + Pthrottle valve = 0,19 + 0,173 = 0,364 kW

Pthrottle alve =

2( 57 - 30) = 0,09 kW 600

Prelief valve + Pthrottle valve = 0,19 + 0,09 = 0,28 kW

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© Festo Didactic GmbH & Co. KG 574166

Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

0,4 0,35 0,3 0,25 Useful Energy

0,2

Wasted Energy

0,15 0,1 0,05 0 Task 1

Task 2

Task 3

Task 4

In tasks 1.1and 1.2, the wasted energy is the energy used to overcome the natural flow resistance of the system. But as seen in the tasks 1.3 and 1.4, the wasted energy is the energy used to open the relief valve by a flow control valve to achieve flow division.

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Exercises and solutions TP 801: Exercise 1 – Earth driller (uncompensated pressure)

12

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Exercise 2 Earth driller (compensated pressure) Learning objectives After completing this exercise you will:  Be familiar with the relationship between flow, pressure and energy consumption when the system has pressure compensation (open-centre load sensing)  Be familiar with the amount of energy waste caused by flow sharing when the system has an opencentre load sensing)

Presentation of the application The speed of the motor of an earth driller is to be controlled by a flow control valve. To reduce energy waste, the system is equipped with an open-centre load-sensing unit.

1. 2. 3. 4. 5. 6. 7. 8.

Application The pump has a flow of (qP) 4 l/min. The motor rotates as soon as the pump starts running. The motor rotates in one direction only. The system has an open centre pressure compensator (Load sensing) The speed of the motor is adjusted by turning a flow control valve. The load acting on the motor changes dependent on soil conditions. The pressure acting on the pump is indicated by pressure sensor p1. The pressure of the load is indicated by pressure sensor p2.

Tasks Measure the pressure acting on the pump and calculate the energy consumption of the pump when: 2.1 The motor rotates at full speed without load. (qA = 4 l/min) 2.2 The motor rotates at full speed with a load of 30 bar (3 MPa). (qA = 4 l/min) 2.3 The motor rotates at half speed without load. (qA = 2 l/min) 2.4 The motor rotates at half speed with a load of 30 bar (3 MPa). (qA = 2 l/min) 2.5 Read the flow meter when the load is increased to 40 bar (4 MPa) (qA = ?) 2.6 Compare the results with the results of Exercise 1 graphically; show the wasted energy for each result.

   

Hints All the tasks are the same as the tasks in the Exercise 1, allowing you to compare the results. Build the circuit shown on the next page. For the tasks 2 and 4, the relief valve (1V2) for load simulation should be set to 30 bar (3 MPa) which is indicated on p2. During load simulation setting, the motor should be run at the desired speed (full speed for 2.1 and 2.2, half speed for 2.3 and 2.4) and the setting adjusted so that 30 bar (3 MPa) of load is obtained on p2 in all cases.

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Exercises and solutions TP 801: Exercise 2 – Earth driller (compensated pressure)

 

In the tasks where no load should be acting on the motor, the valve 1V2 should be turned until the minimum pressure is reached. (Note: It is not possible to get 0 bar (0 MPa).) The theory related to this exercise can be found in chapter 4.1, 4.2.

Safety note Limit the pump pressure to 60 bar (6 MPa) by adjusting 0Z1.

Circuit for Exercise 2

p2

p1

14

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Exercises and solutions TP 801: Exercise 2 – Earth driller (compensated pressure)

As the above example shows, the flow that is not needed by the actuator is dumped to tank via the pressure compensator without having to actuate the pressure relief valve.. The pressure compensator maintains a constant pressure drop (p1 – p2 = constant) across the flow control valve. The pressure acting on the pump is always equal to the load pressure plus the spring setting of the compensator. Calculating the energy consumption of the pump: Theoretical energy requirement (P) of the pump can be calculated by the following formula:

Ppump =

qP ∆p 600 η

Ppump =

Hydraulic power (kW)

qp

=

Pump flow (l/min)

p

=

Pressure difference between two points where the hydraulic power is converted into mechanical energy (bar (MPa))



=

Efficiency of the pump (will be neglected for all exercises/ = 1)

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Exercises and solutions TP 801: Exercise 2 – Earth driller (compensated pressure)

2.1 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed without load. qP = 4 l/min qA = 4 l/min p1 = 15 bar (1.5 MPa) p2 = 13 bar (1.3 MPa)

Ppump =

415 = 0,1 kW 600

In this task, the pump flow is 4 l/min and the tank pressure is assumed to be 0 (that is why p is 15-0). So the power required to drive the pump under the given conditions depends on flow of 4 l/min at 15 bar (1.5 MPa) acting on the pump.

2.2 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed with a load of 30 bar (MPa) qP = 4 l/min qA = 4 l/min p1 = 33 bar (3.3 MPa) p2 = 30 bar (3 MPa)

Ppump =

433 = 0,22 kW 600

In this task, the power required to drive the pump under the given conditions depends on flow of 4 l/min and 30 bar (MPa) acting on the pump. As the pressure acting on the pump is more than the double value from the previous exercise, the power required to drive the pump is also proportionally increased.

2.3 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed without load. qP = 4 l/min qA = 2 l/min p1 = 12 bar (1.2 MPa) p2 = 7 bar (0.7 MPa)

Ppump =

412 = 0,08 kW 600

In this task, the flow sharing does not take place between the motor and the pressure relief valve, but between the motor and the pressure compensator. In such a system, the excess flow from the pump is dumped to tank without being having to reach a higher pressure. On the other hand, the compensator maintains a constant pressure drop across the flow control valve which is equal to the spring setting of the compensator (approx. 5 bar (0.5 MPa)).

16

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Exercises and solutions TP 801: Exercise 2 – Earth driller (compensated pressure)

2.4 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed with a load of 30 bar (3 MPa) qP = 4 l/min qA = 2 l/min p1 = 35 bar (3.5 MPa) p2 = 30 bar (3 MPa)

Ppump =

435 = 0,23 kW 600

In this task, there is a load of 30 bar (3 MPa) acting on the motor while the pressure acting on the pump is only 5 bar higher than the load pressure (0.5 MPa) (equal to the spring setting of the compensator).

2.5 Read the flow on the flow meter when the load is increased to 40 bar (4 MPa). qP = 4 l/min qA = 2 l/min p1 = 45 bar (4.5 MPa) p2 = 40 bar (4 MPa) As the compensator maintains a constant pressure drop across the flow control valve (p1 – p2), the flow passing through also remains constant, independent of the load. If we raise or lower the load value, the flow remains constant, because the compensator will always maintain a pressure drop of 5 bar (0.5 MPa) across the flow control valve.

2.6 Calculate the wasted energy for each task and show graphically Make the calculations as they are done in the previous exercise. 2.1

413 Pmotor = = 0,086 kW 600 Ppump  Pmotor = 0,1- 0,086 = 0,014 kW

2.2

430 Pmotor = = 0,2 kW 600 Ppump  Pmotor = 0,22 - 0,2 = 0,02 kW

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Exercises and solutions TP 801: Exercise 2 – Earth driller (compensated pressure)

2.3

27 Pmotor = = 0,023 kW 600 Ppump  Pmotor = 0,08 - 0,023 = 0,057 kW

2.4

230 Pmotor = = 0,1 kW 600 Ppump  Pmotor = 0,23 - 0,1 = 0,13 kW

0,4 0,35 0,3 0,25 Useful Energy

0,2

Wasted Energy

0,15 0,1 0,05

2. 4 Ta sk

1. 4 Ta sk

Ta sk 2. 3

1. 3 Ta sk

2. 2 Ta sk

1. 2 Ta sk

2. 1 Ta sk

Ta sk

1. 1

0

In tasks 1and 2, the wasted energy is the energy used to overcome the natural flow resistance of the system. But as seen in tasks 3 and 4, the wasted energy for Exercise 2 is much smaller compared to Exercise 1. The energy efficiency benefits of an open-centre load-sensing system can clearly be seen.

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Contents Exercises and worksheets TP 801 Exercise 1 Earth driller (uncompensated pressure) _______________________________________________1 Exercise 2 Earth driller (compensated pressure) _________________________________________________9 Exercise 3 Earth driller (proportional control) _________________________________________________ 17 Exercise 4 Earth driller (proportional control) _________________________________________________ 23 Exercise 5 Earth driller (proportional control with open-centre load-sensing system) _________________ 27 Exercise 6 Earth driller (6/3-way proportional valve) ____________________________________________ 31 Exercise 7 Boom cylinder of an excavator _____________________________________________________ 37 Exercise 8 Boom cylinder of an excavator (pilot check valve) _____________________________________ 41 Exercise 9 Boom cylinder of an excavator (counterbalance valve) _________________________________ 45 Exercise 10 Boom cylinder of an excavator (piloted counterbalance valve) __________________________ 51 Exercise 11 Two hydro motor of a crane system (various valve configurations) ______________________ 57

Exercises and worksheets TP 802 Exercise 1 Steering system of an agricultural machine __________________________________________ Exercise 2 Calculating the turning of the steering wheel _________________________________________ Exercise 3 Steering system of an off-road vehicle (shock valve) ___________________________________ Exercise 4 Steering system of a street sweeping vehicle (priority function) __________________________ Exercise 5 Steering system of an articulated vehicle ____________________________________________

63 69 73 77 81

Exercises and worksheets TP 803 Exercise 1 Mobile block on an excavator (with load-sensing unit) _________________________________ 85 Exercise 2 Mobile block on an excavator (with load-sensing unit) _________________________________ 91 Exercise 3 Remote control of mobile valves (hydraulic joystick) ___________________________________ 93 Exercise 4 Earth driller (load and speed) _____________________________________________________ 97 Exercise 5 Earth driller (pre-compensation) __________________________________________________ 101 Exercise 6 Earth driller (Pressure compensation with LS system) _________________________________ 105 Exercise 7 Two hydro motors of a crane system (with LS system) ________________________________ 109 Exercise 8 Two hydro motors of a crane system (pre-compensation) ______________________________ 113 Exercise 9 Two hydro motors of a crane system (mobile block with pre-compensation for each actuator) 117 Exercise 10 Two hydro motor of a crane system (post-compensation) _____________________________ 121

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II

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Exercises and worksheets TP 801 Exercise 1 Earth driller (uncompensated pressure) Learning objectives After completing this exercise you will:  Be familiar with the relationship between flow, pressure and energy consumption.  Be able to calculate the energy consumption of a hydraulic system under various conditions  Be familiar with the energy waste of flow sharing

Presentation of the application The motor speed of an earth driller is to be controlled by a flow control valve.

Layout

Earth driller

Application 1. The constant pump has a flow of (qP) 4 l/min. 2. The motor rotates as soon as the pump starts running. 3. The motor rotates in one direction only.

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Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

4. 5. 6. 7.

The speed of the motor is adjusted by turning a flow control valve. The load acting on the motor changes dependent on soil conditions. The pressure acting on the pump is indicated by pressure sensor p1. The pressure of the load is indicated by pressure sensor p2.

Tasks Measure the pressure acting on the pump and calculate the energy consumption of the pump when: 1.1 The motor rotates at full speed without load. (qA = 4 l/min) 2.1 The motor rotates at full speed with a load of 30 bar. (3 MPa). (qA = 4 l/min) 3.1 The motor rotates at half speed without load. (qA = 2 l/min) 4.1 The motor rotates at half speed with a load of 30 bar (3 MPa). (qA = 2 l/min) Read the flow on the flow meter when 5.1 The load is increased to 40 bar (4 MPa) (qA = ?) 6.1 Calculate the wasted energy for each task and show graphically.

    

Hints Build the circuit shown on the next page. For the tasks 2 and 4, the relief valve (1V2) for load simulation should be set to 30 bar (3 MPa) which is read on p2. During load simulation setting, the motor should be run at the desired speed (full speed for 1 and 2, half speed for 3 and 4) and the setting adjusted so that 30 bar (3 MPa) of load is obtained at p2 in all cases. In the tasks where no load should be acting on the motor, valve 1V2 should be turned until the minimum pressure is reached. (Note: It is not possible to get 0 bar (0 MPa).) The theory related to this exercise can be found in chapter 2.3, 4.1

Safety note Limit the pump pressure to 60 bar (6 MPa) by adjusting 0Z1.

2

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Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

Circuit for Exercise 1

Calculating the energy consumption of the pump: The theoretical energy requirement (P) of the pump can be calculated by the following formula: Ppump =

qP ∆p 600 η

Ppump =

Hydraulic power (kW)

qp

=

Pump flow (l/min)

p

=

Pressure difference between two points where the hydraulic power is converted into mechanical energy (bar (MPa))



=

Efficiency of the pump (will be neglected for all exercises/ = 1)

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Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

1.1 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed without load. qP = 4 l/min qA = 4 l/min Ppump = p1 = p2 = Conclusion:

1.2 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed with a load of 30 bar (3 MPa). qP = 4 l/min qA = 4 l/min Ppump = p1 = p2 = Conclusion:

4

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Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

1.3 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed without load. qP = 4 l/min qA = 2 l/min ppump = p1 = p2 = Conclusion:

1.4 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed with a load of 30 bar (3 MPa). qP = 4 l/min qA = 2 l/min ppump = p1 = p2 = Conclusion:

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Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

1.5 Read the flow on the flow meter when the load is increased to 40 bar (4 MPa). qP = 4 l/min p2 = 40 bar (4 MPa) ppump = qA = p1 = Conclusion:

1.6 Calculate the wasted energy for each task and show graphically. The useful energy of the system is the energy that is converted into mechanical energy by the hydro motor. The pressure to calculate this energy can be read on pressure sensor p2. The wasted energy is the difference between the pump power and the motor power. Or in other words, there is waste whenever there is a pressure difference without conversion to mechanical energy. (In tasks 4 and 5 such energy waste is caused by the pressure relief valve and the flow control valve.) 1.1 Pmotor = Ppump – pmotor =

1.2 Pmotor = Ppump – pmotor =

1.3 Pmotor = Ppump – pmotor =

6

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Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

Another way to calculate the wasted energy is: prelieff valve = pthrottle valve = prelieff valve + pthrottle valve = 1.4 Pmotor = Ppump – pmotor =

Another way to calculate the wasted energy is: prelieff valve = pthrottle valve = prelieff valve + pthrottle valve =

Example 0,4 0,35 0,3 0,25 Useful Energy

0,2

Wasted Energy

0,15 0,1 0,05 0 Task 1

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Task 2

Task 3

Task 4

7

Exercises and worksheets TP 801: Exercise 1 – Earth driller (uncompensated pressure)

Conclusion:

8

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Exercise 2 Earth driller (compensated pressure) Learning objectives After completing this exercise you will:  Be familiar with the relationship between flow, pressure and energy consumption when the system has pressure compensation (open-centre load sensing)  Be familiar with the amount of energy waste caused by flow sharing when the system has an opencentre load sensing)

Presentation of the application The speed of the motor of an earth driller is to be controlled by a flow control valve. To reduce energy waste, the system is equipped with an open-centre load-sensing unit.

1. 2. 3. 4. 5. 6. 7. 8.

Application The pump has a flow of (qP) 4 l/min. The motor rotates as soon as the pump starts running. The motor rotates in one direction only. The system has an open centre pressure compensator (Load sensing) The speed of the motor is adjusted by turning a flow control valve. The load acting on the motor changes dependent on soil conditions. The pressure acting on the pump is indicated by pressure sensor p1. The pressure of the load is indicated by pressure sensor p2.

Tasks Measure the pressure acting on the pump and calculate the energy consumption of the pump when: 2.1 The motor rotates at full speed without load. (qA = 4 l/min) 2.2 The motor rotates at full speed with a load of 30 bar (3 MPa). (qA = 4 l/min) 2.3 The motor rotates at half speed without load. (qA = 2 l/min) 2.4 The motor rotates at half speed with a load of 30 bar (3 MPa). (qA = 2 l/min) 2.5 Read the flow meter when the load is increased to 40 bar (4 MPa) (qA = ?) 2.6 Compare the results with the results of Exercise 1 graphically; show the wasted energy for each result.

   

Hints All the tasks are the same as the tasks in the Exercise 1, allowing you to compare the results. Build the circuit shown on the next page. For the tasks 2 and 4, the relief valve (1V2) for load simulation should be set to 30 bar (3 MPa) which is indicated on p2. During load simulation setting, the motor should be run at the desired speed (full speed for 2.1 and 2.2, half speed for 2.3 and 2.4) and the setting adjusted so that 30 bar (3 MPa) of load is obtained on p2 in all cases.

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Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

 

In the tasks where no load should be acting on the motor, the valve 1V2 should be turned until the minimum pressure is reached. (Note: It is not possible to get 0 bar (0 MPa).) The theory related to this exercise can be found in chapter 4.1, 4.2.

Safety note Limit the pump pressure to 60 bar (6 MPa) by adjusting 0Z1.

Circuit for Exercise 2

p2

p1

10

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Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

As the above example shows, the flow that is not needed by the actuator is dumped to tank via the pressure compensator without having to actuate the pressure relief valve.. The pressure compensator maintains a constant pressure drop (p1 – p2 = constant) across the flow control valve. The pressure acting on the pump is always equal to the load pressure plus the spring setting of the compensator. Calculating the energy consumption of the pump: Theoretical energy requirement (P) of the pump can be calculated by the following formula:

Ppump =

qP ∆p 600 η

Ppump =

Hydraulic power (kW)

qp

=

Pump flow (l/min)

p

=

Pressure difference between two points where the hydraulic power is converted into mechanical energy (bar (MPa))



=

Efficiency of the pump (will be neglected for all exercises/ = 1)

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11

Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

2.1 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed without load. qP = 4 l/min qA = 4 l/min ppump = p1 = p2 = Conclusion:

2.2 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at full speed with a load of 30 bar (MPa) qP = 4 l/min qA = 4 l/min ppump = p1 = p2 = Conclusion:

12

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Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

2.3 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed without load. qP = 4 l/min qA = 2 l/min ppump = p1 = p2 = Conclusion:

2.4 Measure the pressure acting on the pump and calculate the energy consumption of the pump when the motor rotates at half speed with a load of 30 bar (MPa) qP = 4 l/min qA = 2 l/min p2 = 30 bar (3 MPa) ppump = p1 = Conclusion:

2.5 Read the flow on the flow meter when the load is increased to 40 bar (4 MPa). qP = 4 l/min PA = P1 = p2 = 40 bar (4 MPa)

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Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

Conclusion:

2.6 Calculate the wasted energy for each task and show graphically Make the calculations as they are done in the previous exercise. 2.1 Pmotor = Ppump – pmotor =

2.2 Pmotor = Ppump – pmotor =

2.3 Pmotor = Ppump – pmotor =

2.4 Pmotor = Ppump – pmotor =

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Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

Example 0,4 0,35 0,3 0,25 Useful Energy

0,2

Wasted Energy

0,15 0,1 0,05

2. 4 Ta sk

1. 4 Ta sk

Ta sk 2. 3

1. 3 Ta sk

2. 2 Ta sk

1. 2 Ta sk

Ta sk

Ta sk

1. 1

2. 1

0

Conclusion:

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Exercises and worksheets TP 801: Exercise 2 – Earth driller (compensated pressure)

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Fundamentals 1

Definition________________________________________________________________________ I-1

2 2.1 2.2 2.3

Hydraulics fundamentals ___________________________________________________________ Pressure, flow, relief valve fundamentals ______________________________________________ Pressure drop (p) ________________________________________________________________ Heat generation due to the pressure drop _____________________________________________

3 3.1 3.2 3.3 3.4 3.5

Closed Hydrostatic System ________________________________________________________ Building a closed hydraulic circuit ___________________________________________________ Closed-circuit hydraulic pumps _____________________________________________________ Charge pumps ___________________________________________________________________ Flushing valves __________________________________________________________________ Shock valves ____________________________________________________________________

4 4.1 4.2

Load-sensing systems ____________________________________________________________ I-19 Pump performance _______________________________________________________________ I-19 Load sensing systems: ____________________________________________________________ I-27

5

Variable-displacement pumps ______________________________________________________ I-34

6

Unloading system ________________________________________________________________ I-39

7 7.1 7.2 7.3

Flow-dividing valves ______________________________________________________________ 50/50 spool flow divider __________________________________________________________ Rotary flow divider _______________________________________________________________ Rotary flow divider (pressure intensification) __________________________________________

I-41 I-42 I-44 I-45

8 8.1 8.2 8.3 8.4 8.5 8.6

Mobile valve blocks ______________________________________________________________ Definition_______________________________________________________________________ Symbol for a mobile valve _________________________________________________________ Valve construction _______________________________________________________________ Directional control valve spool _____________________________________________________ Side modules ___________________________________________________________________ Shock and anti-cavitation valve _____________________________________________________

I-47 I-47 I-49 I-50 I-52 I-55 I-60

9 9.1 9.2 9.3 9.4 9.5

6/3-way directional control valves and valve configurations ______________________________ Definition of 6/3-way valves _______________________________________________________ Parallel configuration _____________________________________________________________ Tandem configuration ____________________________________________________________ Series configuration ______________________________________________________________ Basic design of a 6/3-way valve ____________________________________________________

I-63 I-63 I-64 I-65 I-66 I-67

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I-3 I-3 I-7 I-9

I-10 I-10 I-13 I-16 I-16 I-18

I-I

I-II

10 10.1 10.2

Pressure compensation ___________________________________________________________ I-70 Pre-compensation________________________________________________________________ I-75 Post compensation _______________________________________________________________ I-77

11 11.1 11.2 11.3 11.4

Load holding and motion control ____________________________________________________ Delockable non return valve _______________________________________________________ Direct-acting counterbalance valve __________________________________________________ Delockable counter balance valve (overcentre valve) ___________________________________ Pilot ratio _______________________________________________________________________

12

Hydraulic joystick ________________________________________________________________ I-87

13 13.1 13.2 13.3

Priority valve ____________________________________________________________________ 3-way compensator ______________________________________________________________ Priority valve ____________________________________________________________________ Dynamic priority valve ____________________________________________________________

14 14.1 14.2 14.3 14.4 14.5

Steering ________________________________________________________________________ I-95 The steering unit _________________________________________________________________ I-95 Open-centre systems _____________________________________________________________ I-98 Closed-centre systems ___________________________________________________________ I-100 Reaction and non-reaction steering units ____________________________________________ I-101 Steering types __________________________________________________________________ I-102

I-79 I-79 I-80 I-83 I-86

I-90 I-90 I-92 I-93

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1 Definition

The biggest difference between industrial and mobile systems is that mobile systems do not follow a regular work cycle or use as many actuators. The number of actuators etc. used varies depending on the conditions such as load, working surface, ground slope, cycle time, and pressure. And the machine must meet certain demands even in the worst combination of conditions. The most important consideration in mobile systems is efficiency: As mobile machines move independently from any power source, the power supply is limited. In most cases, the only power source is an internal combustion engine with a limited fuel supply. For this reason, components of a mobile system – including reservoirs, pumps, motors and valves – should be as small and light possible. But higher pressures are required to achieve the same performance from a smaller component,. Higher pressures require smaller clearances to prevent leakages – and smaller clearances require cleaner, cooler oil. In industrial hydraulic systems, the force acting on an actuator is usually constant but in mobile applications this force varies a lot. If the pressure relief valve should be set for the highest load, a lot of power will be wasted for the smaller loads, this problem should also be overcome in mobile systems. As the figure below shows, a mobile hydraulic system comprises three main parts: 1. Closed hydrostatic transmission 2. Working hydraulics 3. Steering There are two different types of pumps when classified by duty: The main pump supplies hydraulic power for the motors which make the wheels turn. It is an axial piston pump which allows the direction of flow to be reversed and the flow rate to be adjusted. . The other pumps are called auxiliary pumps. Depending on the application, there may be one auxiliary pump for both working hydraulics and steering or one pump for each application. The working hydraulics parts of mobile applications that resemble industrial hydraulics comprise various valves (mostly proportional) and actuators. The directional control valves used in mobile hydraulics are special designs for mobile applications and are called mobile valve blocks instead of directional control valves.

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I-1

I-2

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2 Hydraulics fundamentals 2.1 Pressure, flow, relief valve fundamentals It is helpful to remember some basic hydraulics fundamentals before proceeding to mobile hydraulics. Some of the fundamentals play an important role in understanding the logic behind the mobile applications. The diagram below shows a mobile application. In mobile applications, the pump is either a gear pump as shown below or a variable-displacement pump. The flow control valve below represents a proportional directional control valve where the oil is throttled to adjust the speed of the actuator.

Relief valve setting = 275 bar (27.5 MPa)

Cracking pressure = 250 bar (25 MPa)

Pump flow= 100 l/min

Let’s find answers to some basic questions: 1. What makes the value of p2 to rise or fall? 2. What is the pressure difference between p1 and p2 if the flow control valve is fully open? 3. What happens to the actuator speed if the flow control valve is partially closed to raise p1 to 200 bar (20 MPa)? 4. What is the pressure at p1 if the cylinder is extending at half speed? 5. What happens if the load on the cylinder increases? 6. What happens at p2 if the load changes position? The answers to these questions can be as follows: 1. The pressure value of p2 is determined by the load weight, its position, and the piston diameter. Any change in the load weight or position will affect the pressure of p2. 2. The pressure difference between two points is determined by the resistance between them. If the flow control valve is fully open it creates no resistance and if we neglect the friction due to oil flow, the value of p1 and p2 is same.

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3. The speed of the actuator depends on oil flow into the cylinder. If the pressure value of p1 is higher than the cracking pressure, the relief valve opens partially or fully and the cylinder only receives part or none of the flow, causing the piston to slow or stop. In the question, the pressure value of p2 is below the cracking pressure, so the cylinder is moving at full speed (if we neglect the leakage losses due to high pressures). The circuit diagram shows that the relief valve is closed, and the cylinder is receiving the full flow from the pump. 4. If the cylinder is extending less than full speed, the pressure at p1 is above the cracking pressure value and below the fully open pressure and 50 l/min is dumped to tank via the relief valve. Half-speed corresponds to a point between these two which is approximately 260 bar (26 MPa). This value may also be found on the characteristics curve of the relevant relief valve. As seen in the illustration below, the relief valve is open due to high pressure and a part of the pump flow is dumped to tank via the relief valve. The flow control valve or the proportional directional control valve plays an important role here, such as changing the resistance value to change the pressure value of p2 .

The opening characteristic curve of a relief valve is shown below. The curve corresponds to a flow rate which passes through the relief valve at certain pressures at a certain preset value (at certain oil viscosity and a certain temperature). When we examine the curve, we can see that the 50 l/min flows to tank when the pressure at the relief valve inlet port is as high as 260 bar (26 MPa) when the maximum setting is 275 bar (27.5 MPa).

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q: Nominal flow

p: Operating pressure

pc: Cracking pressure

pf: Fully open pressure

A curve from a catalogue is shown below. In the catalogue the curve’s starting point is the cracking pressure as the useful area is that between the cracking pressure and fully open points. In the catalogue curve we can also find the corresponding pressure for 50 l/min flow: It is also 260 bar (26 MPa).

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5. If the load on the cylinder increases, this eventually increases the pressure of p2. The increased pressure affects the whole system and P1 also rises. The increasing pressure of p1 causes the relief valve to open further. This causes the relief valve will dump more oil to tank, which slows the piston travel. We can also see this in the curve.

pA = 260 bar (26 MPa) - Before the load increased, the cylinder was travelling at half speed. pB = 270 bar (27 MPa) - Because of the increasing load, p2 rises to 270 bar (27 MPa) and the relief valve opens further and dumps 70 l/min to tank.

6. If the load changes position, this will have the same effect as question 5. As the load is moved away from the boom cylinder, the lever arm becomes longer, applying greater force to the piston. The further away the load, the slower the cylinder moves.

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2.2 Pressure drop (p) The term pressure drop plays an important role for the sound understanding of hydraulic systems. The symbol for pressure drop is p and it simply refers to a pressure difference between two points. Although it is mainly used to ensure proper valve sizing and selection, it is also used to explain some hydraulics fundamentals.

In the above example, there is a valve with a certain resistance value and p1 is the valve inlet port and p2 is the outlet port. Because of the resistance applied by the valve (this may be a flow control valve, an orifice, a directional control valve or a proportional directional control valve), the pressure rises at the inlet of the valve (assuming that the flow direction is always from inlet to outlet). Back pressure is the pressure found at the outlet port of the valve (p2). This is caused by downstream restrictions such as load on a cylinder, filter, reducing fittings – anything that can cause a pressure drop.

The above system represents a hydraulic system with a flow control valve (it may also be a proportional directional control valve) with a certain p and the PRV is not yet open. Let’s examine the variables with the above given condition (pump efficiency neglected): 1. What happens if the flow control valve is gradually closed? If the flow control valve is gradually closed, p1 will increase, but p2 will remain the same. Consequently the p will increase, but this will have no effect on the flow. 2. What happens when the back pressure at point 2 increases? If P2 increases due to back pressure, p1 will increase by the same amount. Consequently p will remain the same and this will also have no effect on the flow. 3. What happens if the flow increases? If flow increases, the resistance at the throttling point increases. As a result, p will rise as seen in the formulas or in the diagram below.

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4. What happens if the viscosity decreases (due to heating of the oil)? Oil with lower viscosity creates lower resistance which causes p1 to decrease, resulting in a decrease in p. This will have no effect on the flow. In the above given conditions, the change in p has no effect on flow, but the change in the flow has a large effect on p (as also seen in the curve). In a flow controlled condition, the change in the flow causes a great change in p. Let’s examine the condition where the PRV is already open to make flow sharing. Now, the changes in p will affect the flow.

1. What happens if the flow control valve is gradually closed? This will increase p1, consequently the p will increase. The increasing value of p1 causes the PRV to open further, reducing flow to the system. 2. What happens when the back pressure at point 2 increases? If p2 increases due to a back pressure, p1 also increases (but only a small amount, this can be found in the characteristic curve of the pressure relief valve), which causes the PRV to open further. In this case, p decreases and the flow also decreases. 3. What happens if the viscosity increases? This causes less resistance in the flow control, consequently, p1 will fall. The fall in p1 causes the PRV to close gradually, so flow through the system increases. Consequently, in a fixed throttling condition (fixed resistance), keeping a constant p in the system will result in a constant flow.

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2.3 Heat generation due to the pressure drop The pressure drop between two components results in wasted energy which is converted into heat when there is no useful movement (such as extending of a cylinder or rotating of a hydraulic motor).

For example, the pressure relief valve is the greatest heat generator in the system, because the oil flows to tank without doing any useful work. Also a flow control valve, a proportional valve which throttles the oil, any orifice or any leakage causes heating of the oil due to a big pressure drop. The power transferred to a fluid causing heat generation can be calculated by the formula given below. P= Hydraulic power (kW) q= Flow (l/min) p= Pressure difference between two points (bar)

P=

q∆p 600

The energy which is converted into heat in a pressure relief valve is:

P=

qprv (p1 - p0 ) 600

The energy which is converted into heat in a flow control valve is: P=

qfc (p1 - p2 ) 600

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3 Closed Hydrostatic System 3.1 Building a closed hydraulic circuit In many vehicles it is preferred to have hydrostatic transmission rather than mechanical transmission despite its lower efficiency. This is because the benefits of hydrostatic transmission outweigh the disadvantages: Hydrostatic transmission provides superior maneuvering abilities and continuously variable speed control between minimum and maximum speeds. The direction of rotation of the wheels is easily reversed by using the features of the variable mobile pump. In mechanical transmission systems, the wheels of a vehicle are driven by shafts and gears from an internal combustion engine. But in hydrostatic transmission, the combustion engine drives a pump and the pump supplies oil for the hydraulic motor(s) which make the wheels turn. The pump’s flow rate is proportional to input speed. But due to limitations of mobile hydraulics such as reservoir size, weight of components etc, the pump and the motor(s) do not operate as they do in a conventional hydraulic system, which is known as an open circuit. Instead, they operate in a closed circuit where both inlet and outlet ports of the pump and motor are connected. In other words, the oil delivered by the pump makes the hydraulic motor rotate and the oil out of the motor goes to the suction port of the pump again. The oil circulates between the pump and the motor in a continuous loop.

The above diagram is the basic representation of a closed hydraulic circuit. The oil circulates in the circuit making the motor(s) rotate. The flow direction is reversed and the flow is adjusted by tilting the swash plate of the pump. For various reasons, the circuit in the diagram cannot operate. 1. There are leakages in both pump and the motor. So the oil in the circuit gets less as the system runs, which make the system impossible to operate. (These leakages are necessary for cooling and cleaning functions.) 2. The suction side of the pumps in closed circuits must always have positive pressure, otherwise the system may not run. But due to load on the motor, this pressure may be too low to make the system run. 3. The oil in the system gets hotter due to extreme working conditions and pressures and has to be replaced with fresh oil.

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In closed circuits, a charge pump is necessary to overcome internal and auxiliary leakages, provide flow for cooling and maintain pressure in the line which acts as a suction side (where the flow is into the pump). As the flow direction should be reversible in closed circuits, two non return valves are used to ensure the supply pump oil flows only in the suction side of the main pump as the non return valve connected to the pressure line will close due to higher pressure and the one on the suction side will open due to lower pressure. In most mobile pumps (variable axial piston pump), the charge pump is in the main pump block and is driven by the same shaft connected to internal combustion engine. No oil can circulate in the main pump when it is idle. For this reason, the charge pump also lubricates the main pump. The charge pump is usually placed in the main pump block, but is separate in some applications. As the diagrams show, the leakage lines from both pump and motors are connected to tank. These leakages are intentionally built in for cooling and lubricating purposes, but as the system wears out, the amount of leakage increases to undesirable levels. As the clearances in mobile applications are smaller, the cleanliness of the oil affects the life of the components as well as the amount of leakage.

Because of higher pressures, smaller clearances and limited cooling sources, the oil very soon gets hot in mobile systems. And because of the smaller clearances, the oil should always maintain a certain level of cleanliness.

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To ensure efficient cooling and removal of contamination, instead of running all the available oil in the circuit, oil which is not required by the pump is flushed to tank, ensuring that the charge pump always receives fresh oil from the tank. The flushing is better done as soon as the oil gets hot. This is done by flushing valve as seen in the circuit. The flushing valve receives relatively hot and contaminated oil from the outlet of the motor and immediately diverts it to the tank. The flushing valve is pressure-operated by the pressure in the pressure line. In most applications, the flushing valve is in the motor block as it has to be placed very near to where the oil gets hot. There are cases where the flushing valve is in the pump block. The setting of the pressure relief valve connected to the flushing valve determines the pressure of the suction line. The last important detail which was missing in the previous diagrams was the shock valve. As the wheels connected to hydraulic motors can receive sudden and severe iMPacts as the result of surface conditions, the pressure in the lines may go over the circuit’s limitations. For this reason, a relief valve is placed in each line to protect the system pressure against spikes resulting from external effects. And these relief valves are called shock valves and set at a higher value than the pump’s pressure setting.

The major components in a closed transmission system are: 1 Main pump 2 Charge pump 3 Charge pump pressure relief valve 4 Non return valve 5 Flushing valve 6 Flush valve pressure relief valve 7 Shock valve 8 Hydraulic motor 9 Heat exchanger 10 Reservoir 11 Filter

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Some of these components can also be seen on the following diagram.

10 11 9

3

8

2 1

7

5

3.2 Closed-circuit hydraulic pumps These are usually axial piston hydrostatic pumps. The flow rate and the flow direction are controlled by varying the swash plate angle. There are two basic types of displacement control: direct displacement control and servo control. Direct displacement control In this type of control, a mechanical lever is attached to the swash plate which the operator controls via a linkage assembly. The force required to engage a direct displacement control is a function of the system pressure that the swash plate is working against. Because the required control force is a function of system pressure, direct displacement controls are generally only suitable to light-duty applications. Servo control This is an interface between the operator and the swash plate. The operator input is a low-force device which ports charge pressure to the servo piston to move the swash plate. As a result, the operator input is not dependent on system pressure. Standard servo controls also have a mechanical feedback mechanism which ensures that the swash plate position stays where set.

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Main components in a closed-circuit pump

3 4

2

1

5 6

1

Charge pump

2

Valve plate

3

Cylinder block

4

Cradle swash plate

5

Piston

6

Ball bearing

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3.3 Charge pumps A charge pump is also a vital element for every closed hydrostatic circuit which is usually a gear or a gerotor pump. It carries out the following functions: 1. Replenishing the fluid lost through the volumetric efficiency of the pump and the motors. 2. Replenishing the fluid lost through the flushing valve. 3. Providing make-up fluid in the loop for load-induced bulk modulus effects. 4. Providing constant flow for the pilot solenoids for the pump control. 5. Providing flow to the servo control actuator. 6. Providing flow for the auxiliary functions such as parking brakes or servo motor displacement. 7. Maintaining low loop pressure to ensure sufficient hold-down forces on the rotating groups. 8. Controlling flushing flow by means of the pressure differential between the charge relief valve and the flushing relief valve.

3.4 Flushing valves

There are two primary fluid loops in a closed-circuit transmission The main system power loop (which is delivered by the main pump, received by the motor(s) and again received by the suction side of the main pump) The charge/cooling loop (a portion of the main loop charged to tank via flushing valves). Quality oil is required in the power loop, but the quality of the fluid is controlled by the charge/cooling loop. The flushing valve removes a portion of the flow from power loop; this must be replenished by the charge pump. The removed fluid is considered dirty and hot, while the power loop requires cool, clean oil. Removing heat and contamination is the utmost priority for system operation, especially in hydrostatic systems with small clearances. Although some systems may function without a flushing valve, adding one increases the service life of components. Flushing valves can either be separate or integrated to the body of a motor or a pump. They always divert the low-pressure side of the power loop to the tank. The pressure side of the loop causes the flushing valve to actuate and this causes the low-pressure side to be diverted to tank. There is always a pressure relief valve in the same block which determines the pressure of the low-pressure side (suction side). The flushing valve can be symbolised as below. The diagram shows that the flushing valve directs oil through the motor’s drain line where the flushing valve is in the motor housing.

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The spool valve housing contains a spool which is centred by the spring. The oil coming from 1 or 2 is transmitted to 4 to be dumped to the tank, depending on which way the spool is actuated by the higherpressure side.

1, 2: System ports

4: Drain port

The oil coming from 2 at high pressure pushes the spool to the left causing the flow through 1 (low-pressure side) being dumped into the tank.

The oil coming from 1 at high pressure pushes the spool to the left causing the flow through 2 (low-pressure side) being dumped into the tank.

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3.5 Shock valves Shock valves (also called as crossover relief valves) are direct-acting pressure relief valves to protect the circuit from sudden rises in pressure. The advantage of using a direct-acting relief valve is that they respond much faster, they are contaminant-tolerant, and have less leakage.

In most closed circuits, bidirectional relief valves are used. The bidirectional poppet is a dual crossover relief in a single cartridge. When pressure at 1 exceeds the nominal setting, the lower poppet acts as a directacting relief valve and opens flow from 1 to 2. When pressure at 2 exceeds the nominal setting, the upper poppet acts as a differential area relief valve and opens flow from 2 to 1 . Note that the valve is designed so that the crack pressure is the same in either direction.

Bidirectional shock valve

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