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Student Manual|3030794000000I~30794-00

Hydraulics Fundamentals

Fluid Power

Fluid Power

Hydraulics Fundamentals

Student Manual30794-00

A

First EditionPublished March 2013

© 1996 by Lab-Volt Ltd.Printed in CanadaAll rights reserved

ISBN 978-2-89289-349-6 (Printed version) ISBN 978-2-89640-651-7 (CD-ROM)

Legal Deposit – Bibliothèque et Archives nationales du Québec, 1996Legal Deposit – Library and Archives Canada, 1996

No part of this publication may be reproduced, stored in a retrieval system, or transmitted inany form by any means, electronic, mechanical, photocopied, recorded, or otherwise, withoutprior written permission from Lab-Volt Ltd.

Information in this document is subject to change without notice and does not represent acommitment on the part of Lab-Volt. The Lab-Volt® materials described in this document arefurnished under a license agreement or a nondisclosure agreement.

The Lab-Volt® logo is a registered trademark of Lab-Volt Systems.

Lab-Volt recognizes product names as trademarks or registered trademarks of theirrespective holders.

All other trademarks are the property of their respective owners. Other trademarks and tradenames may be used in this document to refer to either the entity claiming the marks andnames or their products. Lab-Volt disclaims any proprietary interest in trademarks and tradenames other than its own.

III

Safety and Common Symbols

Safety and Common Symbols

IV

V

Foreword

The Lab-Volt Hydraulics Training System is a modularized presentation of theprinciples of hydraulic energy and its controlled application. The Hydraulics TrainingSystem consists of an introductory and an advanced training program.

The introductory program is based on two manuals: Volume 1,Hydraulics – Fundamentals, covers the basic principles of hydraulics; Volume 2,Hydraulics – Electrical Control, covers electrical circuits and ladder diagrams forhydraulics applications. Both manuals are intended to be used with the Lab-VoltHydraulics Trainer.

The advanced training program expands upon the introductory course withhydraulics applications using programmable controllers, sensors, proportionalcontrols, and servo controls. The covered applications are based on thoseencountered in the industry. The introductory program is a prerequisite for theadvanced program.

This manual, volume 1 of the Hydraulics series, introduces students to the basicprinciples of hydraulics. Subjects covered are the theory, generation, storage, andusage of hydraulic energy. The creation of pressure by applying force to a confinedliquid is discussed. The usefulness of fluid pressure and velocity is examined, andthe relationship between flow rate, velocity, and power is defined. The basic typesof hydraulic circuits are introduced. Identification and operation of basic hydrauliccomponents is also covered. Finally, a methodical approach to troubleshooting isoutlined, based on the first principles of hydraulics.

The Lab-Volt Instructor’s Guide for Hydraulics – Fundamentals (P/N 30794-10)provides answers to all procedure steps and review questions found in each exercisein this manual. We also recommend that you use the Parker-Hannifin’s manualIndustrial Hydraulic Technology as a reference.

VI

Acknowledgments

We wish to thank Mr Patrick Quirion, Mech. Eng., CEFP, MGI, forhis participation in the elaboration of the hydraulics courseware. MrQuirion teaches fluid power classes in Montreal, Canada.

VII

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . XI

Unit 1 Introduction to Hydraulics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1

An introduction to hydraulic circuit. Safety rules to follow when using theLab-Volt Hydraulics Trainer.

Exercise 1-1 Familiarization with theLab-Volt Hydraulics Trainer . . . . . . . . . . . . . . . . . . . . 1-5

Identification of the various system components. Safety rulesto follow when using the Lab-Volt Hydraulics Trainer.

Exercise 1-2 Demonstration of Hydraulic Power . . . . . . . . . . . . . 1-27

Lifting up the hydraulic Power Unit using a small-borecylinder. Investigation of a basic hydraulic circuit.

Unit 2 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

Basic concepts of hydraulics. Creation of pressure by applying force to aconfined fluid. Relationship between flow rate, velocity, and power.

Exercise 2-1 Pressure Limitation . . . . . . . . . . . . . . . . . . . . . . . . . . 2-3

Design and operation of a relief valve. Determining the oilflow path in a circuit using a relief valve. Connection andoperation of a circuit using a relief valve.

Exercise 2-2 Pressure and Force . . . . . . . . . . . . . . . . . . . . . . . . . 2-17

Verifying the formula F = P x A using a cylinder and a loadspring. Discovering what happens to a cylinder when equalpressures are applied to each side of its piston. Pressuredistribution in a cylinder in equilibrium of forces. Measuringthe weight of the hydraulic Power Unit given the pressurerequired to lift it.

Exercise 2-3 Flow Rate and Velocity . . . . . . . . . . . . . . . . . . . . . . . 2-35

Design and operation of a flow control valve. Relationshipbetween flow rate and velocity. Connection and operation ofmeter-in, meter-out, and bypass flow control circuits.

Exercise 2-4 Work and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-57

Definition of the terms “work” and “power”. Relationshipbetween force, work, and power. Calculating the work,power, and efficiency of the circuit used to lift the hydraulicPower Unit.

Table of Contents

VIII

Unit 3 Basic Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-1

Connection and operation of simple, practical hydraulic circuits. Design andoperation of a directional control valve.

Exercise 3-1 Cylinder Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-3

Control of the direction, force, and speed of a cylinder.Design and operation of a directional control valve. Effect ofa change in system pressure and flow rate on the force andspeed of a cylinder.

Exercise 3-2 Cylinders in Series . . . . . . . . . . . . . . . . . . . . . . . . . . 3-19

Description of the operation of a series circuit. Starting andstopping two cylinders at the same time by connecting themin series. Demonstration of pressure intensification in a seriescircuit.

Exercise 3-3 Cylinders in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . 3-31Description of the operation of a parallel circuit. Extensionsequence of parallel cylinders having differing bore sizes.Synchronizing the extension of parallel cylinders using a flowcontrol valve.

Exercise 3-4 Regenerative Circuits . . . . . . . . . . . . . . . . . . . . . . . . 3-41

Design and operation of a regenerative circuit. Effect ofregeneration on cylinder force and speed.

Unit 4 Functional Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1

Connection and operation of functional hydraulic circuits usingaccumulators, hydraulic motors, pressure reducing valves, and remotelycontrolled pressure relief valves.

Exercise 4-1 Accumulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-3

Description of the general types of accumulators. Howaccumulators can be used in auxiliary power, emergencypower, leakage compensation, and shock suppression.Safety requirements for accumulator circuits.

Exercise 4-2 Hydraulic Motor Circuits . . . . . . . . . . . . . . . . . . . . . 4-19

Design and operation of a hydraulic motor. Calculating thetorque and speed of a hydraulic motor. Effect of a change inflow rate or pressure on motor operation.

Table of Contents

IX

Exercise 4-3 Pressure Reducing Valves . . . . . . . . . . . . . . . . . . . . 4-35

Design and operation of a pressure reducing valve.Connection and operation of a clamp and bend circuit usinga pressure reducing valve.

Exercise 4-4 Remotely Controlled Pressure Relief Valves . . . . . 4-53

How to control a pressure relief valve remotely. Connectionand operation of a circuit using a remotely controlled valve tocontrol the tonnage of a press cylinder.

Unit 5 Troubleshooting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-1

Developing a methodical approach for testing the main components of ahydraulic system, based on the manufacturer specifications and on the firstprinciples of hydraulics. Observing the effects of temperature changes onthe operating characteristics of a hydraulic system.

Exercise 5-1 Hydraulic Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-3

Basic operation of a hydraulic pump. Using manufacturerpump specifications to test a pump. The effects of oiltemperature on flow rate and volumetric efficiency.

Exercise 5-2 Directional Valve Testing . . . . . . . . . . . . . . . . . . . . . 5-19

Showing normal leakage of a directional valve. Evaluating thecondition of a directional valve according to the amount ofleakage flow.

Exercise 5-3 Flowmeter Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . 5-31

Verifying the accuracy of a flowmeter. Determining the effectof temperature on flowmeter accuracy.

Exercise 5-4 Effects of Temperature on System Operation . . . . 5-41

The effects of temperature changes on pressure drop andcircuit flow rate.

Appendix A Equipment Utilization ChartB Care of the Hydraulics TrainerC Conversion FactorsD Hydraulics and Pneumatics Graphic Symbols

Bibliography

We Value Your Opinion!

X

XI

Introduction

The basic principles of fluid power date back to Pascal’s research and the inventionof the piston, but only recently has fluid power become a large scale industry. Thegrowing use of hydraulics in industry comes from the need for fast, low cost meansof production with better quality, less waste, and increased power.

Hydraulic systems provide many other advantages. A few of these are spark- andburnout-resistance, fine control, and compact size. This means that almost allmanufactured products have been formed, treated, or handled by fluid power atsome time.

This manual, Hydraulics – Fundamentals, provides basic training in hydraulics. Itcovers the theory, generation, storage, and usage of hydraulic energy.

The manual is divided into five units:

– Units 1 and 2 present the basic concepts of hydraulics. Unit 1 introducesstudents with the Lab-Volt Hydraulics Trainer. Unit 2 discusses the creation ofpressure and defines the relationship between flow rate, work, and power.

– Units 3 and 4 introduce basic and functional hydraulic circuits.

– Unit 5 presents the basic troubleshooting techniques used in troubleshootinghydraulic circuits.

These five units provide a complete course in hydraulics. They lay a solid foundationfor the study of Volume 2 of the courseware series, Electrical Control of HydraulicSystems.

The exercises in this manual provide a systematic and realistic means of learning thesubject matter. Each exercise contains

– A clearly defined Exercise Objective.– A Discussion of the theory involved.– A list of Equipment Required.– A Procedure Summary which provides a bridge between the theoretical

Discussion and the laboratory Procedure.– A detailed step-by-step laboratory Procedure in which the student observes and

measures important phenomena. Illustrations facilitate connecting the modulesand guide the student’s observations. Well-organized tables help in performingcalculations. Questions direct the student’s thinking process and help inunderstanding the principles involved.

– A Conclusion to confirm that objective has been reached.– Review Questions which verify that the material has been well assimilated.

XII

1-1

Unit 1

Introduction to Hydraulics

UNIT OBJECTIVE

When you have completed this unit, you will be able to identify the Hydraulics Trainercomponents and to safely operate the trainer. You will demonstrate your ability byconstructing simple hydraulic circuits.

DISCUSSION OF FUNDAMENTALS

Introduction

The intensive use of hydraulics in today’s industry comes from the many advantagesprovided by hydraulic systems. With hydraulic power, very little energy is requiredto control and transmit tremendous amounts of power. For example, 1.5-kW (2-hp)electric motors can be used to actuate hydraulic hoists lifting up to 4000 kg (8800 lb),as Figure 1-1 shows.

Figure 1-1. Hydraulic hoist.


1-2

Gigantic rockets that hurl satellites into orbit around the earth, and that carry menand women to the moon and other planets also depend on hydraulic power to controltheir flight. Only hydraulic power systems have the “muscle” and power to controlwith the delicacy of a feather touch, the millions of horsepower released by rocketengines and direct the payload to its destination.

Aviation is another industry that presently places a heavy demand on hydraulics. Thehydraulic power used in aircraft travels anywhere a pipe or tube can be run. Aircrafthydraulic systems are lightweight and compact, yet powerful enough to move thecontrol surfaces of the largest planes.

Another industry that relies heavily on hydraulics is robotics. The hydraulic systemsof robots, like those used by automobile manufacturers, are simpler than comparableelectrical systems. In general, the easy speed control, minimum vibration, and designversatility of hydraulics will keep hydraulic power with industry for a long time tocome.

Hydraulics basic principles

Hydraulics is the technology or study of liquid pressure and flow. Liquids arematerials which pour and conform to the shape of their containers. Example ofliquids are oil and water.

Because liquids are not very compressible, they permit to transfer and multiplyforces. Figure 1-2 illustrates this basic property of liquids. The force applied to theinput piston produces a pressure on the liquid. The liquid then exerts the sameamount of pressure equally in all directions. As a result, the pressure applied to theinput piston transfers to the output piston.

Figure 1-2. Direct transfer of force.

Now what happens if the pistons are of different sizes, as in Figure 1-3? The inputpiston is the same size as in the previous example (6.5 cm2), but the output pistonis now 26 cm2. Since the liquid exerts the same amount of pressure equally in alldirections, the force transferred to the output piston now equals 1780 N, whichprovides a mechanical advantage in force of 4:1.


1-3

Figure 1-3. Multiplication of force.

Pressure is the amount of force exerted by a liquid on a unit of area. Pressure ismeasured in kilopascals (kPa) in the S.I. system, in bars (bar) in the metric system,and in pounds per square inch (psi) in the English system. 1 kPa is equal to 0.01 baror 0.145 psi. 1 psi is equal to 6.895 kPa or 0.069 bar. The pressure of a liquid canbe measured by using a pressure gauge, or manometer.

Operation of a basic hydraulic circuit

A hydraulic circuit is a path for oil to flow through hoses and components. Figure 1-4shows a basic hydraulic circuit.

– The reservoir holds the oil.

– The pump “pushes” the oil, attempting to make it flow through the circuit.

– The directional control valve allows the operator to manually control the oil flowto the cylinder.

– The cylinder converts fluid energy into linear mechanical power.

– The relief valve limits system pressure to a safe level by allowing oil to flowdirectly from the pump back to the reservoir when the pressure at the pumpoutput reaches a certain level.


1-4

Figure 1-4. Basic hydraulic circuit.

With the directional control valve in the condition shown in Figure 1-4 (a), thepumped oil flows to the cap end of the cylinder. Since the oil is under pressure fromthe pump, it pushes the piston inside the cylinder, causing the piston rod to extend.The oil on the rod end of the cylinder is drained back to the reservoir through thedirectional control valve.

With the directional control valve in the condition shown in Figure 1-4 (b), thepumped oil flows to the rod end of the cylinder, causing the piston rod to retract. Theoil on the cap end of the cylinder is drained back to the reservoir through thedirectional control valve.

1-5

Exercise 1-1

Familiarization with theLab-Volt Hydraulics Trainer

EXERCISE OBJECTIVE

C To become familiar with the Lab-Volt Hydraulics Trainer;C To identify the various system components;C To be aware of the safety rules to follow when using the Lab-Volt Hydraulics

Trainer.

DISCUSSION

The Lab-Volt Hydraulics Trainer

The Lab-Volt Hydraulics Trainer consists of a work surface, hydraulic componentsand instruments, hoses, and a power unit.

Work surface

The work surface consists of a main perforated panel hinged to an oil catching trayon which hydraulic components can be mounted either horizontally or vertically. Themain panel can be tilted to facilitate the mounting of the components. Two additionalperforated panels, respectively covering a third and two thirds of the main panelsurface, can be mounted on the main panel to increase the work surface area. Anynumber of work surfaces can be positioned side by side and components be bridge-mounted across adjacent work surfaces.

Hydraulic components

Each hydraulic component is attached to a base plate that allows the component tobe secured to the work surface using either push-lock fasteners or the Quick-LockSystem. Each component has its symbol and part number indicated on a stickeraffixed on the component body or on the component base plate.

Figure 1-5a shows how a component can be secured to the work surface whenpush-lock fasteners are used. The component base plate has four identical push-lock fasteners. To secure a component to the work surface, align the four push-lockfasteners with the work surface perforations, then firmly push on the fasteners.


1-6

Figure 1-5a. Securing a component to the work surface with push-lock fasteners.

Figure 1-5b shows how a component can be secured to the work surface when theQuick-Lock System is used. With this system, each component base plate has fourfasteners: three fixed (black) fasteners and one twist-lock fastener (fastener with ayellow knob and a red tab).

(1) First, ensure that the yellow rotating knob of the twist-lock fastener is turnedfully, so that the red tab (pointed by the arrow) of this fastener is fully visible.

On some components, the yellow knob must be turned fully counterclockwise,while on other components, the yellow knob must be turned fully clockwise.

(2) Align the red pins of the four fasteners with the work surface perforations, thenpress the component base plate gently into the work surface.

(3) Lock the component into place by turning the yellow knob fully in the requireddirection, depending on the component.

Note: To secure components to the work surface, the yellow knob mustbe turned fully clockwise on some components, or fully counterclockwiseon other components.

(4) Ensure that the red tab of the twist-lock fastener is not visible, which indicatesthat the component is safely locked into place.


1-7

Figure 1-5b. Securing a component to the work surface with the Quick-Lock System.

To remove the component from the work surface, unlock the component by turningthe yellow knob fully in the required direction so that the red tab of the twist-lockfastener becomes fully visible, then withdraw the component.

Note: Throughout this manual, the components are shown withquick-lock fasteners.


1-8

Hoses

The trainer components and hoses use quick connect fittings. This type of fittingallows you to easily and quickly connect and disconnect circuits. Quick connectfittings have check valve on their end to prevent oil from running out of the hose orcomponent when the hoses are disconnected. Note, however, that these fittingsshould only be connected and disconnected when they are not under pressure.

A hose rack is provided to store the trainer hoses. The rack has a slotted top forhanging hoses, and a drip pan bottom to catch oil from the hose connectors.

Power Unit

The Power Unit supplies oil under pressure to the system. It mainly consists of anoil reservoir, a hydraulic pump, a pressure relief valve, and a filter. Figure 1-6 showsthe Power Unit, as well as its symbol.

Figure 1-6. Power Unit.

The return line filter, connected between the return line port and the reservoir (seeFigure 1-6), keeps dirt and indissoluble contaminants from entering the reservoir.This filter is equipped with a Delta-P gauge measuring the drop in pressure throughthe filter. When the pressure drop is too high, the filter must be replaced. The gaugehas a safety valve which will allow the oil to flow unfiltered into the reservoir if thefilter becomes clogged.


1-9

Figure 1-7 shows an inside view of the Power Unit. The reservoir holds the oil. Thehydraulic pump is connected directly to the electric motor shaft. It convertsmechanical power from the motor into fluid power to supply oil under pressure to thecircuit. A pressure relief valve limits system pressure and working forces to a safelevel by allowing oil to flow directly from the pump output back to the reservoir whenthe pressure at the valve reaches a certain level. This level has been factory set to6200 kPa (900 psi) @ 22°C (72°F). The maximum circuit pressure you will usethroughout this manual is about 4100 kPa (600 psi).

Figure 1-7. Inside view of the Power Unit.

Safety rules

The Lab-Volt Hydraulics Trainer has been designed with safety as a primaryconcern. However, the instructor and student must be aware of certain potentialhazards that exist when using the Hydraulics Trainer.

a. The Power Unit must be connected to an appropriate ac outlet with safetyground. The ground connection must never be removed from the end of thePower Unit line cord. If the cord does not fit your receptacle, call an electrician.The electric cord should be inspected periodically to ensure that the insulationhas not deteriorated.

b. The pressure relief valve on the Power Unit should NEVER be tampered withor readjusted.

c. Hoses, components, and other devices that are not part of the trainer should notbe used with the trainer because they may burst and injure the operator.


1-10

d. Avoid stretching or twisting the hoses. Also, avoid sharp bends which couldpinch or weaken the hose.

e. Leaks on hydraulic equipment should never be tightened while there is pressurein the system. Stop the Power Unit and release the pressure, then repair theleak.

f. Should a component or a system develop a leak that sprays or shoot a streamof fluid, do not try to cover the leak. Immediately turn off the Power Unit. Thereason for this is that high pressure oil can be forced through your skin andcause serious problems. Numerous fluid power personnel have been injectedwith fluid. An awareness of this industrial hazard will help you protect yourselfand the others from injury. Should you be injected with any fluid, get immediatemedical attention.

g. Use caution whenever any part of the trainer is under pressure. It is easy toforget that immobile components may be pressurized to as much as 4100 kPa(600 psi) or more. Make sure that the Power Unit is off whenever connecting ordisconnecting hoses.

h. Hydraulic cylinders produce tremendous forces. Never place the cylinders in aposition where they may become wedged or confined between rigid parts of thetrainer. Damage to the operator and the unit could result.

Figure 1-8. Component safety.

i. Cylinders may pinch fingers. Do not get your hands close to cylinders whenoperating the unit.


1-11

j. When using the flywheel with the hydraulic motor, be sure it is free of sharpedges or burrs. Do not allow the flywheel to turn in your hand. Always wearleather gloves when holding the flywheel. Be sure the flywheel is tight on theshaft.

k. Oil spilled on the trainer or on the floor should be cleaned immediately. Userags or towels. Granular floor-dry should be avoided in the hydraulic laboratorybecause it powders and gets into the hydraulic equipment.

l. Always wear facility approved safety glasses whenever the Hydraulics Traineris being used.

m. Before disassembling your circuits, move the lever of the directional control valvethrough all positions. This will release the pressure in the components and makehose coupling and uncoupling easier.

n. Keep the trainer and its components clean and in good working order. Cleanplastic components with mild soap and water. Harsh cleanser can cause crazing.Inspect components and other equipment for damage. Any damaged equipmentshould not be used until further inspection indicates they are safe for operation.

Following the above safety rules allows you to use the Hydraulics Trainer withoutinjury.

Procedure summary

In the first part of the exercise, you will identify the various components of yourHydraulics Trainer.

In the second part of the exercise, you will configure your work surface.

In the third part of the exercise, you will measure the pressure setting of the reliefvalve in your Power Unit.

In the fourth part of the exercise, you will verify the condition of the return line filteron your Power Unit.

EQUIPMENT REQUIRED

Refer to the Equipment Utilization Chart, in Appendix A of this manual, to obtain thelist of equipment required to perform this exercise.

PROCEDURE

Identifying the trainer components

G 1. Inspect your hydraulic Power Unit. To do so, identify the variouscomponents of the unit by writing the appropriate names in the blank spacesof Figure 1-9. Then, physically locate each component on your Power Unit.


1-12

Figure 1-9. Identifying the Power Unit components.

G 2. The components illustrated in Figure 1-10 are supplied with your HydraulicsTrainer. Get these components from their storage location, then look at thesymbol drawn on the sticker affixed on each component. Draw the symbolof each component in Figure 1-10.

G 3. Examine the Pressure Gauges. These instruments convert pressure intorotary motion which translates to a dial reading. Each Pressure Gauge isequipped with three quick connect fittings. These fittings are interconnected,so that the hoses connected to a gauge are also connected together.

The Pressure Gauges are calibrated in metric units of bars (bar) and inenglish units of pounds per square inch (psi). They measure pressuresbetween 0 and 69 bar (0 and 6900 kPa) or 0 and 1000 psi. Based on thePressure Gauge dials, how many bar is 300 psi?

G 4. How many psi is 3500 kPa?

Note: 1 bar equals 100 kPa.


1-13

Figure 1-10. Identifying the Hydraulics Trainer components.


1-14

G 5. Examine the two 5-ports Manifolds. These devices are identical. Eachmanifold has five quick connect fittings. These fittings are interconnected,so that the hoses connected to a manifold are also connected together.

One of the two 5-ports Manifolds is used as a supply manifold. It receivesthe oil under pressure directly from the Power Unit and supplies it to thecircuit. The other 5-ports Manifolds is used as a return manifold. It receivesthe oil from the circuit and returns it to the Power Unit reservoir through thefilter.

To which port on the Power Unit must the supply manifold be connected?

To which port on the Power Unit must the return manifold be connected?

G 6. Examine the various valves of your trainer. Valves are used in hydraulics tocontrol pressure and flow. Some valves have two ports. Other valves havemore. List the number of ports on each trainer valve in Table 1-1.

TYPE OF VALVE NUMBER OF PORTS

Flow Control Valve

Directional Control Valve

Relief Valve

Pressure Reducing Valve

Table 1-1. Identifying the trainer valves.

G 7. Examine the cylinders of your trainer. Cylinders are actuators that convertfluid energy into linear mechanical power. The cylinders of your trainer areof double-acting type because they work in both the extension and retractionstroke of their piston rod. List the number of ports on each cylinder inTable 1-2.

TYPE OF CYLINDER NUMBER OF PORTS

Double-acting, 2.54-cm (1-in) bore x 1.59-cm(0.625-in) rod x 10.16-cm (4-in) stroke cylinder

Double-acting, 3.81-cm (1.5-in) bore x 1.59-cm(0.625-in) rod x 10.16-cm (4-in) stroke cylinder

Table 1-2. Identifying the trainer cylinders.


1-15

Configuring the work surface

G 8. Install your work surface on a work table or on a support bench, if any. Makesure the work surface is secured to the work table or support bench toensure that it will not move or fall down. If you use a support bench, makesure the four caster brakes are locked.

G 9. Figure 1-11 shows different ways of configuring your work surface. Themain panel can be tilted to facilitate component mounting. Additional panelscan be mounted on the main panel to increase the work surface area. Theyboth can be tilted and used as control panels by mounting your hydraulicinstruments (Pressure Gauges and Flowmeter) on them.

Configure your work surface according to your preferences:

– To help you lift and tilt the panels, lift handles have been supplied withyour trainer kit. To fasten a lift handle to a panel, align the fasteners ofthe handle base plate with the panel perforations, then lock the handleinto place with the fasteners, as shown in Figure 1-12.

– To tilt a panel, slowly lift it until the desired inclination is obtained, thenhold the panel in place using the two legs on the back of the panel.Fasteners on the legs and perforations on each side of the panel allowyou to tighten down the legs, as shown in Figure 1-13. Tighten thesedown.

CAUTION!

When using tilted surfaces, always check their legs to makesure they are secure before turning on the Power Unit.Failure to this important step may result in panels orcomponents coming loose from the trainer. The result can bepersonal injury or equipment damage.


1-16

Figure 1-11. Various work surface configurations.


1-17

Figure 1-12. Fastening lift handles to a panel.

Figure 1-13. Tightening the panel legs.


1-18

Measuring the pressure setting of the Power Unit relief valve

G 10. Set up the basic circuit shown in Figure 1-14. To do so, perform thefollowing steps:

a. Mount the supply manifold (5-ports Manifold) and the Pressure Gaugeon the work surface. Secure these components to the work surface:align the fasteners on the component base plate with the perforationsof the work surface, then lock the components into place with thefasteners.

Note: Do not mount the supply manifold too near of the edge ofthe work surface. This will prevent oil from dripping onto the floorwhen you disconnect hoses from the supply manifold.

b. Connect a hose between the pressure line port of the Power Unit andthe input port of the supply manifold, as Figure 1-14 shows. Connect asecond hose between one of the four remaining ports on the supplymanifold and one of the three ports on the Pressure Gauge.

To connect a hose, pull the knurled collar back over the hose end (seeFigure 1-15), push the hose onto the fitting until it seats firmly, thenrelease the collar. To make sure a hose is firmly connected, pull on thehose. If it holds, it is correctly connected. Avoid stretching or twisting thehoses. Also, avoid sharp bends which could pinch or weaken the hose.

To disconnect a hose, push the hose toward the fitting while pulling theknurled collar back over the hose, then pull the hose off the fitting.


1-19

Figure 1-14. Basic circuit to mount.


1-20

Figure 1-15. Connecting and disconnecting a hose.

G 11. Before starting the Power Unit, perform the following start-up procedure:

a. Make sure your hoses are firmly connected.b. Check the level of the oil in the reservoir as indicated by the

temperature/oil level indicator on the Power Unit. The red line indicatesthe low oil level and the black line indicates the full oil level. With thePower Unit off, oil should cover, but not be over, the black line abovethe indicator, as Figure 1-16 shows.

Figure 1-16. Oil should cover but not be over the black line.


1-21

Fresh oil must be added to the reservoir periodically becausedisconnecting quick-connect fittings spills some oil. If required, add oilby unscrewing the reservoir breather/filler cap and by filling the reservoirup to the black line. Use one of the fluids listed on the informationsticker on the front of the reservoir. Spilled or drained oil should NOT bere-used. If re-use is imperative, the oil must be carefully strained orfiltered as it is returned to the reservoir.

c. Put on safety glasses.d. Make sure the power switch on the Power Unit is set to the OFF

position.e. Plug the Power Unit line cord into an appropriate ac outlet.

G 12. Turn on the Power Unit by setting its power switch to ON. Since the oil flowis blocked at the Pressure Gauge because there is no return path to thereservoir, all the pumped oil is now flowing through the pressure relief valveinside the Power Unit.

The Pressure Gauge reading corresponds to the pressure setting of thepressure relief valve and to the pressure at the pressure line port of thePower Unit. Record the Pressure Gauge reading below.

Gauge pressure: kPa or psi

Note: If you are working with S.I. units, multiply the measuredpressure in bar by 100 to obtain the equivalent pressure in kPa.

G 13. Turn off the Power Unit.

Verifying the condition of the return line filter

G 14. Disconnect the end of the hose connected to the input port of the supplymanifold, then connect it to the return line port of the Power Unit, asFigure 1-17 shows. This circuit allows all the pumped oil to return directly tothe reservoir through the Power Unit return line filter.


1-22

Figure 1-17. Modified circuit.

G 15. Turn on the Power Unit.

G 16. Evaluate and record the reading of the Delta-P gauge on the return linefilter. This is the drop in pressure across the return line filter, in psi.

Delta-P gauge pressure: kPa or psi

If the pressure drop is greater than 70 kPa (10 psi), the filter needs to bereplaced. Does the filter need to be replaced?

G Yes G No

G 17. Turn off the Power Unit. If the filter needs to be replaced, get instructor’sattention. Appendix B of this manual indicates how to replace the filter.


1-23

G 18. Disconnect all hoses and return them to the hose rack. The loose ends ofthe hoses must be kept inside the drip pan to prevent oil from dripping onthe floor. Wipe off any hydraulic oil residue.

G 19. Remove all components from the work surface and wipe off any hydraulicoil residue. Return all components to their storage location.

G 20. Clean up any hydraulic oil from the floor and the trainer. Properly disposeof any paper towels and rags used to clean up oil.

CONCLUSION

In this exercise, you identified the various trainer components. You connected abasic circuit restricting the system pressure through the pressure relief valve tomeasure the valve pressure setting. Next, you connected the pressure line port tothe return line port on the Power Unit and verified that the pressure drop across thereturn line filter was lower than 70 kPa (10 psi).

REVIEW QUESTIONS

1. Which port on the Power Unit provides oil under pressure to the circuit?

2. How many ports are there on the input manifold?

3. What is the purpose of the return manifold?

4. What does the Delta-P gauge on the return line filter measure?

5. Why is it necessary to have a pressure relief valve in a hydraulic circuit?


1-24

6. In the circuit shown in Figure 1-18, what should be the pressure gauge reading?Explain.

Figure 1-18. Circuit for review question 6.


1-25

7. Study the graphic diagram shown below and identify each of the letteredsymbols.

A. E.

B. F.

C. G.

D. H.

Figure 1-19. Symbol identification.

1-26

1-27

Exercise 1-2

Demonstration of Hydraulic Power

EXERCISE OBJECTIVE

C To raise a heavy load using a small hydraulic actuator;C To investigate a basic hydraulic circuit.

DISCUSSION

Hydraulic power is often called “the muscle of industry”. Hydraulic power can beused to lift entire buildings, or to move huge heavy loads. One of the most commonhydraulic power applications is to raise various objects.

Figure 1-20. Directional control valve lever moved outward from the valve body.

Figure 1-20 shows a typical hydraulic circuit using a cylinder to lift and lower a heavyload. A directional control valve controls the direction of oil flow in the system and,therefore, the direction of motion of the cylinder piston. The valve has four ports,labelled P, T, A, and B. P and T stand for pressure and tank (or reservoir), and A andB are output ports. The valve can be operated in three different positions.


1-28

When the directional control valve lever is moved outward from the valve body, asin Figure 1-20, the oil from the pump flows through path P-B of the directional controlvalve to the lower end of the cylinder. Since the oil is under pressure, it pushes upon the piston inside the cylinder, which lifts the attached load. As the piston movesupward, it forces the oil at the upper end of the cylinder to exit the cylinder. This oildrains the cylinder to the reservoir through path A-T of the valve.

When the directional control valve lever is moved toward the valve body, as inFigure 1-21, the oil from the pump flows through path P-A of the valve to the upperend of the cylinder. The oil pushes the piston downward, which lowers the attachedload. At the same time, the oil at the lower end of the cylinder flows back to thereservoir through path B-T of the directional control valve.

Figure 1-21. Directional control valve lever moved toward the valve body.

When the directional control valve lever is released, the valve automatically returnsto the center (neutral) position, as shown in Figure 1-22. In this position, all four portsare blocked and oil cannot escape from either side of the cylinder. This stops themovement of the piston and causes oil to flow from the pump back to the reservoirthrough the pressure relief valve.


1-29

Figure 1-22. Directional control valve lever centred.

Procedure summary

In the first part of the exercise, you will try to lift the hydraulic Power Unit by yourself.

In the second part of the exercise, you will set up and operate a hydraulic circuitusing a small-bore cylinder to raise and lower the Power Unit.

EQUIPMENT REQUIRED


PROCEDURE

Estimating the weight of the Power Unit

G 1. Make sure the Power Unit line cord is disconnected from the wall outlet.Make sure there are no hoses connected to the Power Unit.


1-30

G 2. Try to lift the Power Unit to feel how heavy it is. Be careful not to tip or dropthe Power Unit. Can the Power Unit be easily lifted?

G Yes G No

G 3. How much do you think the Power Unit weighs?

Setup

G 4. Get the 2.54-cm (1-in) bore cylinder from your storage location. Remove thecylinder from its adapter by unscrewing its retaining ring, as shown inFigure 1-23. Make sure the cylinder tip (bullet) is removed from the cylinderrod end.

Figure 1-23. Unscrew the retaining ring and remove the cylinder.

G 5. Insert the cylinder rod into the cylinder hole in the Power Unit lifting frame,as shown in Figure 1-24 a). Fasten the cylinder to the lifting frame bytightening its retaining ring securely.


1-31

Figure 1-24. Power Unit installation.

G 6. Make sure the Power Unit is near of your work surface. Position the liftingframe over the Power Unit, with its open side at the rear of the Power Unit,as Figure 1-24 b) shows.


1-32

CAUTION!

For safety purposes, the base of the lifting frame has threefull sides used to prevent anybody from placing his feetunder the Power Unit when it is lifted. Therefore, make surethe open side of the lifting frame is at the rear of the PowerUnit.

G 7. Get an extra-long hose (1.52 m/60 in) from your storage location and fill itwith oil. To do so, connect one end of the hose to the pressure line port onthe Power Unit and the other end to the return port of the Power Unit.

Before starting the Power Unit, perform the following start-up procedure:

a. Check the level of the oil in the reservoir. Add fresh oil if required.b. Put on safety glasses.c. Make sure the power switch on the Power Unit is set to the OFF

position.d. Plug the Power Unit line cord into an ac outlet.

Turn on the Power Unit by setting its power switch to ON; this will fill thehose with oil. Turn off the Power Unit.

Remove the hose filled with oil and fill a second extra-long hose with oil byrepeating the same procedure. Turn off the Power Unit and remove thesecond hose.

G 8. Connect the two ports of the cylinder to each other by using one of theextra-long hoses you filled with oil. Slowly pull the piston rod of the cylinderout until it touches the lifting attachment on the Power Unit, asFigure 1-24 c) shows.

G 9. Fasten the cylinder to the Power Unit by screwing the lifting attachment ontothe threaded end of the cylinder rod, as Figure 1-24 d) shows. Then,disconnect the hose from the cylinder.

G 10. Connect the circuit shown in Figure 1-25. Use the two extra-long hoses filledwith oil to connect the cylinder to ports A and B of the Directional ControlValve. Relate each hose connection of this circuit to the pictorial diagram inFigure 1-26.

Note: For ease of connection, the Directional Control Valvesupplied with your Hydraulics Trainer is bolted to a sub-plate towhich the hoses can be connected. The arrangement of ports P,T, A, and B on the valve subplate does not follow the symbol forthe directional valve appearing on the manufacturer nameplate ontop of the valve and on the Lab-Volt symbol sticker. Thus, port Pactually faces port B on the subplate, while port T faces port A.Therefore, always refer to the letters stamped on the valvesubplate when connecting the valve into a circuit.


1-33

Figure 1-25. Setup used for lifting the Power Unit.


1-34

Figure 1-26. Connection diagram of the circuit in Figure 1-25.

G 11. Have your instructor verify your setup. Do NOT proceed to the next stepuntil your setup has been approved.

Lifting the Power Unit using a small cylinder

G 12. Before starting the Power Unit, perform the following steps:

a. Make sure the hoses and the Power Unit line cord will not becomewedged between rigid parts of the trainer when the Power Unit is lifted.

b. Make sure the Relief Valve is connected correctly. The Pressure (P)port must be connected to the supply manifold. The Tank (T) port mustbe connected to the return manifold. The Vent (V) port must be leftunconnected.

c. Make sure the hoses are firmly connected.d. Put on safety glasses.


1-35

G 13. Pull the Relief Valve adjustment knob and turn it fully counterclockwise,then turn it 3 turns clockwise. Use the vernier scale on the knob foraccurate adjustment.

G 14. Turn the Flow Control Valve adjustment knob fully clockwise, then turn it1 turn counterclockwise. Use the vernier scale on the knob for accurateadjustment.

G 15. Make sure that all persons are standing clear of the Power Unit. Turn on thePower Unit. The cylinder rod should not move yet.

G 16. Retract the cylinder rod by moving the lever of the Directional Control Valveoutward from the valve body, as Figure 1-27 shows. Keep the lever shifteduntil the rod is fully retracted, then release it. What happens to the PowerUnit?

CAUTION!

Do not place any part of your body under the Power Unitwhile it is hanging from the lifting frame.

Figure 1-27. Directional Control Valve lever positions.

G 17. Extend the cylinder rod by moving the lever of the Directional Control Valvetoward the valve body. What happens to the Power Unit?


1-36

G 18. Move the lever of the Directional Control Valve outward from the valve body,then release it while the cylinder is retracting and in midstroke. Does thecylinder rod stop or does the Power Unit begin to move downward?

G 19. Turn off the Power Unit. Does the Power Unit begin to move downward?

G Yes G No

G 20. Move the lever of the Directional Control Valve toward the valve body. Theweight of the Power Unit will push the cylinder down. Keep the lever shifteduntil the rod is fully extended and the Power Unit has returned to the ground,then release it.

G 21. Open the Relief Valve completely by turning its adjustment knob fullycounterclockwise.

G 22. Disconnect the line cord of the Power Unit from the wall outlet. Disconnectall hoses. Wipe off any hydraulic oil residue.

Note: If you experience difficulty to disconnect equipment, movethe directional valve lever back and forth to relieve static pressurethat might be trapped in the A and B cylinder lines.

G 23. Unscrew the cylinder from the lifting attachment on the Power Unit. Unscrewthe ring retaining the cylinder to the lifting frame. Remove the cylinder fromthe lifting frame. Reinstall the cylinder on its adapter by fastening itsretaining ring securely.


G 25. Clean up any hydraulic oil from the floor and from the trainer. Properlydispose of any paper towels and rags used to clean up oil.

CONCLUSION

In the first part of this exercise, you tried to lift the Power Unit by yourself and sawthat it was quite heavy. You then set up and operated a circuit using a smallhydraulic cylinder to lift and lower the Power Unit. The small cylinder easily lifted andlowered the Power Unit. Hydraulic circuits are often used as non-flowing or staticcircuits. Static circuits transmit power by pushing on a confined liquid, as opposedto dynamic circuits, which transmit power by using the energy associated with motionof a liquid.


1-37

REVIEW QUESTIONS

1. What caused the cylinder rod to retract during this exercise?

2. How do you explain that such a small cylinder can lift such a heavy load?

3. Figure 1-28 shows the circuit you used in this exercise (without the Relief Valveand Flow Control Valve ). Draw arrow heads on the lines in Figure 1-28 to showthe flow of fluid in your circuit. Use arrow heads as illustrated in the diagram.


1-38

2-1

Unit 2

Fundamentals

UNIT OBJECTIVE

When you have completed this unit, you will be able to state the laws governinghydraulics, and perform simple calculations involving force, pressure, area, velocity,and rate of flow.


Hydraulic equipment is found in industry, on agricultural machinery, and onconstruction machinery, as Figure 2-1 shows. While each machine or job may usea different style of components in its circuits, the concept or basic ideas behind thesecomponents is the same.

This unit demonstrates how flowing oil behaves and shows the difference betweenflowing and non-flowing systems. This may be your first exposure to physical lawssuch as the Pascal’s Law. As you proceed, you will be introduced to several otherlaws which you will test and use. You will find that these laws are very important inthe field of hydraulics.

Fundamentals

2-2

Figure 2-1. Hydraulic applications.

2-3

Exercise 2-1

Pressure Limitation

EXERCISE OBJECTIVE

C To introduce the operation of a pressure relief valve.C To establish the oil flow path in a circuit using a pressure relief valve.C To connect and operate a circuit using a pressure relief valve.

DISCUSSION

Pressure limitation

Pressure is the amount of force exerted against a given surface. Flow is themovement of fluid caused by a difference in pressure between two points. Fluidalways flows from a higher pressure point to a lower pressure point. The citywaterworks, for example, builds up a pressure greater than the atmospheric pressurein our water pipes. As a result, when we turn on a water tap, the water is forced out.

When two parallel paths of flow are available, fluid will always take the path of leastresistance. An example of this in the everyday life would be a garden hose branchinginto two sections, as Figure 2-2 shows. One section is blocked, while the othersection allows water to move freely in it. All the water will flow through the unblockedsection since it is less restrictive than the blocked section. The input pressure willrise just enough for the water to flow through the unblocked section. The pressurein the blocked section will not build up beyond the level required to make the waterflow in the unblocked section. The pressure gauges in Figure 2-2, therefore, willindicate low, equal pressures.

Figure 2-2. Unrestricted flow path.

Pressure Limitation

2-4

Now what happens if we squeeze the unblocked section so that water is restrictedbut not completely confined, as Figure 2-3 shows? All the water will flow through thesqueezed section since it is still less restrictive than the blocked section. The inputpressure will rise to the level necessary to flow through the restricted path. Thepressure in the blocked section will not build up beyond the needs of the squeezedsection. The pressure gauges in Figure 2-3, therefore, will indicate high, equalpressures.

Figure 2-3. Restricted flow path.

So we see that the pressure in the blocked section can never be higher than thepressure in the unblocked section. In fact, these pressures will always be equal. Ifthe restricted section were closed completely, confining water instead of merelyrestricting it, the pressure in both sections would equal the maximum pressureavailable at the input.

In a hydraulic circuit, flow is produced by the action of a pump, which continuouslydischarges the oil at a certain flow rate. Pressure is not created by the pump itselfbut by resistance to the oil flow. When the oil is allowed to flow with no resistancethrough a hydraulic circuit, the pressure in that circuit is theoretically zero. When theflow is resisted, however, the circuit pressure increases to the amount necessary totake the easier path.

Relief valves

Figure 2-4 shows a hydraulic circuit consisting of two parallel paths of flow. The oilfrom the pump can pass through a relief valve or through a hydraulic circuitconsisting of a directional control valve and a cylinder.

The relief valve can be compared to the hand in the hose example previouslydescribed. It limits the maximum pressure in the system by providing an alternateflow path to the reservoir whenever the oil flow to the circuit is blocked, as when thedirectional valve is in the blocked center position or when the cylinder is fullyextended or retracted.

Pressure Limitation

2-5

The relief valve is connected between the pump pressure line and reservoir. It isnormally non-passing. It is adjusted to open at a pressure slightly higher than thecircuit requirement and divert the pumped oil to the reservoir when this pressure isreached.

In Figure 2-4, for example, all the oil from the pump flows through the circuit as longas the cylinder is not fully extended because the circuit provides an easier path thanthe relief valve. While the cylinder is extending, the pressure rises only to the amountnecessary to force oil on the rod side of the cylinder into the reservoir (here 700 kPa,or 100 psi).

Figure 2-4. Oil flows through the circuit.

Once the cylinder is fully extended, the cylinder circuit becomes blocked and thepumped oil can no longer flow through it. The system pressure climbs to 3450 kPa(500 psi), then the relief valve opens and the oil is dumped back to the reservoir atthe relief valve pressure setting of 3450 kPa (500 psi), as Figure 2-5 shows.Thereafter, no flow occurs throughout the circuit and the pressure is equalthroughout. The circuit pressure, therefore, cannot build up beyond the relief valvepressure setting.

Pressure Limitation

2-6

Figure 2-5. Oil flows through the relief valve.

Hydraulics Trainer relief valves

Your Hydraulics Trainer contains two relief valves. One of these valves, called mainrelief valve, is located inside the Power Unit. The other valve, called secondaryrelief valve, is supplied with your kit of hydraulic components. The two valves areidentical. However, you will operate the secondary valve only. The main valve isfactory-set at a higher pressure than the secondary valve. It is used as an additionalsafety device for backing up the secondary valve. It should not be readjusted ortampered with.

Figure 2-6 illustrates the relief valve supplied with your kit of hydraulic components.This valve is of pilot-operated type. The valve body has three ports: a pressure (P)port, which is to be connected to the pump pressure line, a tank (T) port, which is tobe connected to the reservoir, and a vent (V) port, which is used for control of thevalve from a remote point by external valve. The use of the vent port will bediscussed in detail in Exercise 4-4. When not used, this port should be leftunconnected.

By sensing the upstream pressure on the P port of the valve, an internal spoolcontrols the flow of oil through the valve by acting on a large spring. The pressurelevel where the spool is wide open and all the pumped oil passes through the valveis called relieving pressure, or full-open pressure.

The relieving pressure can be set by using the adjustment knob on the valve body.Turning the knob clockwise increases the compression of a small spring locatedabove the valve spool, which increases the relieving pressure and allows higherpressures to build up in the circuit. Notice that the knob must first be pulled beforeit can be turned. When the knob is released, a spring forces the knob to engage afixed spline. This prevents vibrations and shocks from changing the adjustment.

Pressure Limitation

2-7

The pressure at which the relief valve begins to open is called cracking pressure.This pressure is below the valve relieving pressure. At cracking pressure, the valveopens just enough to let the first few drops of oil through. Pressure override is thepressure difference between the cracking pressure and the relieving pressure.

Before turning on the Power Unit, the valve should always be completely open(adjustment knob turned fully counterclockwise) to allow the pump to start under thelightest load and to prevent the system components from being subjected topressure surges. Once the Power Unit is running, the relief valve can be closedgradually until the desired pressure is reached.

REFERENCE MATERIAL

For detailed information on pilot-operated relief valves, refer to the chapter entitledPilot Operated Pressure Control Valve in the Parker-Hannifin’s manual IndustrialHydraulic Technology.

Pressure Limitation

2-8

Figure 2-6. The trainer Relief Valve.

Procedure summary

In the first part of the exercise, you will measure the cracking pressure of the ReliefValve supplied with your kit of hydraulic components. You will adjust the valverelieving pressure by modifying the compression of its spring.

In the second part of the exercise, you will test the effect of pressure limitation on abasic hydraulic circuit.

Pressure Limitation

2-9

EQUIPMENT REQUIRED


PROCEDURE

Relief valve operation

G 1. Connect the circuit shown in Figure 2-7. Refer to the connection diagramshown in Figure 2-8 to make your connections.

Note: As Figure 2-7 shows, the vent (V) port of the Relief Valveis unused in this circuit. Therefore, leave this port unconnected.

Figure 2-7. Schematic diagram of the circuit for adjusting the Relief Valve.

Pressure Limitation

2-10

Figure 2-8. Connection diagram of the circuit for adjusting the Relief Valve.


a. Make sure the hoses are firmly connected.b. Check the level of the oil in the reservoir. Add oil if required.c. Put on safety glasses.d. Make sure the power switch on the Power Unit is set to the OFF

position.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve completely. To do so, pull the valve adjustment

knob and turn it fully counterclockwise.

G 3. Turn on the Power Unit by setting its power switch to ON. Since the oil flowis blocked at gauge A, all the pumped oil is now being forced through theRelief Valve.

The pressure reading of gauge A is the minimum pressure required todevelop an oil flow through the valve (cracking pressure). It corresponds tothe pressure required to counteract the resistance of the spring inside thevalve. Record below the pressure reading of gauge A.

Cracking pressure = kPa or psi

Pressure Limitation

2-11

Note: The Trainer Pressure Gauges provide “bar” and “psi”readings. Since bar is a metric unit of measurement forpressures, students working with S.I. units must multiply themeasured pressure in bars by 100 to obtain the equivalentpressure in kilopascals (kPa).

G 4. Now, compress the spring of the Relief Valve by turning its adjustment knobclockwise 2 turns. Use the vernier scale on the knob for the adjustment.What is the reading of gauge A?

Pressure = kPa or psi

G 5. Why does the pressure reading increase as the spring compression isincreased?

G 6. Turn the Relief Valve adjustment knob fully clockwise while watching thereading of gauge A. Can the pressure level be increased beyond 6200 kPa(900 psi)? Why?


G 8. Based on the cracking pressure recorded in step 3, at which pressure willthe Relief Valve start to open if the relieving pressure is set to 3450 kPa(500 psi)?

Limiting system pressure

G 9. Modify the existing circuit in order to obtain the circuit shown in Figures 2-9and 2-10. Make sure to mount the 3.81-cm (1.5-in) bore cylinder in aposition where its rod can extend freely.

Pressure Limitation

2-12

Note: For ease of connection, the Directional Control Valvesupplied with your Hydraulics Trainer is bolted to a subplate towhich the hoses can be connected. The arrangement of ports P,T, A, and B on the valve subplate does not follow the symbol forthe directional valve appearing in Figure 2-9 and on themanufacturer nameplate on top of the valve. Thus, port P actuallyfaces port B on the subplate, while port T faces port A. Therefore,always refer to the letters stamped on the valve subplate whenconnecting the valve into a circuit.

Figure 2-9. Schematic diagram of the cylinder actuation circuit.

G 10. Make sure the hoses are firmly connected. Open the Relief Valvecompletely by turning its adjustment knob fully counterclockwise.


G 12. Turn the Relief Valve adjustment knob clockwise until gauge A reads1400 kPa (200 psi).

G 13. Stay clear of the cylinder rod. Move the lever of the Directional ControlValve toward the valve body, which should extend the cylinder rod. Then,move the lever outward from the valve body, which should retract the rod.

Pressure Limitation

2-13

Figure 2-10. Connection diagram of the cylinder actuation circuit.

G 14. While watching the reading of gauge A, move the lever of the directionalvalve toward the valve body to extend the cylinder rod. What is the pressureat gauge A during the extension stroke of the rod?


G 15. What is the pressure at gauge A when the cylinder is fully extended?


G 16. Move the lever of the directional valve outward from the valve body toretract the cylinder rod.


Pressure Limitation

2-14

G 18. While watching the reading of gauge A, move the lever of the directionalvalve toward the valve body to extend the cylinder rod. What is the pressureat gauge A during the extension stroke of the cylinder rod?


G 19. What is the pressure at gauge A when the cylinder rod is fully extended?



G 21. Turn off the Power Unit. Open the Relief Valve completely by turning itsadjustment knob fully counterclockwise.

G 22. Explain the reason for the nearly identical pressures registered duringcylinder extension at the two relief valve pressure settings.

G 23. Why does the circuit pressure increase when the cylinder rod is fullyextended?

G 24. Disconnect all hoses. It may be necessary to move the directional valvelever back and forth to relieve static pressure; the quick connects can thenbe removed. Wipe off any hydraulic oil residue.



Pressure Limitation

2-15

CONCLUSION

In the first part of the exercise, you measured the minimum pressure setting of arelief valve by connecting the valve between the pump pressure line and reservoirand by opening the valve completely.

You then modified the valve relieving pressure by increasing the compression of itsinternal spring, which increased the circuit pressure.

In the second part of the exercise, you tested the effect of pressure limitation on abasic hydraulic circuit. You learned that pressure changes depend on the movementof oil through the circuit. When the cylinder rod extends or retracts, the circuitpressure rises only to the amount required to force oil out of the cylinder back intothe reservoir. When the cylinder rod becomes fully extended or retracted, however,the circuit pressure rises to the relief valve pressure setting.

Up to that point, we have seen that pilot-operated relief valves provide pressurecontrol by sensing pressure upstream on their input line. Pilot-operated relief valvescan also sense pressure in another part of the system or even in a remote systemby means of a vent line. This type of operation is identified as remote control and isachieved through the use of the relief valve vent port. Remote control of a relief valvewill be described in detail in Exercise 4-4.

REVIEW QUESTIONS

1. What is the purpose of a relief valve?

2. Explain the difference between the main relief valve in the Power Unit and theRelief Valve supplied with your kit of hydraulic components (secondary reliefvalve)?

3. What type of relief valve is used in your Hydraulics Trainer?

Pressure Limitation

2-16

4. What might happen to a hydraulic system if the tank port of the relief valve is notconnected to the power unit return line port?

5. Define the term cracking pressure.

6. In the circuit of Figure 2-11, what will be the pressure reading of gauge A duringcylinder extension and when the cylinder is fully extended if the relief valvepressure setting is changed from 3400 kPa (500 psi) to 6900 kPa (1000 psi)?

Note: The pressure required to extend the cylinder rod is 600 kPa(85 psi).


2-17

Exercise 2-2

Pressure and Force

EXERCISE OBJECTIVE

C To verify the formula F = P x A using a cylinder and a load spring;C To discover what happens to a cylinder when equal pressure is applied to each

side of its piston;C To explain the concept of pressure distribution in a cylinder in equilibrium of

forces;C To determine the weight of the Power Unit given the pressure required to lift it.

DISCUSSION

Pascal’s Law

Pascal’s Law states that pressure applied on a confined fluid is transmittedundiminished in all directions, and acts with equal force on equal areas, andat right angles to them.

Figure 2-12 illustrates this basic properties of fluids. The bottle in this example iscompletely filled with a non-compressible fluid. When a stopper is placed in the topof the bottle, and a force is applied to the top of the stopper, the fluid inside the bottleresists compression by pushing with an equal pressure in all directions.

Figure 2-12. Force applied to a confined fluid.

The generated pressure is equal to the force applied to the top of the stopper dividedby the area of the stopper. In equation form:

Pressure and Force

2-18

S.I. units:

English units:

As you can see, pressure is measured in “Newtons per 10 square centimeters (kPa)”in S.I. units, or in “pounds per square inch” (psi) in English units. In physics formula,the word “per” can be rewritten as a division sign.

Memorizing the pyramid in Figure 2-13 will make rearranging of formula P = F/Aeasier. In the pyramid, the letter on the top row equals the product of the bottom twoletters. A letter on the bottom row equals the top letter divided by the other bottomletter.

Figure 2-13. Rearranging formulas.

Hydraulic pressure versus cylinder force

In a hydraulic circuit, the force that pushes the oil, attempting to make it flow, comesfrom a mechanical pump. When the oil pushed on by the pump is confined withina restricted area, as in the body of a cylinder, there is a pressure build-up, and thispressure can be used to do useful work.

As you can see, pressure is not created by the pump but by resistance to theoil flow. The amount of pressure created in a circuit will only be as high as requiredto counteract the least resistance to flow in the circuit. Resistance to flow mainlycomes from three sources: resistance to motion of the load attached to the cylinder,frictional resistance of the cylinder seals, and frictional resistance of the inner wallof the hoses.

Pressure and Force

2-19

In Figure 2-14, the oil from the pump is confined in the cap end of the cylinder. Asa result, pressure develops in the cap end of the cylinder. This pressure is exertedevenly over the entire surface of the cap end of the cylinder. It acts on the piston,resulting in a mechanical force to push the load.

Figure 2-14. Cylinder pushing a load.

To find the amount of force generated by the piston during its extension, we canrewrite the formula P = F/A as F = P x A. Therefore, the generated force is equal tothe pressure in the cap end of the cylinder times the piston area being acted upon.This area is called full area, or “face” area.

In Figure 2-15, the oil from the pump is confined in the rod end of the cylinder. Asa result, pressure develops in the rod end of the cylinder. This pressure is exertedevenly over the entire surface of the rod end of the cylinder. It acts on the piston,resulting in a mechanical force to pull the load.

Pressure and Force

2-20

Figure 2-15. Cylinder pulling a load.

This time, however, the generated force is lower because the piston area availablefor the pressure to act on is reduced by the fact that the cylinder rod covers a portionof the piston. This area is called annular area, or “donut” area. Therefore, the systemmust generate more pressure to pull than to push the load.

Conversion factors

Table 2-1 shows the conversion factors used to convert measurements of force,pressure, and area from S.I. units to English units and vice versa.

Pressure and Force

2-21

Force

Newtons (N) x 0.225 = Pounds-force(lb; lbf)

x 4.448 = Newtons (N)

Pressure

Kilopascals (kPa) x 0.145 = Pounds-forceper square inch(psi; lb/in2; lbf/in2)

x 6.895 = Kilopascals (kPa)

Area

Square centimeters(cm2)

x 0.155 = Square inches(in2)

x 6.45 = Square centimeters(cm2)

Table 2-1. Conversion factors.

For example, the pressure generated by the fluid in Figure 2-12 is 10 kPa, inS.I. units, or 1.45 psi, in English units, as demonstrated below:

S.I. units:

English units:

REFERENCE MATERIAL

For additional information on the relationship between force and pressure, refer tothe chapters entitled Hydraulic Transmission of Force and Energy and HydraulicActuators in the Parker-Hannifin’s manual Industrial Hydraulic Technology.

Procedure summary

In the first part of the exercise, you will verify the formula F = P x A by measuring thecompression force of a cylinder on a loading device.

In the second part of the exercise, you will predict and demonstrate what happenswhen equal pressure is applied to both sides of a piston.

In the third part of the exercise, you will determine how much pressure there is ineach side of a cylinder in equilibrium of forces.

Pressure and Force

2-22

In the fourth part of the exercise, you will measure the pressure required to lift thePower Unit in order to determine its weight.

EQUIPMENT REQUIRED


PROCEDURE

Conversion of pressure to force

G 1. What is the formula for determining the force in a hydraulic system?

G 2. If the bore diameter, D, of a cylinder piston equals 3.81 cm (1.5 in), calculatethe full area, Af, of this piston. Use the formula shown in Figure 2-14. It isgiven below for your convenience.

G 3. Using this area and the formula from step 1, calculate the theoretical forceof the cylinder for the pressure levels in Table 2-2. Record your calculationsin Table 2-2 under “THEORETICAL”.

PRESSURE APPLIED ONFULL PISTON AREA

THEORETICAL CYLINDERFORCE

ACTUAL CYLINDER FORCE

3500 kPa (500 psi)

2800 kPa (400 psi)

2100 kPa (300 psi)

Table 2-2. Cylinder force versus pressure.

G 4. Remove the 3.81-cm (1.5-in) bore cylinder from its adapter by unscrewingits retaining ring. Make sure the cylinder tip (bullet) is removed from thecylinder rod end.

G 5. As Figure 2-16 (a) shows, screw the cylinder into the Loading Device untilthe load piston inside the Loading Device begins to push on the spring andthe cylinder fittings point upwards. Do not use a tool to turn the cylinder!

Pressure and Force

2-23

Figure 2-16. Loading device assembly.

Note: If the 3.81-cm (1.5-in) bore cylinder is not fully retracted,do not try to screw the cylinder into the Loading Device. Insteadconnect the cylinder actuation circuit of Figure 2-10. Turn theknob of the relief valve fully counterclockwise, then turn on thepower unit. Actuate the lever of the directional control valve toretract the cylinder rod fully, then turn off the power unit.Disconnect the circuit. Now screw the cylinder into the LoadingDevice as shown in Figure 2-16 (a).

Pressure and Force

2-24

G 6. Clip the NEWTON/LBF-graduated ruler to the Loading Device, and align the“0 Newton” or “0 lbf” mark with the colored line on the load piston.Figure 2-16 (b) shows ruler installation for measurement of forces inNewtons (N). The ruler must be installed on the other side of the LoadingDevice in order to measure forces in pounds (lbf or lb).

G 7. Connect the circuit shown in Figures 2-17 and 2-18.

Figure 2-17. Schematic diagram of the circuit for measuring the output force of a cylinder.


a. Make sure the hoses are firmly connected.b. Check the level of the oil in the reservoir. Add oil if required.c. Put on safety glasses.d. Make sure the power switch on the Power Unit is set to the OFF pos-

ition.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve by turning its adjustment knob fully

counterclockwise.

Pressure and Force

2-25

Figure 2-18. Connection diagram of the circuit for measuring the output force of a cylinder.


G 10. Move the lever of the directional valve toward the valve body to directthe pumped oil toward the cap end of the cylinder. While keeping the valvelever shifted, turn the Relief Valve adjustment knob clockwise until thepressure at gauge A equals 4100 kPa (600 psi). Observe that the appliedpressure caused the cylinder to compress the spring in the Loading Device.

G 11. With the lever of the directional valve still shifted toward the valve body, turnthe Relief Valve adjustment knob counterclockwise to decrease thepressure at gauge A at 3500 kPa (500 psi). Note the force reading on theLoading Device, and record this value in Table 2-2 under “ACTUAL”.

Note: To counteract hysteresis of the spring load device andobtain a more accurate force reading at 3500 kPa (500 psi), thepressure applied to the cylinder piston must first be set at4100 kPa (600 psi) and then decreased at 3500 kPa (500 psi).

Pressure and Force

2-26

G 12. By shifting the directional valve lever and adjusting the knob on the ReliefValve to decrease the pressure at gauge A in steps, measure the forcereading for the pressure levels in Table 2-2. Record your results inTable 2-2.

G 13. When you have finished, move the directional valve lever outward from thevalve body to retract the rod, then turn off the Power Unit. Open the ReliefValve completely (turn knob fully counterclockwise).

G 14. Compare the actual forces you obtained in the experiment with thetheoretical forces in Table 2-2. Are these values within 10% of each other?

G Yes G No

G 15. Does force increase or decrease as pressure increases?

Applying equal pressure on both sides of a piston

G 16. Connect the circuit shown in Figure 2-19. Use the 2.54-cm (1-in) borecylinder.

Figure 2-19. Applying equal pressure on both sides of a piston.

Pressure and Force

2-27

Note: If the 2.54-cm (1-in) bore cylinder is not fully retracted, donot connect the circuit of Figure 2-19. Instead connect thecylinder actuation circuit shown in Figure 2-10. Open the reliefvalve completely, then turn on the power unit. Actuate the leverof the directional valve to retract the cylinder rod fully, then turnoff the power unit. Disconnect the circuit. Now, connect the circuitof Figure 2-19.

G 17. Examine the circuit of Figure 2-19. This circuit applies equal pressure to thefull and annular sides of the piston. However, the piston area available forthe pressure to act on is less on the annular side because the cylinder rodcovers a portion of the piston. Given that force is equal to pressuremultiplied by area, predict which side of the piston will develop the mostforce.

G 18. What do you think will happen to the cylinder rod?


G 20. Turn the Relief Valve adjustment knob clockwise until the circuit pressureat gauge A equals 2100 kPa (300 psi).

G 21. While observing the cylinder rod, move the lever of the directional valvetoward the valve body so that the pumped oil is directed toward both sidesof the cylinder piston. In which direction does the rod move? Why?

G 22. Turn off the Power Unit. Open the relief valve completely (turn knob fullycounterclockwise).

Pressure distribution in a cylinder in equilibrium of forces

G 23. Disconnect the 2.54-cm (1-in) bore cylinder from the circuit. Connect the twoports of this cylinder to the return manifold. Slowly push the piston rod inuntil it is retracted half way. Disconnect the cylinder from the returnmanifold, then connect the circuit shown in Figure 2-20 (a).

Pressure and Force

2-28

Figure 2-20. Determining pressure distribution in a cylinder.

Pressure and Force

2-29

G 24. Examine the circuit in Figure 2-20 (a). The oil in the rod side of the cylinderis captured because there is no return path to the reservoir. The pistoncannot move because the oil cannot be compressed. The pressures in thecap and rod sides build until the forces exerted on both sides of the pistonare exactly equal.

– The force on the full area of the piston is:

– The force on the annular area of the piston is:

– Since these forces are equal:

Based on the above formulas, predict which gauge in Figure 2-20 (a) willread the most pressure. Explain.

G 25. Turn on the Power Unit. Turn the relief valve adjustment knob clockwiseuntil the input pressure at gauge A (Pf) is 1400 kPa (200 psi). Then, recordthe output pressure at gauge B (Pa) in Table 2-3.

PART A

INPUT PRESSUREAPPLIED ONFULL PISTON

AREA

INPUT PRESSUREAT GAUGE A

(Pf)

OUTPUT PRESSURE ATGAUGE B

(Pa)

INPUT/OUTPUTPRESSURE RATIO

(Pf/Pa)

RECIPROCAL OFAREA RATIO (Aa/Af)

1400 kPa(200 psi)

2100 kPa(300 psi)

PART B

INPUTPRESSURE

APPLIED ONANNULAR

AREA

INPUT PRESSUREAT GAUGE A (Pa)

OUTPUT PRESSURE ATGAUGE B (Pf)

INPUT/OUTPUTPRESSURE RATIO

(Pa/Pf)

RECIPROCAL OFAREA RATIO (Af/Aa)

1400 kPa(200 psi)

2100 kPa(300 psi)

Table 2-3. Pressure distribution in the cylinder of Figure 2-20.

G 26. Increase the Relief Valve pressure setting until the input pressure atgauge A is 2100 kPa (300 psi) and again record the output pressure atgauge B in Table 2-3.

G 27. Turn off the Power Unit. Open the Relief Valve completely (turn knob fullycounterclockwise).

G 28. Switch the two hoses connected to the ports of the cylinder with each otherso that the rod end is connected to gauge A and the cap end is connectedto gauge B, as Figure 2-20 (b) shows.

Pressure and Force

2-30

G 29. Examine the circuit in Figure 2-20 (b). Predict which gauge will read themost pressure. Explain why.

G 30. Turn on the Power Unit. Turn the Relief Valve adjustment knob clockwiseuntil the input pressure at gauge A (Pa) is 1400 kPa (200 psi). Record theoutput pressure at gauge B (Pf) in Table 2-3.

G 31. Increase the Relief Valve pressure setting until the input pressure atgauge A is 2100 kPa (300 psi) and again record the output pressure atgauge B in Table 2-3.


G 33. Complete the fourth column of Table 2-3, “INPUT/OUTPUT PRESSURERATIO”, using the input and output pressures registered in parts A and B ofthe experiment.

G 34. Complete the fifth column of Table 2-3, “RECIPROCAL OF AREA RATIO”,given that the piston diameter, D, is 2.54 cm (1 in) and the rod diameter, d,is 1.59 cm (0.625 in). Use the formulas below to determine Af and Aa.

Af = D² x 0.7854

Aa = (D² ! d²) x 0.7854

G 35. In a cylinder in equilibrium of forces, the ratio of input to output pressure istheoretically equal to the reciprocal (inverse) of the area ratio.

Compare the input/output pressure ratios Pf/Pa in Table 2-3, part A, with thereciprocal of area ratio, Aa/Af. Also, compare the input/output pressure ratiosPa/Pf in Table 2-3, part B, with the reciprocal of area ratio, Af/Aa. Are thepressure ratios approximately equal to the reciprocal of area ratio?

G Yes G No

Pressure and Force

2-31

Measuring the weight of the Power Unit

G 36. Disconnect the Power Unit line cord from the wall outlet.

G 37. Disconnect the 2.54-cm (1-in) bore cylinder from the circuit, then remove thecylinder from its adapter.

G 38. Insert the cylinder rod into the cylinder hole in the Power Unit lifting frame,then fasten the cylinder to the lifting frame by tightening its retaining ringsecurely. Position the lifting frame over the Power Unit, with its open side atthe rear of the Power Unit.

G 39. Connect the two cylinder ports together using a hose full of oil, then pull thepiston rod out until it touches the lifting attachment on the Power Unit.Fasten the cylinder to the Power Unit by screwing the lifting attachment ontothe threaded end of the cylinder rod. Then, disconnect the hose from thecylinder.

G 40. Connect the circuit shown in Figure 2-21.

CAUTION!

Make sure the hoses and Power Unit line cord will notbecome wedged between rigid parts of the trainer when thePower Unit is lifted.

G 41. Plug the Power Unit line cord into the wall outlet, then turn on the PowerUnit.

G 42. Move the lever of the directional valve outward from the valve body andslowly turn the Relief Valve adjustment knob clockwise until the Power Unitbegins to rise. Then, release the valve lever.

G 43. According to gauge A, how much pressure is currently applied on theannular area of the cylinder piston?


G 44. Move the lever of the directional valve toward the valve body to return thePower Unit to the ground.

Pressure and Force

2-32

Figure 2-21. Circuit used to lift the Power Unit.


G 46. Based on the annular pressure recorded in step 43, determine the weight(mass) of the Power Unit in both S.I. and English units.

Note: 1 Newton is equal to 0.1019 kilogram.

G 47. Disconnect the Power Unit line cord from the wall outlet, then disconnect allhoses. Wipe off any hydraulic oil residue.

G 48. Unscrew the cylinder from the Power Unit lifting attachment. Unscrew thering retaining the cylinder to the lifting frame. Remove the cylinder from thelifting frame. Reinstall the cylinder on its adapter by fastening its retainingring securely.


Pressure and Force

2-33


CONCLUSION

In this exercise, you learned that the force exerted on a given surface is directlyproportional to the pressure applied on this surface. Since the relationship betweenforce and pressure is linear, it is possible to predict the force exerted by the cylinderat any pressure setting.

You also learned what happens when equal pressure is applied to both sides of apiston. Since the working area on the rod side of the cylinder is less than the workingarea on the cap side of the cylinder, the piston has a tendency to extend when equalpressure is applied to each end.

You then determined the pressure distribution in a cylinder in equilibrium of forces.The cylinder was blocked and the oil was captured in the cylinder, so the pressuresin the cap and rod sides had to build until the forces exerted on both sides of thepiston were exactly equal. The rod side of a cylinder in equilibrium of forces mustbuild more pressure than the cap side because the working area on the rod side(annular area) is less than the working area on the cap side (full area).

Finally, you measured the pressure required to lift the Power Unit using the 2.54-cm(1-in) bore cylinder. You then calculated the force exerted on the annular area of thepiston using the measured pressure and the formula F = P x A. This force wascorresponding to the weight (mass) of the Power Unit.

REVIEW QUESTIONS

1. What is the formula for calculating the force in a hydraulic system? How can yourewrite this formula to calculate pressure?

2. What is the formula for calculating the surface area of a piston?

3. How much pressure must be applied to the cap end of a 2.54-cm (1-in) borecylinder in order to compress a spring 5.08 cm (2 in), if the spring rate is728 N/cm (416 lb/in)?

Pressure and Force

2-34

4. In the circuit of Figure 2-20 (a), what will be the pressure at gauge B if thepressure at gauge A pressure is raised to 3500 kPa (500 psi)?

5. In the circuit of Figure 2-20 (b), what will be the pressure at gauge B if thepressure at gauge A pressure is raised to 3500 kPa (500 psi)?

2-35

Exercise 2-3

Flow Rate and Velocity

EXERCISE OBJECTIVE

C To describe the operation of a flow control valve;C To establish the relationship between flow rate and velocity;C To operate meter-in, meter-out, and bypass flow control circuits.

DISCUSSION

Flow rate is the volume of fluid passing a point in a given period of time. Flow rateis often measured in liters per minute (l/min) in metric units. It is usually measuredin US gallons per minute [gal(US)/min] in English units. 1 l/min equals0.264 gal(US)/min.

Figure 2-22 shows an example. If 100 liters (26.4 US gallons) of water flow past thebridge within one minute, then the river has a flow rate of 100 l/min[26.4 gal(US)/min].

Figure 2-22. River flowing under a bridge.

Velocity is the average speed of a particle of fluid past a given point. In hydraulics,velocity is often measured in centimeters per min (cm/min) in metric units, or ininches per min (in/min) in English units.


2-36

In a hydraulic line, the rate of oil flow is equal to the oil velocity multiplied by the linecross-sectional area. In equation form:

Metric units:

Note: A liter is 1000 cm3. Therefore, divide the number of cubic centimetersper minute (cm3/min) by 1000 to obtain flow rates in l/min.

English units:

Note: A US gallon is 231 in3. Therefore, divide the number of cubic inches perminute (in3/min) by 231 to obtain flow rates in gal(US)/min.

The above formulas tell us that a constant flow rate will result in a higher velocitywhen the cross-sectional area decreases or a lower velocity when the cross-sectional area increases. In fact, the velocity of oil is inversely proportional to thecross-sectional area. Figure 2-23 shows an example, in which a constant flow rateis pumped through two pipes of different diameters. The cross-sectional area ofpipe B is twice as large as the cross-sectional area of pipe A. Oil velocity in pipe B,then, is only half as fast as oil velocity in pipe A.

Figure 2-23. Relationship between oil velocity and cross-sectional area.

Flow rate and rod speed

The speed at which a cylinder rod moves is determined by how fast the pump canfill the volume behind the cylinder piston. The more flow the cylinder receives, themore quickly the volume behind the piston will fill with oil and the faster the rod willextend or retract.


2-37

The speed of a cylinder rod (V) is calculated by dividing the oil flow rate (Q) by thepiston area (A) being acted upon. In equation form:

The extension speed of a cylinder rod, then, is equal to the oil flow rate divided bythe full piston area, as Figure 2-24 shows. The flow rate and piston area aremultiplied by multiplication constants for correct numerical results. Figure 2-24 alsoshows the formula for calculating the extension time of the cylinder rod, which is avariation of the formula used to calculate the extension speed.

Figure 2-24. Rod speed during extension.

The retraction speed of a cylinder rod is equal to the oil flow rate divided by theannular piston area, as Figure 2-25 shows. Since there is less volume to fill duringretraction, the rod will retract faster than it extends for any given flow rate.


2-38

Figure 2-25. Rod speed during retraction.

Flow measurement

The rate of oil flow is measured with an instrument called flowmeter. Figure 2-26shows the Flowmeter provided with your Hydraulics Trainer. Inside the Flowmeteris a red mark on a white indicating ring. The ring slides over a graduated cylinder,indicating the amount of flow. The Flowmeter must be connected for the direction offlow to be measured, the input port being at the bottom of the scale.

The trainer Flowmeter is graduated in liters per minute (lpm) only. As we said at thebeginning of the exercise, liters per minute is a metric unit of measurement for flowrates. When working with English units, the measured flow rate in liters per minutemust be multiplied by 0.264 for determining the equivalent flow rate in US gallons perminute [gal(US)/min].

Note: The trainer Flowmeter provides a “lpm” reading. Lpm means exactly thesame as l/min, that is, “liters per minute”. Since, however, l/min is the metric unitcommonly used for measuring flow rates, flow values in liters per minute will beexpressed in l/min throughout the manual.


2-39

Figure 2-26. Trainer Flowmeter.

Flowmeters are designed to accurately read the rate of flow at a specific oiltemperature. At lower temperatures, the oil is thick, which places extra pressure onthe internal parts of the flowmeter and causes the flowmeter reading to be slightlyhigher than the actual flow rate. As the oil warms and becomes thinner, theflowmeter reading gets closer to the actual flow value.

Flow control valves

A flow control valve is an adjustable resistance to flow that operates very much likea faucet. By adjusting the resistance, or opening, of this valve, you can modify therate of oil flow to a cylinder and, therefore, the speed of its piston rod.

Since the flow control valve increases the circuit resistance, the pump must apply ahigher pressure to overcome this resistance. This may open the relief valve partially,causing some part of the pumped oil to return to the reservoir through the reliefvalve, and less oil to go to the flow control valve and cylinder.

Figure 2-27 shows an example. The pump in this figure has a constant flow rate of3.0 l/min [0.8 gal(US)/min]. The sum of the flow rates in the two parallel paths of flow,then, will always be equal to 3.0 l/min [0.8 gal(US)/min]. Decreasing the opening ofthe flow control valve will cause more oil to go to the relief valve and less oil to go tothe cylinder. Conversely, increasing the opening of the flow control valve will causeless oil to go to the relief valve and more oil to go to the cylinder.


2-40

Figure 2-27. Parallel flow paths.

A basic rule of hydraulics states that whenever oil flows through a component, thereis a pressure difference, or pressure drop ()P) across the component, due tofrictional resistance, or opposition to the oil flow, of the component. The pressuredrop increases as the component resistance increases. In Figure 2-27, for example,the smaller the flow control valve opening, the higher the resistance of the valve, andthe greater the pressure drop across the valve. When the valve is open completely,the valve resistance to oil flow is minimum, so the pressure drop across the valve isalso minimum.

Figure 2-28 shows the Flow Control Valve supplied with your kit of hydrauliccomponents. It consists of a needle valve and a check valve integrated in onepackage.

The needle valve is an adjustable orifice restricting the oil flow from the input to theoutput port. The check valve allows the oil to flow freely from the output to the inputport, however it keeps the oil from flowing in the other direction. Turning the FlowControl Valve knob counterclockwise increases the needle valve orifice and allowsmore oil to pass through the valve, which increases the cylinder speed.

The trainer Flow Control Valve is of non-compensated type. This means that thevalve does not compensate for pressure changes in the system, resulting in adifferent flow rate through the needle valve for the same needle setting.

Some flow control valves compensate for pressure changes in the system byadjusting the pressure drop across the needle valve, which maintains a constant flowrate through the needle valve for the same needle setting. These valves are ofpressure-compensated type.


2-41

Figure 2-28. Non-compensated Flow Control Valve.

Flow control circuits

There are three ways to meter the oil flow in order to control the speed of a cylinder,which are: meter-in, meter-out, and bypass.

With the meter-in method, the flow control valve is connected in series between thepump and the cylinder, as Figure 2-29 (a) shows. It restricts the working oil flow tothe cylinder. The extra flow delivered by the pump is drained back to the reservoirthrough the relief valve. This method is useful to control cylinders having a load thatresists to the pump delivery, as cylinders raising a load.


2-42

Figure 2-29. Basic flow control circuits.

With the meter-out method, the flow control valve is connected in series betweenthe cylinder and the reservoir, as Figure 2-29 (b) shows. It restricts the flow awayfrom the cylinder. The extra flow delivered by the pump is drained back to thereservoir through the relief valve. This method is useful to slow down cylindershaving a load that tends to run away, as cylinders lowering a load.

With the bypass method, the flow control valve is connected between the pump andthe reservoir, as Figure 2-29 (c) shows. The extra flow is diverted directly to thereservoir through the flow control valve. This method is more energy efficient thanthe meter-in and meter-out methods because the extra flow returns to the reservoirat the load pressure rather than at the relief valve pressure. However, this methodis less accurate because it does not provide direct control of the working flow to thecylinder.

Conversion factors

Table 2-4 shows the conversion factors used to convert measurements of flowrate, velocity, and area from S.I. (or metric) units to English units, and vice versa.


2-43

FLOW RATE

Liters per minute(l/min)

x 0.264 =US gallons perminute [gal(US)/min]

x 3.79 =Liters per minute(l/min)

VELOCITY

Centimeters perminute (cm/min)

x 0.394 =Inches per minute(in/min)

x 2.54 =Centimeters perminute (cm/min)

AREA

Squarecentimeters (cm2)

x 0.155 = Square inches (in2) x 6.45 =Squarecentimeters (cm2)


REFERENCE MATERIAL

For additional information on flow control valves and flow control circuits, refer to thechapter entitled Flow Control Valves in the Parker-Hannifin’s manual IndustrialHydraulic Technology.

Procedure summary

In this exercise, you will test the operation of meter-in, meter-out, and bypass flowcontrol circuits while noting the actuation times and pressure drops across the trainerFlow Control Valve. Then, you will test the effect of a meter-out circuit on an over-running load.

EQUIPMENT REQUIRED


PROCEDURE

Meter-in flow control circuit

G 1. What is the formula for calculating the extension time, t, of a piston rod?(Refer to Figure 2-24.)


2-44

G 2. Using this formula, calculate the extension time of the 3.81-cm (1.5-in) borecylinder rod for the flow rates in Table 2-5. The stroke length, L, is 10.16 cm(4 in). Record your calculations in Table 2-5 under “THEORETICAL”.

G 3. Connect the circuit shown in Figures 2-30 and 2-31. This circuit meters theoil flow going to the cylinder.

Notice that the Flow Control Valve must be connected so that the arrowpoints away from the pump.

Figure 2-30. Schematic diagram of a meter-in flow control circuit.


2-45

Figure 2-31. Connection diagram of a meter-in flow control circuit.



position.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve completely (turn knob fully counterclockwise).

G 5. Close the Flow Control Valve completely by turning its adjustment knob fullyclockwise.


G 7. Move the lever of the directional valve toward the valve body and observethe pressure reading at gauge A. Since the Flow Control Valve is fully


2-46

closed, the pumped oil is blocked at the Flow Control Valve and is nowbeing forced through the relief valve, so gauge A indicates the minimumpressure setting of the relief valve. While keeping the directional valve levershifted, turn the relief valve adjustment knob clockwise until gauge A reads2100 kPa (300 psi).

G 8. With the lever of the directional valve still shifted toward the valve body,open the Flow Control Valve, 1 turn counterclockwise, to extend the rod.You should observe that the extension speed increases as you increase theopening of the Flow Control Valve.

G 9. Move the lever of the directional valve outward from the valve body toretract the rod. Did the rod retract faster than it extended?

G Yes G No

G 10. Move the lever of the directional valve toward the valve body to extend thecylinder rod. As the rod extends, close the Flow Control Valve completelyby turning its adjustment knob fully clockwise. Does the Flow Control Valveprovide direct control of the rod speed?

G Yes G No

G 11. Retract the rod by moving the lever of the directional valve outward from thevalve body. Observe that the rod still retracts at full speed even though theFlow Control Valve is completely closed. Explain why.

G 12. Open the Flow Control Valve 1 turn counterclockwise.

G 13. Now adjust the Flow Control Valve to 1.5 l/min [0.40 gal(US)/min]*. To doso, move the lever of the directional valve toward the valve body to extendthe cylinder rod. As the cylinder extends, observe the Flowmeter reading.Adjust the Flow Control Valve so that the Flowmeter reads 1.5 l/min[0.40 gal(US)/min], then retract the rod. Accurate adjustment may requirethat the cylinder be extended and retracted several times.

Note: The trainer Flowmeter provides a “lpm” reading. Lpmmeans exactly the same as l/min, that is, “liters per minute”.Since, however, l/min is the metric unit commonly used formeasuring flow rates, flow values in liters per minute will beexpressed in l/min throughout this manual.


2-47

G 14. Measure the time required for the rod to extend fully using a stopwatch orthe second hand on a watch. Record this value in Table 2-5 under“ACTUAL”. Also record the readings of gauges A and B while the rod isextending. When you have finished, retract the rod.

FLOW RATE TOCYLINDER

THEORETICALEXTENSION

TIME

ACTUALEXTENSION

TIMEGAUGE A GAUGE B

)P(GAUGE A !GAUGE B)

1.5 l/min0.40 gal(US)/min



Table 2-5. Meter-in flow control circuit data.

G 15. Repeat steps 13 and 14 for the other flow rates in Table 2-5.


G 17. Compare the actual and theoretical extension times registered in Table 2-5.Are these values within 10% of each other?

G Yes G No

G 18. Does the rod speed increase or decrease as the flow rate decreases?

G 19. Calculate the pressure drop (GAUGE A ! GAUGE B) across the FlowControl Valve for each flow rate in Table 2-5. Record your results inTable 2-5 under “)P”.

G 20. According to Table 2-5, does the pressure drop across the valve increaseor decrease as the opening of the valve is increased? Why?


2-48

Meter-out flow control circuit

G 21. Connect the circuit shown in Figure 2-32. This circuit meters the oil flowgoing out of the cylinder.

Notice that the Flow Control Valve must be connected so that the arrowpoints toward the pump.

Figure 2-32. Meter-out flow control circuit.



G 24. Move the lever of the directional valve outward from the valve body toretract the cylinder rod completely. With the cylinder rod fully retracted, allthe oil from the pump now flows through the Relief Valve and gauge Bindicates the minimum pressure setting of the Relief Valve. While keepingthe directional valve lever shifted, turn the Relief Valve adjustment knobclockwise until gauge B reads 2100 kPa (300 psi).

G 25. Move the lever of the directional valve toward the valve body to extend thecylinder rod. As the rod extends, open the Flow Control Valve 1 turncounterclockwise. You should observe that the extension speed of the rodincreases as you increase the Flow Control Valve opening. Retract the rod.


2-49

G 26. Move the lever of the directional valve toward the valve body to extend thecylinder rod. As the rod extends, close the Flow Control Valve completely(turn knob fully clockwise). Does the Flow Control Valve provide directcontrol of the rod speed?

G Yes G No

G 27. Retract the rod. Does the setting of the Flow Control Valve have an effecton the retraction speed?

G Yes G No

G 28. Adjust the Flow Control Valve so that the Flowmeter reads 1.5 l/min[0.40 gal(US)/min] as the rod extends, then retract the rod.

G 29. Measure the extension time of the cylinder rod. Record this value inTable 2-6 under “EXTENSION”. Also record the readings of gauges A andB while the rod is extending.

FLOW RATE FROMCYLINDER

EXTENSIONTIME

GAUGE A GAUGE B)P (GAUGE A !

GAUGE B)




Table 2-6. Meter-out flow control circuit data.



G 32. Calculate the pressure drop (GAUGE A ! GAUGE B) across the FlowControl Valve for each flow rate in Table 2-6. Record your results inTable 2-6 under “)P”.


2-50

G 33. According to Table 2-6, does the pressure drop across the valve increaseor decrease as the opening of the valve is increased? Why?

G 34. Are the pressure drops in Table 2-6 for a meter-out circuit similar to thepressure drops in Table 2-5 for a meter-in circuit? Why?

Bypass flow control circuit

G 35. Connect the circuit shown in Figure 2-33. This circuit diverts the extra flowdirectly to the reservoir through the Flow Control Valve. Notice that the FlowControl Valve must be connected so that the arrow points toward the pump.



G 38. Move the lever of the directional valve toward the valve body to extend thepiston rod completely. With the Flow Control Valve closed and the cylinderrod fully extended, all the oil from the pump now flows through the ReliefValve and gauge A indicates the pressure setting of the Relief Valve. Whilekeeping the directional valve lever shifted, adjust the Relief Valve so thatgauge A reads 2100 kPa (300 psi). Then, retract the rod completely.


2-51

Figure 2-33. Bypass flow control circuit.

G 39. Move the lever of the directional valve toward the valve body to extend therod fully. Since there is no bypass flow path between the pump and thereservoir (i.e., since the flow control valve is fully closed), all the oil flowgoes to the cylinder and the rod extension time is minimum. Retract the rod,then extend it and measure the rod extension time. Also, note the flowmeterreading while the rod extends. Record your results in the first row ofTable 2-7.

BYPASS FLOW PATH ROD EXTENSION TIME FLOW RATE

None (flow control valvefully closed)

Minimum:

Open (flow control valvefully open)

Maximum:

Partially open (flowcontrol valve partiallyopen)

Intermediate:

Table 2-7. Bypass flow control circuit data.

G 40. Retract the rod, then open the Flow Control Valve completely by turning itsadjustment knob fully counterclockwise. Extend the rod fully. Since thebypass flow path is open (i.e., since the Flow Control Valve is open), the oilflow is diverted directly to the reservoir and the rod extension time is


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maximum. Retract the rod, then extend it and measure the rod extensiontime. Also, note the flowmeter reading while the rod extends. Record yourresults in the second row of Table 2-7.

G 41. Retract the rod. Calculate the difference ()t) between the maximum andminimum rod extension times recorded in Table 2-7. Then, determine theintermediate rod extension time and record your result in the last row ofTable 2-7.

)t = maximum extension time (with bypass) - minimumextension time (no bypass) = s

Intermediate extension time = minimum extension time (without bypass)

+ )t/2 = s

G 42. Extend and retract the rod several times and adjust the flow control valve sothat the rod extension time is equal to the intermediate value obtained instep 41. Note and record the Flowmeter reading when the rod extends atthe intermediate speed in the last row of Table 2-7.

G 43. Continue to experiment with bypass flow control by varying the opening ofthe Flow Control Valve and observing the effect that this has on the rodextension time. Record your observations.

Controlling the speed of an over-running load


G 45. Remove the 2.54-cm (1-in) bore cylinder from its adapter. Make sure thecylinder tip (bullet) is removed from the cylinder rod end.

G 46. Insert the cylinder rod into the cylinder hole in the Power Unit lifting frame,then fasten the cylinder to the lifting frame by tightening its retaining ringsecurely. Position the lifting frame over the Power Unit, with its open side atthe rear of the Power Unit.

G 47. Connect the two cylinder ports together using a hose full of oil, then pull thepiston rod out until it touches the lifting attachment on the Power Unit.Fasten the cylinder to the Power Unit by screwing the lifting attachment ontothe threaded end of the cylinder rod. Then, disconnect the hose from thecylinder.


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G 48. Connect the circuit used to lift the Power Unit shown in Figure 2-34. TheFlow Control Valve will restrict the flow going out of the cylinder.

CAUTION!

Make sure the hoses and the Power Unit line cord will notbecome wedged between rigid parts of the trainer when thePower Unit is lifted.

Figure 2-34. Meter-out circuit with over-running load.

G 49. Open the Flow Control Valve completely by turning its adjustment knob fullycounterclockwise.

G 50. Plug the Power Unit line cord into a wall outlet, then turn on the Power Unit.


G 52. Move the lever of the directional valve outward from the valve body to lift thePower Unit. Then, move the lever of the directional valve toward the valvebody. Does the Power Unit returns to ground uncontrolled?

G Yes G No


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G 53. Move the lever of the directional valve outward from the valve body to lift thePower Unit, then release the lever. Close the Flow Control Valve completelyby turning its adjustment knob fully clockwise.

G 54. Move the lever of the directional valve toward the valve body. Does thePower Unit return to ground? Why?

G 55. Move the lever of the directional valve toward the valve body and slowlyopen the Flow Control Valve c turn. Does the Power Unit return to groundsuddenly or smoothly? Why?

G 56. Once the Power Unit has returned to ground, move the lever of theDirectional Control Valve outward from the valve body to lift the Power Unit.Is the lifting speed controlled by the Flow Control Valve? Explain why.

G 57. Move the lever of the directional valve toward the valve body until the PowerUnit has returned to ground.



G 60. Remove all components from the work surface and wipe any hydraulic oilresidue. Return all components to their storage location.



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CONCLUSION

By now, you are familiar with flow control valves. The circuits you studied in thisexercise show the three basic flow control techniques. As you tested the flow controlcircuits, you saw how well hydraulic devices can be controlled under differentconditions.

In the first part of this exercise, you used meter-in control to change the cylinderextension speed. At the same time, you observed that the setting of the Flow ControlValve had little effect on the speed of retraction. This was due to the check valveinside the valve. The meter-in control is straight-forward. This control works bestagainst a load which does not change direction. Hydraulic presses and positioningequipment are good examples of such a load.

In the second part of the exercise, you used meter-out control to change the cylinderextension speed. The meter-out circuit is useful in controlling loads that mightsuddenly start pulling on the actuator and tend to run away. Earth moving equipmentis designed to lift and dump loads. At the point where the load begins to drop quickly,a meter-out control circuit keeps flow from moving out of the rod end of the cylinderand generates a back pressure to keep the bucket from falling uncontrolled.

In the third part of the exercise, you tested a bypass flow control circuit. It is fairlydifferent from the meter-in and meter-out circuits. The extra flow is diverted directlyto the reservoir through the Flow Control Valve. This method is more energy efficientthan the meter-in and meter-out controls because the extra flow returns to thereservoir at the load pressure rather than at the relief valve pressure. However, thismethod is less accurate because it does not provide direct control of the working flowto the cylinder.

Finally, you tested the effect of a meter-out circuit on an over-running load. Withouta meter-out circuit, the Power Unit falls uncontrolled to the ground. With a meter-outcircuit, the Power Unit smoothly returns to the ground. The speed at which it returnsto the ground can be controlled by modifying the Flow Control Valve opening.

REVIEW QUESTIONS

1. What happens to the speed of a piston rod as the diameter of the pistonincreases but the flow rate remains the same? Explain.

2. Find two ways to decrease the speed at which a cylinder rod extends or retracts.


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3. What flow rate is required to make a 10.16-cm (4-in) bore x 3.81-cm (1.5-in) rodx 30.48-cm (12-in) stroke cylinder extend in 6 seconds?

4. Describe the route of hydraulic oil moved by the pump and not metered throughthe flow control valve in either a meter-in or meter-out circuit.

5. What type of metering circuit is used to control cylinders having a load thatresists to the pump delivery, as cylinders raising a load?

6. What type of metering circuit is used to slow down cylinders having a load thattends to run away, as cylinders lowering a load?

7. Name one advantage and one disadvantage of a bypass flow control circuit overthe meter-in and meter-out circuits.

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Exercise 2-4

Work and Power

EXERCISE OBJECTIVE

C To define the terms “work” and “power”;C To establish the relationship between force, work, and power;C To calculate the work, power, and efficiency of a hydraulic system.

DISCUSSION

Work

Work is the motion of a load through a distance that results in something usefulbeing done. Work is expressed in units of force multiplied by distance. In hydraulicsystems, work is measured in Joules (J) or Newton-meters (N@m) in S.I. units, andin feet-pounds (ft@lb) in English units.

When the exerted force is constant throughout the motion of the load, the amountof performed work is equal to the force exerted multiplied by the distance moved. Inequation form:

S.I. units:

English units:

In Figure 2-35, for example, if the cylinder exerted a force of 100 N (22.5 lb) over avertical distance of 1 m (3.28 ft), then 100 J (73.8 ftAlb) of work was accomplished.

Work and Power

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Figure 2-35. Performed work: 100 J (73.8 ft@lb).

Power

Power is the rate at which work is done. It is measured in watts (W) in S.I. units, orin horsepower (hp) in English units. Power is equal to the amount of work performedin a given period of time. In equation form:

S.I. units:

English units:

Note: A horsepower (hp) is 550 ft@lb per second. Therefore, divide the numberof feet-pounds (ft@lb) by 550 to obtain power in hp.

Work and Power

2-59

In Figure 2-35, for example, if the 100 J (73.8 ft@lb) work were done in 2 seconds, therate of doing work would be

S.I. units:

English units:

Conversion of power in a hydraulic system

A hydraulic system operates through a two-step power conversion process, asFigure 2-36 shows.

Figure 2-36. Conversion of power in a hydraulic system.

The first step is for the pump to convert the mechanical power given to it by themotor into fluid power. The second step is for the cylinder to convert the fluid powerback into mechanical power to move the load.

Pump output power

The amount of fluid power generated by the pump is equal to the circuit flow ratemultiplied by the circuit pressure. In equation form:

S.I. units:

English units:

Work and Power

2-60

Thus, an increase in either circuit pressure or flow rate will increase the fluid powergenerated by the pump. Conversely, if system pressure or flow rate decreases, thefluid power generated by the pump will decrease.

Dissipated power

As Figure 2-37 shows, not all the fluid power from the pump is converted intomechanical power at the cylinder. Some power is lost as heat by frictional resistanceto oil flow in the hoses, valves, and cylinder seals.

Figure 2-37. Power distribution in a hydraulic system.

The amount of power dissipated as heat by any component is equal to the rate of oilflow through the component multiplied by the pressure drop across it. In equationform:

S.I. units:

English units:

Work and Power

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Efficiency

The electric motor which drives the hydraulic pump in your Power Unit consumeselectrical power. Electrical power is measured in watts. If the electric motor, thepump, the transmission hoses, and all components were 100% efficient, the electricmotor would consume the same amount of power as the cylinder in Figure 2-36.However, because some energy is always lost as heat from friction, the electricmotor will consume more power than the cylinder.

Often, the overall efficiency of a hydraulic system must be calculated to know howmuch power is actually being used. The formula for overall system efficiency as apercentage is:

Notice that the output and input power values for this equation must be stated in thesame kind of units (watt, horsepower, etc.)

Sometimes, however, we are more interested in knowing the efficiency of thehydraulic circuit itself, that is, the amount of pump output power actually being usedby the cylinder. The formula for hydraulic circuit efficiency as a percentage is:

Conversion factors

Table 2-8 shows the conversion factors used to convert measurements of workand power from S.I. units to English units, and vice versa.

Work

Joules (J) x 0.738 = Foot-pounds (ft@lb) x 1.355 = = Joules (J)

Power

Watts (W) x 0.0013 = Horsepower (hp) x 745.7 = = Watts (W)


REFERENCE MATERIAL

For additional information on work and power, refer to the chapter entitled ThePhysical World of a Machine in the Parker-Hannifin’s manual Industrial HydraulicTechnology.

Work and Power

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Procedure summary

In the first part of the exercise, you will lift the Power Unit using the small cylinder.You will measure the retraction time and the pressure required at the cylinder piston,then you will calculate the cylinder work and power.

In the second part of the exercise, you will restrict the oil flow from the pump with theFlow Control Valve. You will measure the pump output power and the amount ofpower dissipated by the valve at several different pressures.

In the third part of the exercise, you will calculate the efficiency of the circuit used tolift the Power Unit, using the data collected in the first and second parts of theexercise.

EQUIPMENT REQUIRED


PROCEDURE

Cylinder work and power


G 2. Remove the 2.54-cm (1-in) bore cylinder from its adapter by unscrewing itsretaining ring. Make sure the cylinder tip (bullet) is removed from thecylinder rod end.

G 3. Insert the 2.54-cm (1-in) bore cylinder rod into the cylinder hole in the PowerUnit lifting frame. Fasten the cylinder to the lifting frame by tightening itsretaining ring securely. Position the lifting frame over the Power Unit, withits open side at the rear of the Power Unit.

G 4. Connect the two cylinder ports together using a hose. Pull out the piston roduntil it touches the lifting attachment on the Power Unit. Fasten the cylinderto the Power Unit by screwing the lifting attachment onto the threaded endof the cylinder rod. Then, disconnect the hose from the cylinder.

G 5. Connect the circuit shown in Figures 2-38 and 2-39.

Work and Power

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Figure 2-38. Schematic diagram of the circuit for measuring cylinder work and power.

Work and Power

2-64

Figure 2-39. Connection diagram of the circuit for measuring cylinder work and power.

CAUTION!

Make sure the Power Unit line cord will not become wedgedbetween the Power Unit and the lifting frame when the PowerUnit is lifted.

G 6. Before turning on the Power Unit, perform the following start-up procedure:


position.e. Plug the Power Unit line cord into an appropriate outlet.f. Open the Relief Valve completely (turn knob fully counterclockwise).

G 7. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it ¾ turn.

Work and Power

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G 9. Turn the Relief Valve adjustment knob clockwise until the pressure reaches4200 kPa (600 psi) at gauge A.

G 10. Move the lever of the directional valve outward from the valve body to raisethe Power Unit. Measure the time the cylinder takes to retract fully asaccurately as possible. Record the retraction time in Table 2-9.

RETRACTIONTIME

ANNULARPRESSURE

DEVELOPEDFORCE

CYLINDERWORK

CYLINDERPOWER

Table 2-9. Cylinder work and power.


G 12. Perform the following steps to accurately measure the pressure required atthe cylinder to lift the Power Unit:

– Turn the Relief Valve adjustment knob fully counterclockwise to set thepressure to minimum.

– Move the directional valve lever outward from the valve body and slowlyincrease the Relief Valve pressure setting (turn knob clockwise) until thePower Unit begins to rise. Then, release the valve lever.

– Record in Table 2-9 the annular pressure required to raise the powerunit, as indicated by gauge B.

Note: The pressure reading at gauge B may slowly drop after thedirectional valve lever is released, due to internal leakage in thedirectional valve. Take your reading immediately after the valvelever is released.

G 13. Move the lever of the directional valve toward the valve body to return thePower Unit to the ground. Turn off the Power Unit. Open the Relief Valvecompletely (turn knob fully counterclockwise).

G 14. Using the formula F = P x A, calculate the force developed by the cylinderto lift the Power Unit, based on the annular pressure registered in Table 2-9.Record your calculated value in Table 2-9.

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G 15. Calculate the amount of work accomplished by the cylinder to lift the PowerUnit, based on the force value registered in Table 2-9, and on a moveddistance of 0.102 m (0.333 foot). Record your calculated value in Table 2-9.

Note: The moved distance corresponds to the cylinder strokelength. This length is 10.16 cm (4 in), or 0.102 m (0.333 foot).

G 16. Calculate the amount of power developed at the cylinder when the PowerUnit is lifted, based on the retraction time and work values registered inTable 2-9. Record your calculated value in Table 2-9.

G 17. Disconnect all hoses and wipe off any hydraulic oil residue. Unscrew thecylinder from the Power Unit lifting attachment. Unscrew the ring retainingthe cylinder to the lifting frame. Remove the cylinder from the lifting frame.Reinstall the cylinder on its adapter by fastening its retaining ring securely.

Pump output power and power dissipation


Figure 2-40. Circuit for measuring the pump output power and dissipated power.


Work and Power

2-67


G 21. With the Flow Control Valve fully closed, all the oil from the pump is flowingthrough the Relief Valve and gauge A indicates the minimum pressuresetting of the relief valve. Turn the relief valve adjustment knob clockwiseuntil gauge A reads 4200 kPa (600 psi).

G 22. Open the Flow Control Valve completely by turning its adjustment knob fullycounterclockwise. The full pump flow should now be going through the FlowControl Valve. Record below the reading of the Flowmeter.

Full pump flow = l/min or gal(US)/min

Note: The trainer Flowmeter is graduated in liters per minuteonly. If you are working with English units, multiply the measuredflow rate in liters per minute by 0.264 for determining theequivalent flow rate in gallons (US) per minute.

G 23. Now reduce the Flow Control Valve opening until the circuit pressure atgauge A is 1400 kPa (200 psi). Since the system pressure is below theRelief Valve pressure setting, the flow rate indicated by the Flowmeter is themaximum flow available from the pump at this pressure. Record theFlowmeter reading in Table 2-10 under “FLOW”. Also, record the pressurereading at gauge B.

CIRCUITPRESSURE(GAUGE A)

FLOW RATEPRESSURE

AT GAUGE BPUMP OUTPUT

POWER

POWERDISSIPATEDBY VALVE

1400 kPa(200 psi)

2100 kPa(300 psi)

2800 kPa(400 psi)

3500 kPa(500 psi)

Table 2-10. Power dissipation versus pressure drop.

G 24. Repeat step 23 for the other pressures listed in Table 2-10.


G 26. Using the values recorded in Table 2-10, plot the pump flow rate versuspressure data on the graph of Figure 2-41.

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Figure 2-41. Flow rate versus circuit pressure.

G 27. According to Figure 2-41, does the flow rate decrease as the systempressure increases? Explain.

G 28. Calculate the pump output power for each circuit pressure listed inTable 2-10, based on the registered flow values. Record your calculationsin Table 2-10 under “PUMP OUTPUT POWER”.

Work and Power

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G 29. Does the pump output power increase as the Flow Control Valve openingis decreased? If so, explain why.

G 30. Calculate the amount of power dissipated by the Flow Control Valve foreach circuit pressure listed in Table 2-10, based on the registered flow rateand on the pressure drop (gauge A - gauge B) across the valve. Recordyour calculations in 2-10 under “POWER DISSIPATED BY VALVE”.

G 31. Does the dissipated power increase as the Flow Control Valve opening isdecreased? If so, explain why.

Efficiency

G 32. Record below the annular pressure required to lift the Power Unit, asregistered in Table 2-9 of this exercise.

G 33. Based on the curve plotted in Figure 2-41, what is the pump flow rate at thepressure level recorded in step 32?

G 34. Calculate the pump output power at the pressure and flow rate recorded insteps 32 and 33. Record your calculated value in Table 2-11 under “PUMPOUTPUT POWER”.

PUMP OUTPUT POWERCYLINDER OUTPUT

POWEREFFICIENCY

Table 2-11. Circuit efficiency.

G 35. Record in Table 2-11 the amount of power developed at the cylinder whenthe Power Unit is lifted, as registered in Table 2-9.

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G 36. Calculate the efficiency of the circuit used to lift the Power Unit, based onthe data registered in Table 2-11. Record your calculated value inTable 2-11.

G 37. Was the circuit used to lift the Power Unit 100% efficient? Why?




CONCLUSION

In the first part of the exercise, you used formulas to calculate work and power fromthe results of your tests. You were able to find the amount of work performed by thecylinder to lift the Power Unit a given distance based on the amount of force exertedand on the cylinder stroke length. You also calculated the cylinder output power byusing rod retraction time measurements since power is work divided by time.

In the second part of this exercise, you observed that fluid power is converted intoheat when fluid flows through a restricted orifice such as a flow control valve. Thegreater the pressure drop across the flow control valve, the greater the dissipatedpower. You also learned that the flow rate of a pump decreases as the circuitpressure increases, because of the increased internal leakage of the pump. Therelationship between pump flow rate and circuit pressure will be studied in detail ina later exercise.

In the third part of this exercise, you learned about efficiency in hydraulic systems.You were able to calculate the efficiency of the circuit used to lift the Power Unitbased on the pump and cylinder power measurements previously performed.

Work and Power

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REVIEW QUESTIONS

1. What happens to the power dissipated by a directional control valve when theflow rate across the valve doubles?

2. A 1300-N load is moved 1 m by a 3-cm bore cylinder. The cylinder is thenreplaced with a 6-cm bore cylinder and the load is again moved 1 m. Whichcylinder accomplished the greatest amount of work?

3. A 5-in bore, 3-feet stroke cylinder must lift a 5000-lb load through its stroke in4 seconds. Calculate the amount of power required at the cylinder if thehydraulic circuit is 100% efficient.

4. In a hydraulic system where the combined efficiency of the electric motor, thepump, and the hoses is 75%, how many watts would the electric motor draw tosatisfy the power requirement from review question 3?

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3-1

Unit 3

Basic Circuits

UNIT OBJECTIVE

When you have completed this unit, you will be able to operate and test simple,practical hydraulic circuits. You will also be able to describe the operation of adirectional control valve.


Hydraulic systems perform a variety of tasks, ranging from the very simple to thevery complex. Controlling cylinders is one of the most important aspects ofhydraulics. For example, two cylinders may be required to operate at the samespeed, or a cylinder may need to extend rapidly under no-load conditions.

Exercise 3-1 shows how a directional control valve alter the flow paths, and itdiscusses cylinder speed and force control. Although previous exercises have madeuse of directional control valves, this will be your first chance to learn about theirdesign.

Exercises 3-2 and 3-3 discuss cylinder synchronization. The synchronized cylinderson the front end loader shown in Figure 3-1 extend and retract at the same speed,preventing the bucket from twisting and jamming.

Figure 3-1. Front end loader.

Regenerative circuits, discussed in Exercise 3-4, provide a means for rapidextension of a cylinder. By using the cylinder output flow to supplement the pumpflow to the cylinder input port, the cylinder will extend rapidly until it meets a load. Forexample, the piston of a hydraulic wood splitter extends quickly until it comes incontact with the piece of wood.

3-2

3-3

Exercise 3-1

Cylinder Control

EXERCISE OBJECTIVE

C To learn how to control the direction, force, and speed of a cylinder;C To introduce the operation of a directional control valve;C To describe the effect a change in system pressure or flow rate has on the speed

of a cylinder;C To describe the effect a change in system pressure or flow rate has on the force

exerted by a cylinder.

DISCUSSION

This exercise introduces the three types of cylinder control functions: directioncontrol, force control, and pressure control:

C The direction of motion of a cylinder is controlled by selecting the direction ofthe flow of oil through the cylinder, using a directional control valve.

C The force output of a cylinder is controlled by modifying the amount of pressureavailable at its piston, using a pressure control valve.

C The extension or retraction speed of a cylinder is controlled by modifying theflow rate to the cylinder, using a flow control valve.

Controlling the direction of motion of a cylinder

The direction of motion of a cylinder is controlled by using a directional valve. Adirectional valve is a valve that stops, diverts, or reverses the oil flow. Directionalvalves are found as two-way, three-way, and four-way types, as Figure 3-2 shows.It is important to note that the terminology “two-way, three-way, etc.” does not trulydescribe the number of ways, or flow paths, provided by the valve. It rather refers tothe number of port connections of the valve. A four-way valve, for example, has fourport connections.

Cylinder Control

3-4

Figure 3-2. Types of directional valves.

Cylinder Control

3-5

C Two-way directional valves allow or block flow through a line. They serve as“on” or “off” device to isolate various system parts, as Figure 3-2 (a) shows.

C Three-way directional valves provide true directional control. They consist of apressure port, a tank port, and a cylinder port. They are used to power cylindersthat operate in one direction (single-acting cylinders) in either extension orretraction stroke, as Figure 3-2 (b) shows. The rod of such cylinders are returnedby non-hydraulic forces.

C Four-way directional valves consist of a pressure port, a tank port, and twocylinder ports. They are used to alternately extend and retract cylinders thatoperate in two directions (double-acting cylinders), as Figure 3-2 (c) shows. Thisis the type of directional valve you have used to extend and retract cylinders inmost exercises since Exercise 1-1.

Operation of the trainer directional valve

Figure 3-3 shows the directional valve supplied with your Hydraulics Trainer. Thecomplete description of this valve is a four-way, three-position, spring-centered,closed-center, lever-operated directional valve.

Number of ways (ports)

The trainer directional valve is a four-way valve because it has four connectionports, as Figure 3-3 shows:

C The “P” (PRESSURE) port is connected to the pump and supplies the oil underpressure to the valve;

C The “T” (TANK) port is connected to the reservoir;

C The “A” and “B” ports are connected to the cylinder. They alternately serve as theoil supply and return ports as the cylinder is moved in one direction and then theother.

The designations on the external connections of a 4-way valve vary considerablybetween manufacturers. Common designations for the pump and reservoirconnections are “P” and “T”, or “Inlet” and “Outlet”. Common designations for thecylinder connections are “A” and “B”, or “Cylinder 1” and “Cylinder 2”.

Cylinder Control

3-6

Figure 3-3. Four-way, three-position, spring-centered, closed-center, lever-operated directionalvalve.

Number of positions

The trainer directional valve is a 3-position valve because it has two extremepositions and a center position, as Figure 3-3 shows. A closely fitting spool is slidback and forth to align passageways to direct the oil.

C Moving the valve lever toward the valve body causes the oil from the pump toenter the valve through port “P” and to exit the valve through port “A”. The oil thenmoves to the cylinder to extend or retract the rod, depending on whether port Ais connected to the rod or cap end of the cylinder. The oil forced out of thecylinder returns to port “B” of the directional valve and flows to the reservoirthrough port “T”. This position is referred to as the “straight-through” position.

Cylinder Control

3-7

C Moving the valve lever outward from the valve body causes the oil from thepump to enter the valve through port “P” and to exit the valve through port “B”.The oil then moves to the cylinder to extend or retract the rod. The oil forced outof the cylinder returns to port “A” and flows to the reservoir through port “T”. Thisposition is referred to as the “cross-connected” position.

C Releasing the valve lever automatically returns the valve to the center position,thanks to an internal spring. The center position is the neutral position so thecylinder can be stopped and held at any point in the stroke. Without the centerposition, the cylinder would always be moving through the stroke or be stoppedat one of the extreme positions.

Center conditions

The trainer directional valve is of closed-center type because it blocks flow betweenall ports when it is centered. If ports A and B are connected to a cylinder, the positionof this cylinder will essentially be locked when the valve is in the center position, sothe pump flow will go through the relief valve.

Figure 3-4 shows other center configurations commonly used on 3-positiondirectional valves.

Figure 3-4. Typical center configurations on 3-position directional valves.

C Tandem-center valves lock the load when they are centered, in addition toproviding an alternate flow path for the pump flow through ports “P” and “T”, whichsaves energy.

C Float-center valves connect ports A and B to the reservoir when they arecentered. Since there is no pressure flowing to ports A and B, a cylinder or motorconnected to them would be free to move or “float”. If a cylinder were on ports Aand B, it could be moved manually for set up. If a motor were on such a valve, itwould stop gently without cavitation to protect both the motor and its load.

Cylinder Control

3-8

C Open-center valves connect all the ports together when they are centered. Thesevalves provide a float-center condition, in addition to providing a flow path for thepump flow through ports “P” and “T”, which saves energy.

Valve activation

The trainer directional valve is lever-operated because it is activated by manuallyshifting a lever, as Figure 3-3 shows. Several other actuators can be used to activatedirectional valves:

C Mechanical: a lever moved by a cam or linkage from a machine member;

C Pilot: a piston moved by pressure, controlled by another directional valve;

C Solenoid: a rod moved by magnetic forces in a solenoid.

When reading hydraulic diagrams, it is important to know that all connections to adirectional valve are made from the valve block which shows the oil flow when thevalve is deactivated, or at rest. On three-position spring-centered directional valves,which is the type of valve supplied with your Hydraulics Trainer, circuit connectionsare made to the center block, which is the block showing the oil flow when the valveis at rest. On two-position spring-return directional valves, which is the type of valveused in the circuits of Figure 3-2, circuit connections are made to the block nearestthe spring, which is the block showing the oil flow when the valve solenoid isdeenergized.

Controlling the speed and force output of a cylinder

When working with hydraulic equipment, it is often useful to change the speed of acylinder without affecting its force output, or to change its force output withoutaffecting its speed.

C Flow control affects only the cylinder speed. Hence, a flow control valveprovides easy control of the speed of a cylinder by allowing a given amount of oilto flow through the cylinder.

C Pressure control affects only the cylinder force output. Hence, a relief valveprovides easy control of the maximum force output of a cylinder by allowing agiven amount of pressure to develop at the cylinder piston.

The use of the proper control valve allows you to select the combination of force andspeed that you need.

REFERENCE MATERIAL

For detailed information on directional control valves, refer to the chapter entitledDirectional Control Valves in the Parker-Hannifin’s manual Industrial HydraulicTechnology.

Cylinder Control

3-9

Procedure summary

In the first part of the exercise, you will test a 4-way directional control valve for flowdirection with the valve lever in all three positions.

In the second part of the exercise, you will determine the effect of changing systempressure on the cylinder retraction time.

In the third part of the exercise, you will determine the effect of changing system flowrate on the cylinder retraction time.

In the fourth part of the exercise, you will determine the effect of changing systempressure on the cylinder force.

In the fifth part of the exercise, you will determine the effect of changing system flowrate on the cylinder force.

EQUIPMENT REQUIRED


PROCEDURE

Testing the operation of a 4-way directional valve

G 1. Make sure the Power Unit line cord is disconnected from the wall outlet.

G 2. Remove the 3.81-cm (1.5-in) bore cylinder from its adapter by unscrewingits retaining ring. Make sure the cylinder tip (bullet) is removed from thecylinder rod end.

G 3. Insert the 3.81-cm (1.5-in) bore cylinder rod into the cylinder hole in thePower Unit lifting frame. Fasten the cylinder to the lifting frame by tighteningits retaining ring securely. Position the lifting frame over the Power Unit, withits open side at the rear of the Power Unit.

G 4. Connect the two cylinder ports together using a hose full of oil. Pull thepiston rod out until it touches the lifting attachment on the Power Unit.Fasten the cylinder to the Power Unit by screwing the lifting attachment ontothe threaded end of the cylinder rod. Then, disconnect the hose from thecylinder.

G 5. Connect the circuit used to lift the Power Unit shown in Figure 3-5. Noticethat the same circuit will be used in all five parts of the exercise.

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3-10

CAUTION!

Make sure the hoses and Power Unit line cord will notbecome wedged between rigid parts of the trainer when thePower Unit is lifted.

Figure 3-5. Circuit for determining the effect of directional control, pressure control, and flowcontrol on cylinder operation.




G 7. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it 1 turn.

G 8. Turn on the Power Unit. According to the Flowmeter reading, is the oil fromthe pump flowing through the Flow Control Valve? Why?

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3-11

G 9. Turn the Relief Valve adjustment knob clockwise until the system pressureat gauge A is 3500 kPa (500 psi).

G 10. Move the lever of the directional valve outward from the valve body whileobserving the Power Unit. Based on the direction the Power Unit moved,determine whether the directional valve P (pressure) port is connected toport A or B when the valve lever is moved outward. Also determine whetherthe T (tank) port is connected to port A or B.

Lever moved outward: Port P connected to port

Port T connected to port

G 11. Move the lever of the directional valve toward the valve body whileobserving the Power Unit. Based on the direction the Power Unit moved,determine whether the directional valve P port is connected to port A or Bwhen the valve lever is moved inward. Also determine whether the T port isconnected to port A or B.

Lever moved inward: Port P connected to port

Port T connected to port

G 12. Record which direction the Power Unit moves for each lever position inTable 3-1.

LEVER POSITION ACTION

MOVED TOWARD VALVE BODY

MOVED OUTWARD FROM VALVEBODY

CENTERED

Table 3-1. Power Unit action versus lever position.

G 13. Actuate the lever of the directional valve until the Power Unit has returnedto the ground. Turn off the Power Unit.

G 14. Switch the two hoses connected to the cylinder with each other so that thecap end is connected to port B and the rod end is connected to port A.

Note: If you experience difficulty to disconnect equipment, movethe directional valve lever back and forth to relieve static pressurethat might be trapped in the A and B cylinder lines.

Cylinder Control

3-12


G 16. Raise and lower the Power Unit by shifting the lever of the directional valve.Record which direction the Power Unit moves for each lever position inTable 3-2.

LEVER POSITION ACTION

MOVED TOWARD VALVE BODY

MOVED OUTWARD FROM VALVEBODY

CENTERED

Table 3-2. Power Unit action versus lever position.

G 17. Actuate the lever of the directional valve until the Power Unit has returnedto the ground. Turn off the Power Unit. Open the Relief Valve completely(turn knob fully counterclockwise).

G 18. Can the cylinder motion be easily changed by reversing the cylinderconnections to the directional valve?

G Yes G No

G 19. Switch the two hoses connected to the cylinder in order to re-obtain thecircuit shown in Figure 3-5, then proceed to the next part of the exercise.

Effect of pressure control on cylinder speed

G 20. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it 4 turns. Use the vernier scale on the valve knob to ensure anaccurate valve setting.



G 23. Move the lever of the directional valve outward from the valve body to raisethe Power Unit. Note the pressure reading at gauge A during cylinderretraction and when the cylinder is fully retracted. Record your readings inTable 3-3. Also record the retraction time.

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3-13

SYSTEMPRESSURE

RETRACTIONTIME

PRESSURE

CYL. MOVING CYL. STOPPED

2800 kPa (400 psi)

3500 kPa (500 psi)

Table 3-3. Effect of pressure control on cylinder speed.

G 24. Move the lever of the directional valve toward the valve body to return thePower Unit to the ground, then release the valve lever.


G 26. Move the lever of the directional valve outward from the valve body to raisethe Power Unit. Note the pressure reading at gauge A during cylinderretraction and when the cylinder is fully retracted. Record your readings inTable 3-3. Also record the retraction time.


G 28. According to Table 3-3, does the cylinder rod retract faster as the systempressure is increased? Why?

G 29. Explain the reason for the nearly identical pressure registered at gauge Aduring cylinder retraction at the two relief valve settings.

G 30. Trace the oil flow through the system after the cylinder rod is fully extendedor retracted but before the Directional Control Valve is returned to the centerposition.

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3-14

Effect of flow control on cylinder speed



G 33. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it 1 turn. Use the vernier scale on the valve knob for accurateadjustment.

G 34. Move the lever of the directional valve outward from the valve body to raisethe Power Unit. Measure and record the retraction time in the "1 TURNOPEN" row of Table 3-4.

FLOW CONTROL VALVE SETTING RETRACTION TIME

1 TURN OPEN

3 TURNS OPEN

Table 3-4. Effect of flow control on cylinder speed.


G 36. Open the Flow Control Valve 2 additional turns counterclockwise. Again usethe vernier scale on the knob for accurate adjustment.

G 37. Move the lever of the directional valve outward from the valve body to raisethe Power Unit. Measure and record the retraction time in the "3 TURNSOPEN" row of Table 3-4.


G 39. According to Table 3-4, does the cylinder rod retract faster as the flow rateis increased? Why?

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3-15

Effect of pressure control on cylinder force


G 41. Make sure the Relief Valve is open completely (knob turned fullycounterclockwise). Turn on the Power Unit.

G 42. Move the lever of the directional valve outward from the valve body whileobserving the Power Unit. Is the Power Unit lifted?

G Yes G No

G 43. Release the directional valve lever. Increase the amount of pressureavailable at the cylinder piston. To do so, turn the Relief Valve adjustmentknob clockwise until the system pressure at gauge A is 2100 kPa (300 psi).

G 44. Move the lever of the directional valve outward from the valve body whileobserving the Power Unit. Is the Power Unit lifted?

G Yes G No



G 47. From the observations you made in the previous steps, does increasing therelief valve pressure setting allow the cylinder to lift heavier loads? Why?

Effect of flow control on cylinder force


G 49. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it 3 turns.

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3-16

G 50. Move the lever of the directional valve outward from the valve body andincrease the Relief Valve pressure setting until the Power Unit begins torise, then slowly decrease the valve setting until the cylinder stops.

G 51. While keeping the directional valve lever in the outward position, increasethe Flow Control Valve opening by turning its adjustment knob 1 turncounterclockwise. Note the effect on the cylinder in the “4 TURNS” row ofTable 3-5.

FLOW CONTROL SETTING EFFECT ON CYLINDER

4 TURNS OPEN

5 TURNS OPEN

Table 3-5. Effect of flow control on cylinder force.

G 52. While keeping the directional valve lever in the outward position, open theFlow Control Valve 1 more turn counterclockwise. Note the effect on thecylinder in the “5 TURNS” row of Table 3-5.

G 53. Move the lever of the directional valve lever toward the valve body to returnthe Power Unit to the ground. Turn off the Power Unit. Open the ReliefValve completely (turn knob fully counterclockwise).

G 54. What effect did the adjustment of the Flow Control Valve have on the stalledcylinder? Why?

G 55. Based on what you have learned in this exercise, describe the effect of theRelief Valve and Flow Control Valve on the speed and force of a cylinder.


Cylinder Control

3-17



CONCLUSION

In the first part of the exercise, you tested a four-way directional valve and showedhow the valve ports were interconnected for each of the lever positions. You learnedthat the term way refers to the total number of working ports or connections on thevalve, and that the term position refers to the number of settings for the valve spool.You also learned that the center design determines how the ports are interconnectedwhen the valve spool is in the center or neutral position. The directional valvesupplied with your Hydraulics Trainer is of closed-center type because it blocks theflow between all ports when it is centered.

In the other parts of the exercise, you discovered that hydraulic power can be easilycontrolled. You learned how relief valves and flow control valves affect the cylinderforce and speed. You discovered that pressure control affects only the force, andthat flow control affects only the speed. When working with hydraulic equipment, itis often useful to change one variable without affecting the other. The use of theproper control valve allows you to select the combination of force and speed that youneed.

In conclusion, keep in mind that there are only three control valve categories.Though you will find many special hydraulic components, they will usually comeunder one of these three categories: pressure control, flow control, or directionalcontrol.

REVIEW QUESTIONS

1. What happens to the force output of a cylinder piston when the pressure appliedon the piston decreases?

2. What happens to the speed of a piston rod when the pressure applied on thepiston increases?

3. If the oil flow to a cylinder increases, does the force output increase, decrease,or stay the same?

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3-18

4. If the oil flow to a cylinder decreases, does the speed of the piston rod increase,decrease, or stay the same?

5. What happens to the retraction speed of a cylinder rod if the diameter of the rodincreases but the oil flow to the cylinder stays the same?

6. What happens to the extension speed of a cylinder rod when the rod diameterdecreases, the pressure increases, but the flow rate to the cylinder stays thesame?

3-19

Exercise 3-2

Cylinders in Series

EXERCISE OBJECTIVE

C To describe the operation of a series circuit;C To cause two cylinders to start and stop at the same time by connecting them in

series.C To demonstrate pressure intensification in a series circuit.

DISCUSSION

Cylinders in series

In some hydraulic applications, it is necessary for two cylinders to work together inunison. For example, two cylinders may be required to start and stop extending atthe same time. Cylinders operating in this manner are said to be synchronized.

One method to synchronize two cylinders consists in connecting them in series, sothat the discharge flow from one cylinder serves as the input flow to the secondcylinder.

Figure 3-6 shows two cylinders connected in series. The rod end of one cylinder isconnected to the cap end of the second cylinder. With this circuit, neither cylindercan move unless the other is also moving.

Figure 3-6. Cylinders connected in series.

Shifting the lever of the directional valve in Figure 3-6 will cause the cylinders to startand stop at the same time. If the cylinders are of the same size and stroke, however,cylinder 2 (downstream cylinder) will extend slower, and it will not extend completely

Cylinders in Series

3-20

since the flow out of the rod end of cylinder 1 (upstream cylinder) will be less thanthe flow entering the cap end of cylinder 1.

If the two cylinders are of the same size, the total force of the two cylinders in seriesis equal to the system pressure multiplied by the area of one of the piston.Differences in cylinder size, however, will cause considerable variation.

Pressure intensification in a series circuit

Intensification of the system pressure will occur in a series circuit if the flow from therod end of the upstream cylinder is blocked or severely restricted as by a heavy loadon the downstream cylinder.

Figure 3-7 shows an example. The cylinders in this figure have the same size as thetwo cylinders supplied with your Hydraulics Trainer. The input force exerted on thefull piston area of the upstream cylinder is 1775 N. Since the flow from the rod endof this cylinder is partially blocked by the heavy load on the downstream cylinder, thepressure at the rod end of the upstream cylinder will rise until the force applied onthe annular piston area of the upstream cylinder equals the input force of 1775 N.

Figure 3-7. Pressure intensification in a series circuit.

Due to the difference in exposure area between the full and annular piston area, thepressure at the rod end of the upstream cylinder will intensify by a factor equal to theratio of the full piston area to the annular piston area, Af/Aa, resulting in a pressureof 5763 kPa at the cap end of the downstream cylinder, and in a force of 6570 N atthe output load. Notice that the intensification ratio Af/Aa will hold as long as there isno load on the upstream cylinder.

Cylinders in Series

3-21

REFERENCE MATERIAL

For detailed information on cylinder synchronization, refer to the chapter entitledCheck Valves, Accumulators and Cylinders in the Parker-Hannifin’s manualIndustrial Hydraulic Technology.

Procedure summary

In the first part of the exercise, you will connect two cylinders in series with a loadingdevice on the upstream cylinder. The loading device will be a flow control valve. Youwill gradually increase the load on the upstream cylinder, while noting the effect achange in load has on the synchronization of the cylinders, along with the pressuresapplied to both cylinders.

In the second part of the exercise, you will connect the loading device to thedownstream cylinder. You will repeat the manipulations performed in the first part ofthe exercise in order to observe the effect a change in load position has on theoperation of the series circuit.

EQUIPMENT REQUIRED


PROCEDURE

Series circuit with load on the upstream cylinder

G 1. Connect the circuit shown in Figure 3-8. In this circuit, the Flow ControlValve will act as a loading device on the upstream cylinder [2.54-cm (1-in)bore].

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3-22

Figure 3-8. Series circuit with load on the upstream cylinder.

Note: If the rods of the 2.54-cm (1-in) and 3.81-cm (1.5-in) borecylinders are not fully retracted, do not connect the circuit inFigure 3-8. Instead, retract the rod of each cylinder hydraulicallyby using the cylinder actuation circuit shown in Figure 2-10. Whenboth rods are retracted, disconnect the circuit of Figure 2-10 andconnect the circuit shown in Figure 3-8.





G 4. Open the Flow Control Valve completely (turn knob fully counterclockwise)so that no load is placed on the upstream cylinder.

G 5. Move the lever of the directional valve toward the valve body to extend bothcylinders. While keeping the valve lever shifted, turn the Relief Valveadjustment knob clockwise until the circuit pressure at gauge A is 1720 kPa(250 psi). Retract the cylinders by moving the valve lever outward from thevalve body.

Cylinders in Series

3-23

G 6. While observing the cylinders, move the lever of the directional valve towardthe valve body to extend their rod. Do the cylinders start and stop at thesame time? Does the downstream cylinder [3.81-cm (1.5-in) bore] extendfully? Explain.

G 7. Retract the cylinders.

G 8. Now extend and then retract the cylinders again and observe the pressurereadings at gauges A and B as the cylinders are extending. Record thesepressures in the “NO-LOAD” row of Table 3-6.

LOAD CONDITIONPRESSURE

GAUGE A (UPSTREAMCYLINDER)

GAUGE B (DOWN-STREAM CYLINDER)

NO-LOAD

MEDIUM LOAD

HEAVY LOAD

STALLED

Table 3-6. Circuit pressures with load on the upstream cylinder.

G 9. Put a medium load on the upstream cylinder. To do so, close the FlowControl Valve completely (turn knob fully clockwise), then open it 1 turn.

G 10. Extend and then retract the cylinders. Do they start and stop at the sametime?

G Yes G No

G 11. Extend and then retract the cylinders and observe the pressure readings atgauges A and B as the cylinders are extending. Record these pressures inthe “MEDIUM LOAD” row of Table 3-6.

G 12. Put a heavy load on the upstream cylinder. To do so, close the Flow ControlValve completely (turn knob fully clockwise), then open it ½ turn.

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3-24

G 13. Extend and then retract the cylinders. Do they start and stop at the sametime?

G Yes G No

G 14. From your observations, does a change in load affect the synchronizationof the two cylinders? Explain why.

G 15. Extend and then retract the cylinders and observe the pressure readings atgauges A and B as the cylinders are extending. Record these pressures inthe “HEAVY LOAD” row of Table 3-6.

G 16. Close the Flow Control Valve completely (turn knob fully clockwise) so thatthe upstream cylinder will stall completely.

G 17. Attempt to extend the cylinders, while observing the pressure readings atgauges A and B. Record these pressures in the “STALLED” row ofTable 3-6.

G 18. Turn off the Power Unit. Do not modify the Relief Valve pressure setting.

G 19. According to Table 3-6, did the pressure required to extend the downstreamcylinder remain approximately constant when the various loads were placedon the upstream cylinder? Why?

G 20. Why did the pressure at gauge B drop when the load stalled the cylinders?

G 21. If you were asked to modify the bore size of the usptream cylinder to allowboth cylinders to complete their full stroke (10.16 cm/4 in) at the same time,what bore size would you select?

Cylinders in Series

3-25

Note: Keep the same rod size (1.59 cm/0.625 in) for the upstreamcylinder.

Series circuit with load on the downstream cylinder

G 22. Modify your circuit connections in order to place the loading device (FlowControl Valve) on the downstream cylinder [3.81-cm (1.5-in) bore], asFigure 3-9 shows.

Figure 3-9. Series circuit with load on the downstream cylinder.


G 24. Open the Flow Control Valve completely (turn knob fully counterclockwise)so that no load is placed on the downstream cylinder.

G 25. Extend the cylinders. Do they still start and stop at the same time? Explainwhy.

Cylinders in Series

3-26


G 27. Extend and then retract the cylinders and observe the pressure readings atgauges A and B as the cylinders are extending. Record these pressure inthe “NO-LOAD” row of Table 3-7.

LOAD CONDITION

PRESSURE

GAUGE A (UPSTREAMCYLINDER)

GAUGE B (DOWN-STREAM CYLINDER)

NO-LOAD

MEDIUM LOAD

HEAVY LOAD

STALLED

Table 3-7. Circuit pressures with load on the downstream cylinder.

G 28. Put a medium load on the downstream cylinder. To do so, close the FlowControl Valve completely (turn knob fully clockwise), then open it 1 turn.

G 29. Extend and then retract the cylinders and observe the pressure readings atgauges A and B as the cylinders are extending. Record these pressures inthe “MEDIUM LOAD” row of Table 3-7.

G 30. Put a heavy load on the downstream cylinder. To do so, close the FlowControl Valve completely (turn knob fully clockwise), then open it ½ turn.

G 31. Extend and then retract the cylinders and observe the pressure readings atgauges A and B as the cylinders are extending. Record these pressures inthe “HEAVY LOAD” row of Table 3-7.

G 32. Close the Flow Control Valve completely (turn knob fully clockwise) so thatthe downstream cylinder will stall.

G 33. Attempt to extend the cylinders while observing the pressure readings atgauges A and B. Record these pressures in the “STALLED” row ofTable 3-7.


Cylinders in Series

3-27

G 35. According to Table 3-7, did the pressures at gauge A and B increase as theload on the downstream cylinder was increased? Why?

G 36. According to Table 3-7, is the ratio of gauge A to gauge B pressureapproximately equal to the ratio of annular to full piston area (Aa/Af) of theupstream cylinder for the MEDIUM, HEAVY, and STALLED load conditions?If so, explain why.

G 37. Explain the reason why pressure intensification can occur in a series circuit.

G 38. Disconnect the line cord from the wall outlet, then disconnect all hoses.Wipe off any hydraulic oil residue.



CONCLUSION

This exercise showed some circuit principles which govern two cylinders connectedin series. You observed that neither cylinder could move unless the other was alsomoving. You saw that once the upstream cylinder was stalled, no flow could takeplace into the downstream cylinder and the downstream cylinder would stopimmediately. When you switched the loading device (Flow Control Valve) on thecylinders, you noticed some changes:

Cylinders in Series

3-28

– With the loading device placed on the upstream cylinder, the pressure at the capend of the downstream cylinder remained relatively constant as the load wasincreased. This is because the only load on the downstream cylinder was theresistance of its internal seals and the resistance of the oil flowing back to thereservoir. This load does not depend on load variations in the rest of the circuit.

– With the loading device placed on the downstream cylinder, however, thepressure at the cap end of the downstream cylinder increased as the loadincreased. The input pressure intensified by a factor equal to the ratio of full toannular piston area (Af/Aa) of the upstream cylinder, due to the difference inexposure area between each side of the cylinder piston.

REVIEW QUESTIONS

1. What is meant by "cylinder synchronization"?

2. In a series circuit, can the upstream cylinder be extended after the downstreamcylinder has completed its movement? Why?

3. In a series circuit where the two cylinders are of the same size and stroke, willboth cylinders extend completely? Explain why.

4. In a series circuit where a heavy load is attached on the downstream cylinder,by which amount will the input pressure intensify if the upstream cylinder is a3.81-cm (1.5-in) bore x 1.59-cm (0.625-in) rod cylinder?

Cylinders in Series

3-29

5. Calculate the theoretical force output in the circuit of Figure 3-10.


3-30

3-31

Exercise 3-3

Cylinders in Parallel

EXERCISE OBJECTIVE

C To describe the operation of a parallel circuit;C To describe the extension sequence of parallel cylinders having differing bore

sizes;C To synchronize the extension of parallel cylinders using a flow control valve.

DISCUSSION

Cylinders in parallel

Figure 3-11 shows two cylinders connected in parallel. The rod and cap ends of onecylinder are connected to the corresponding ends of the other cylinder. Since thesecylinders are of the same size, cylinder 1 will extend first because it requires thelowest pressure to move its load. Once cylinder 1 is extended, the systempressure will climb to the level required for cylinder 2 to extend. Once cylinder 2 isextended, the system pressure will climb to the setting of the relief valve.

Figure 3-11. Cylinders connected in parallel.

Synchronization of parallel cylinders

In theory, two cylinders connected in parallel should operate in synchronization ifthey are of identical size and stroke and if they are evenly loaded, since they bothreceive the same flow rate from the same power unit. In practice, however,manufacturing any two cylinders or articles to be exactly identical is impossible.There are always small differences in dimensions, internal friction, surface texture,internal leakage, etc.


3-32

This does not mean that synchronization of parallel cylinders is impossible. Onepopular method of synchronizing parallel cylinders, called mechanical yoke method,is shown in Figure 3-12. In this method, a strong yoke connects the two cylinder rodstogether. The weight of the two loads is distributed evenly between the two cylindersso that the cylinders extend at the same speed, even if the loads are of differentweight.

Figure 3-12. Synchronization of parallel cylinders using a mechanical yoke.

If mechanical synchronization is not possible or practical, parallel cylinders can beapproximately synchronized using the flow control valve method. In this method, aflow control valve is connected in series with the cylinder requiring the lowestpressure to move in order to increase the resistance of this line (circuit path).

Figure 3-13 shows an example. Cylinders 1 and 2 are of the same size, howevercylinder 1 requires 1400 kPa at its cap end to lift the light load, while cylinder 2requires 3500 kPa at its cap end to lift the heavier load. A flow control valve,connected in the line of cylinder 1, is adjusted so that it creates an additionalpressure drop of 2100 kPa in this line when cylinder 1 extends. Since equalpressures of 3500 kPa (500 psi) are required in each cylinder line, the oil from thepump will divide equally between the two lines, causing the cylinders to move at thesame time and speed. The cylinders will operate in unison for a limited number ofcycles. Eventually, they will drift out of synchronization, and the flow control valve willhave to be re-adjusted to synchronize them again. Also, load variations will causethe cylinders to go out of synchronization if the flow control valve is of non-compensated type, because this type of valve does not compensate for pressurechanges in the system.


3-33

Figure 3-13. Synchronization of parallel cylinders using a flow control valve.

REFERENCE MATERIAL

For detailed information on cylinder synchronization, refer to the chapter entitledCheck Valves, Accumulators and Cylinders in the Parker-Hannifin’s manualIndustrial Hydraulic Technology.

Procedure summary

In the first part of the exercise, you will connect two cylinders in parallel with eachother, with a loading device on the larger cylinder. The loading device will be a flowcontrol valve. You will determine which cylinder moves first when a heavy load isplaced on the large cylinder.

In the second part of the exercise, you will connect the flow control valve to thesmaller cylinder. You will determine which cylinder moves first when a heavy loadis placed on the small cylinder. Then you will synchronize the two cylinders and varythe extension sequence by modifying the flow control valve setting.

EQUIPMENT REQUIRED



3-34

PROCEDURE

Cylinders in parallel


Figure 3-14. Cylinders in parallel with a heavy load on the large cylinder.

G 2. Examine the circuit in Figure 3-14. The cylinders are in parallel with eachother. The Flow Control Valve acts as a loading device on the large cylinder[3.81-cm (1.5-in) bore] only, so the cylinders are unevenly loaded. The FlowControl Valve is partially closed to simulate a heavy load on the largecylinder.

When the directional valve is shifted to extend the cylinders, the oil from thepump is directed to the cap side of both cylinders at the same time. Thelarge cylinder must counteract the high resistance offered by the FlowControl Valve before it can extend. The small cylinder [2.54-cm (1-in) bore]must counteract the resistance of the oil flowing back to the reservoir beforeit can extend. The cylinder requiring the lowest pressure to move will extendfirst.


3-35

Predict which cylinder will extend first when the Flow Control Valve ispartially closed to simulate a heavy load on the large cylinder, and explainwhy.




G 4. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it ½ turn.


G 6. Turn the relief valve adjustment knob clockwise until the circuit pressure atgauge A is 2100 kPa (300 psi).

G 7. Move the lever of the directional valve toward the valve body to extend thetwo cylinders and observe them as they extend. Which cylinder extendedfirst? Why?




3-36

Cylinder synchronization using a Flow Control Valve

G 10. Modify your circuit connections in order to place the loading device (FlowControl Valve) on the small cylinder [2.54-cm (1-in) bore], as Figure 3-15shows.

Figure 3-15. Cylinders in parallel with a heavy load on the small cylinder.

G 11. Predict which cylinder will extend first when the Flow Control Valve ispartially closed to simulate a heavy load on the small cylinder.

G 12. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it ¼ turn.



3-37

G 14. Turn the Relief Valve adjustment knob clockwise until the circuit pressureat gauge A is 2100 kPa (300 psi).

G 15. Move the lever of the directional valve toward the valve body to extend thetwo cylinders and observe them as they extend. Which cylinder extendedfirst? Why?


G 17. Adjust the setting of the Flow Control Valve so that both cylinders completetheir full stroke at the same time during extension. Accurate adjustment mayrequire that the cylinders be extended and retracted several times.

G 18. Extend and retract the cylinders several times with the new Flow ControlValve setting. Do the cylinders remain synchronized?

G Yes G No

G 19. Let the system run for about 15 minutes. Do not modify the Flow ControlValve setting.

G 20. Extend and retract the cylinders several times. Did the cylinders stay insynchronization?

G Yes G No

G 21. Does the Flow Control Valve have to be readjusted when the cylinders areoperated over an extended period of time? Explain why.


3-38

G 22. Try to adjust the Flow Control Valve so that the small cylinder completes itsextension approximately 2 seconds after the large cylinder completes itsextension. Can the extension sequence of the cylinders be controlled in aparallel circuit? Explain.

G 23. Make sure the cylinders are fully retracted, then turn off the Power Unit.Open the Relief Valve completely (turn knob fully counterclockwise).




CONCLUSION

This exercise showed some circuit principles which govern two cylinders connectedin parallel. You observed that the individual loads governed cylinder movement, andthe cylinder requiring the lowest pressure to move its load always moved first. Thisis because oil always flows through the path requiring the lowest pressure, and thepressure in a cylinder depends on the load.

REVIEW QUESTIONS

1. In the circuit of Figure 3-16, which cylinder will move first if the flow control valveis open completely? Why?


3-39


2. In the circuit of Figure 3-16, which cylinder will move first if the flow control valveis adjusted so that it creates a pressure drop of 2400 kPa (350 psi)? Explain.

3. Calculate the theoretical pressure drop, )P, required across the flow controlvalve in Figure 3-16 to synchronize the extension of the two cylinders.

4. Describe the extension sequence in the circuit of Figure 3-16 when the flowcontrol valve is open completely and the relief valve pressure setting is 2100 kPa(200 psi).

3-40

3-41

Exercise 3-4

Regenerative Circuits

EXERCISE OBJECTIVE

C To describe the operation of a regenerative circuit;C To describe the effect of regeneration on cylinder speed;C To describe the effect of regeneration on cylinder force.

DISCUSSION

Principle of regeneration

The primary purpose of regenerative circuits is to provide rapid extension speedswith a minimum pump output flow. Regeneration is accomplished by sending theoil which flows out of the rod end of a cylinder back into the cap end of this cylinder.Figure 3-17 shows a regenerative circuit.

C When the directional valve is shifted to extend the cylinder (straight-throughposition), the pumped oil is directed to both sides of the piston at the same time.This results in two opposite forces simultaneously acting on each side of thepiston. Since, however, the cap end of the piston has a larger surface areaexposed to oil pressure than the rod end, a greater force is exerted on the fullpiston area, causing the cylinder rod to extend. The oil forced out from the rodend adds to that coming from the pump and enters the cap end of the cylinder.This extra oil speeds up the cylinder by increasing its flow rate.

C When the directional valve is shifted to retract the cylinder (cross-connectedposition), the pumped oil is blocked at port B of the directional valve, but isallowed to flow directly to the rod end of the cylinder, causing the cylinder toretract.


3-42

Figure 3-17. Regenerative circuit.


3-43

Regeneration can only occur in extension. The reason for this is that the forceacting to extend the rod is greater than the force acting to retract the rod for anygiven amount of pressure, because the piston area at the rod end is less than thatat the cap end.

Cylinder speed during regeneration

When a cylinder extends in regeneration, the oil flowing out of its rod end helps thepump oil to fill its cap end. This reduces the volume of oil required from the pump tocompletely extend the cylinder. The required volume from the pump is equal to thevolume of oil inside the cylinder when it is extended minus the volume of oil insidethe cylinder when it is retracted. This is equal to the volume occupied of the cylinderrod. The extension speed of a cylinder in regeneration, then, is determined by howfast the pump can fill the volume of the cylinder rod. In equation form:

S.I. units:

English units:

So we can see that connecting a cylinder in regeneration increases the rod extensionspeed. In fact, this speed is increased by a factor equal to the ratio of the full pistonarea to the rod area, Af / Arod. For example, an Af / Arod ratio of 2 means that the fullpiston area is twice that of the rod area. This also means that the extension speedin regeneration will be twice as fast as the extension speed in normal mode.

The formula for calculating the amount of time required for a cylinder in regenerationto complete its stroke is the formula for extension speed divided into the strokelength. The formula is as follows:

S.I. units:

English units:

So we can see that connecting a cylinder in regeneration reduces the rod extensiontime. In fact, the extension time is reduced by a factor equal to the ratio of the fullpiston area to the rod area, Af / Arod. For example, an Af / Arod ratio of 2 means thatthe extension time in regeneration will be only half the extension time in normalmode.


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Cylinder force during regeneration

Regenerative circuits have a disadvantage: they reduce the force generated by thecylinder during extension. This is because the force generated in the direction ofextension is diminished by the opposite force generated in the direction ofretraction—remember that both sides of the piston are interconnected andexperience the same pressure. Therefore, the net (effective) area on which the forceis exerted is the rod area. This means that the force generated by the cylinder duringextension is equal to the circuit pressure multiplied by the rod area. In equation form:

S.I. units:

English units:

So we see that connecting a cylinder in regeneration provides a faster extensionspeed but reduces the generated force. In fact, the cylinder force is sacrificed for rodspeed.

Applications

When designing a regeneration circuit, the size of the cylinder rod must be carefullyselected as it determines both the cylinder speed and force. The higher the Af / Arod

ratio, the higher the extension speed, but the lower the force output. An Af / Arod ratioof 2 is often used since it provides approximately equal forces and speeds duringextension and retraction. Varying the Af / Arod ratio too far from 2, however, will resultin reduced force capabilities; the net (effective) force generated may even not besufficient for the cylinder to extend.

Regeneration is often used to only extend the rod to the work load at high speed.When the moment arrives for work to be done, the rod end of the cylinder is drainedback to the reservoir so that full force is applied to the load. 4-position directionalvalves are used to control these two stages of the extension cycle. Figure 3-18shows an example:

C In Figure 3-18 (a), the directional valve is in the center position, and no oil flowsto the cylinder.

C In Figure 3-18 (b), the valve is shifted to the regeneration position. The cylinderrod extends rapidly to the work load.

C Once the rod reaches the load, the valve switches to the normal extensionposition to increase the force to the load, as Figure 3-18 (c) shows.

C When the rod is fully extended, the valve switches to the retraction position, asFigure 3-18 (d) shows. The cylinder rod retracts at normal speed.


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Figure 3-18. Simple regeneration circuit using a four-position directional valve.

REFERENCE MATERIAL

For detailed information on regeneration, refer to the chapter entitled Check Valves,Accumulators and Cylinders in the Parker-Hannifin’s manual Industrial HydraulicTechnology.

Procedure summary

In the first part of the exercise, you will determine the effect of regeneration on theextension time of a cylinder. To do so, you will measure the time required for acylinder to extend in both regenerative and normal modes of operation. You will thencompare the results obtained in each mode.

In the second part of the exercise, you will determine the effect of regeneration onthe force output of a cylinder. To do so, you will measure the force output of acylinder in both regenerative and normal modes of operation. You will then comparethe results obtained in each mode.


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EQUIPMENT REQUIRED


PROCEDURE

Effect of regeneration on cylinder extension time

G 1. Connect the circuit shown in Figure 3-19. In this circuit, the Flow ControlValve will be used to reduce the flow into the circuit so the cylinder speedis easier to time. The valve would not be used in industrial regenerativecircuits of this design.

Note: Do not connect the Loading Device to the cylinder yet. TheLoading Device will be used later in the exercise.




G 3. Open the Flow Control Valve completely (turn knob fully counterclockwise).


G 5. With the directional valve lever in the center position, the pump flow isblocked at the rod end of the cylinder, and gauge B indicates the ReliefValve pressure setting. Turn the Relief Valve adjustment knob clockwiseuntil the circuit pressure at gauge B is 2100 kPa (300 psi).

G 6. Move the lever of the directional valve toward the valve body to extend thecylinder and adjust the Flow Control Valve so that the Flowmeter reads1.5 l/min [0.4 gal(US)/min] during cylinder extension, then retract thecylinder. Accurate adjustment may require that the cylinder be extended andretracted several times.


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Figure 3-19. Cylinder in a regenerative circuit.

G 7. Extend the cylinder and note the extension time and the pressure readingsat gauges A and B as the cylinder extends. Record these in the“REGENERATIVE” row of Table 3-8.

CIRCUIT CONDITION EXTENSION TIMEEXTENSION PRESSURES

GAUGE A GAUGE B

REGENERATIVE

NORMAL

Table 3-8. Cylinder data in regenerative and normal modes.

G 8. Retract the cylinder.


G 10. Change your regenerative circuit into a normal (non-regenerative) circuit. Todo so, disconnect the rod end of the cylinder from the 4-port manifoldinstalled on port P of the directional valve, then connect the rod end toport B of the directional valve, as Figure 3-20 shows.


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Figure 3-20. Cylinder in a normal (non-regenerative) circuit.

Note: Do not connect the Loading Device to the cylinder yet. TheLoading Device will be used later in the exercise.


G 12. Move the lever of the directional valve toward the valve body to extend thecylinder fully. While keeping the valve lever shifted, turn the Relief Valveadjustment knob clockwise until the circuit pressure at gauge A is 2100 kPa(300 psi). Retract the cylinder by moving the lever of the directional valveoutward from the valve body.

G 13. Move the lever of the directional valve toward the valve body to extend thecylinder and adjust the Flow Control Valve so that the Flowmeter reads1.5 l/min [0.4 gal(US)/min] during cylinder extension, then retract thecylinder. Accurate adjustment may require that the cylinder be extended andretracted several times.

G 14. Extend the cylinder and note the extension time and the pressure readingsat gauges A and B as the cylinder extends. Record these in the “NORMAL”row of Table 3-8.

G 15. Retract the cylinder.


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G 17. According to Table 3-8, is the extension time observed in regenerative modeshorter than that observed in normal mode? Why?

G 18. Calculate the theoretical extension time of the 2.54-cm (1-in) bore x 1.59-cm(0.625-in) rod x 10.16-cm (4-in) stroke cylinder in regeneration when theflow rate is 1.5 l/min [0.4 gal(US)/min]. Then, compare your result with theactual extension time recorded in Table 3-8. Are these values approximatelyequal?

G 19. Why did the cylinder extend when both sides of the piston were pressurizedin regenerative mode?

G 20. Explain the reason for the very low pressure required to extend the cylinderin normal mode.

Effect of regeneration on cylinder force output

G 21. Change your circuit into a regenerative circuit. To do so, disconnect the rodend of the cylinder from port B of the directional valve, then connect the rodend to the 4-port manifold installed on port P of the directional valve, asshown in Figure 3-19.


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G 22. Disconnect the 2.54-cm (1-in) bore cylinder from the circuit. Remove thecylinder from its adapter by unscrewing its retaining ring. Make sure thecylinder tip (bullet) is removed from the cylinder rod end. Screw the cylinderinto the Loading Device. Then, reconnect the cylinder into the circuit asshown in Figure 3-19.

G 23. Clip the NEWTON/LBF-graduated ruler to the Loading Device, and align the“0” mark with the colored line on the load piston.



G 26. Move the lever of the directional valve toward the valve body and turn theRelief Valve adjustment knob clockwise until the circuit pressure at gauge Ais 4200 kPa (600 psi), then turn the knob counterclockwise to decrease thecircuit pressure until gauge A reads 3500 kPa (500 psi). Release the valvelever.

G 27. Note and record the force reading on the Loading Device in the“REGENERATIVE” row of Table 3-9.

CIRCUIT CONDITION CYLINDER FORCE OUTPUT

REGENERATIVE

NORMAL

Table 3-9. Effect of regeneration on cylinder force.

G 28. Retract the cylinder, then turn off the Power Unit. Open the Relief Valvecompletely (turn knob fully counterclockwise).

G 29. Change your regenerative circuit into a normal circuit. To do so, disconnectthe rod end of the cylinder from the 4-port manifold installed on port P of thedirectional valve, then connect the rod end to port B of the directional valve,as shown in Figure 3-20.



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G 31. Move the lever of the directional valve toward the valve body and turn theRelief Valve adjustment knob clockwise until the circuit pressure at gauge Ais 4200 kPa (600 psi), then turn the knob counterclockwise to decrease thecircuit pressure until gauge A reads 3500 kPa (500 psi). Release the valvelever.

G 32. Note and record the force reading on the Loading Device in the “NORMAL”row of Table 3-9.

G 33. Retract the cylinder, then turn off the Power Unit. Open the Relief Valvecompletely (turn knob fully counterclockwise).

G 34. According to Table 3-9, is the force observed in regeneration mode reducedfrom that observed in normal mode? Explain why.

G 35. Calculate the theoretical force output of the 2.54-cm (1-in) bore x 1.59-cm(0.625-in) rod x 10.16-cm (4-in) stroke cylinder in regeneration. Then,compare your result with the actual force output recorded in Table 3-9. Arethese values approximately equal?





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CONCLUSION

In this exercise, you learned that a regenerative circuit increases the extensionspeed of a cylinder. You caused a cylinder to extend more rapidly by applying equalpressures to both sides of the piston. The extension time was reduced by a factorequal to the ratio of the full piston area to the rod area, Af / Arod. You also learned thatregenerative circuits decrease the force generated during extension because forceis sacrificed for cylinder speed.

Regenerative circuits offer a solution to a serious problem in hydraulics: slowextension under no load. With regenerative extension, the cylinder rod can extendrapidly until the low force generated by the cylinder is no longer enough for theapplication. At this point, a directional valve can be automatically shifted to allownormal extension with increased force. Four-position directional valve are often usedto control these two stages of the extension cycle.

REVIEW QUESTIONS

1. What is a regenerative circuit?

2. Would a cylinder generate more force in a normal circuit or in a regenerativecircuit?

3. Would a cylinder extend more rapidly in a normal circuit or a regenerativecircuit?

4. What happens to the extension speed and force generated in a regenerativecircuit when the piston rod diameter is decreased?

5. By which amount is the cylinder force output reduced in a regenerative circuitgiving double the normal extension speed?

4-1

Unit 4

Functional Circuits

UNIT OBJECTIVE

When you have completed this unit, you will be able to construct and operatefunctional hydraulic circuits using accumulators, hydraulic motors, pressure reducingvalves, and remotely controlled relief valves.


This unit introduces the design and operation of hydraulic application circuits usingaccumulators, hydraulic motors, pressure reducing valves, and remotely controlledrelief valves.

In many hydraulic systems, there is a need for an alternate power source.Accumulator circuits, discussed in Exercise 4-1, assist the hydraulic pump by storinghydraulic fluid at the system pressure. Accumulator circuits are used in hundreds ofdifferent applications as auxiliary and emergency power sources and for leakagecompensation and shock suppression.

Numerous hydraulic applications, including gear boxes, belts, winches, andproduction machinery, require fluid energy to be converted into mechanical rotaryenergy. Hydraulic motors, discussed in Exercise 4-2, can do this conversion directlywithout any intermediate machinery. Hydraulic motors are instantly reversible, andthey do not burn under excessively heavy loads as electric motors do.

In hydraulic circuits having more than one branch, it may be necessary to have ahigh pressure in one branch, and a low pressure in another. Pressure reducingvalves, discussed in Exercise 4-3, are used to reduce pressure in the low pressurebranch.

Remote operation of a hydraulic circuit is often convenient and sometimes essentialfor safety. Remote control of a relief valve, discussed in Exercise 4-4, allows theoperator to select between several levels of system pressure from a remote controllocation.

4-2

4-3

Exercise 4-1

Accumulators

EXERCISE OBJECTIVE

C To describe the general types of accumulators;C To learn how accumulators can be used in auxiliary power, emergency power,

leakage compensation, and shock suppression;C To understand the safety requirements for accumulator circuits.

DISCUSSION

Hydraulic accumulators

A hydraulic accumulator stores oil under pressure. This potential energy may thenbe converted into working energy to assist the pump.

Figure 4-1 shows the three types of hydraulic accumulators with their correspondingsymbol.

Figure 4-1. General types of accumulators.

A. Weight-loaded accumulator: consists of a weight acting on a piston. Theincoming oil forces the piston to lift the weight, which charges the accumulator.

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4-4

When the accumulator is allowed to discharge, the weight pushes the oil out ofthe accumulator at constant pressure throughout the piston stroke. Theadvantage of this type of accumulator is that it discharges at a constant pressure.The disadvantage is that it is large and bulky in size.

B. Spring-loaded accumulator: consists of a spring acting on a piston. Theincoming oil forces the piston to compress the spring, which charges theaccumulator. When the accumulator is allowed to discharge, the springdecompresses and pushes the oil out of the accumulator. The advantage of thistype of accumulator is that it is less bulky than the weighted unit. Thedisadvantage is that it does not discharge at a constant pressure because theforce level of the spring decreases as it decompresses.

C. Gas-loaded accumulator: consists of a volume of gas exposed to systempressure. The gas chamber is separated from the oil by a piston, a diaphragm,or a bladder. The accumulator is precharged to a predetermined pressure whilethe accumulator is completely empty of oil. Oil can enter the accumulator whenthe oil pressure is higher than the precharge pressure. Oil filling the accumulatorcompresses the gas and raises the pressure in the gas chamber. Higherpressures compress the bladder more than lower pressures, allowing a largervolume of oil to enter the accumulator. When the accumulator is allowed todischarge, the gas decompresses and pushes the oil out of the accumulator. Themain advantage of this type of accumulator is that the pressure at which the oilis stored can be changed simply by modifying the precharge pressure—with aweight-loaded accumulator, this would require you to change the load on thepiston, which is more tricky. The disadvantage is that it does not discharge at aconstant pressure. This type is the most popular in high pressure applications.

Accumulator applications

The most common uses of accumulators are:

C Auxiliary power;C Emergency power;C Leakage compensation;C Shock suppression.

Auxiliary power

Several applications require an auxiliary power source to supplement the pump flow.Instead of using a large pump to generate a high power during a fraction of the cycle,a smaller pump is used to spread power evenly over the cycle. Figure 4-2 shows anexample. The cylinder in this circuit is operated infrequently. During periods when nooil is required for cylinder operation, the pump stores oil in the accumulator. A checkvalve connected between the power unit and the accumulator prevents theaccumulator from discharging through the power unit pressure line port when thepump pressure becomes lower than the accumulator pressure.

Accumulators

4-5

Figure 4-2. Accumulator used as an auxiliary power source.

When the moment arrives for the cylinder to extend, a solenoid-operated directionalvalve is shifted down. Oil under pressure is discharged from the accumulator, joiningthat coming from the pump, which extends the cylinder faster than with pump oilalone.

Emergency power

Some applications require an alternate power source to return the system to a safestatus in case electrical power is lost. On a hydraulic press, for example, it may benecessary to automatically retract a cylinder in the event electrical power is lost inorder to release the press. The accumulator provides the high pressure oil requiredfor this function.

Leakage compensation

Several applications require a cylinder to maintain position and pressure during longstandby periods. However, leakage losses and temperature variations cause thepressure to slowly drop over time. Accumulators can compensate for the decreasein pressure, so the pump does not need to run continuously.

Shock suppression

Sudden stoppage or reversal of high velocity oil causes high pressure surges in ahydraulic circuit. These pressure increases, or shocks, are caused by the inertia ofoil when it stops quickly. The accumulator cushions the oil by compressing the gasin gas-loaded units or by compressing the spring in spring-loaded units. Forexample, an accumulator may be used to absorb some of the shock produced whenthe pump flow is suddenly stopped or changes direction as a directional valve isshifted.

Accumulators

4-6

Safety considerations

The primary safety rule to follow when using an accumulator circuit is always havethe accumulator completely discharged before removing it from the circuit. Neverattempt to service or disassemble an accumulator without special training and allproper tools.

The reason for this is that pressurized accumulators can take off like a rocket if ahose or component in the accumulator line is disconnected. In Figure 4-2, forexample, the accumulator is placed in a blocked circuit to allow oil storage for lateruse. When the pump is turned off, the check valve prevents the accumulator fromdischarging through the pump so that oil under pressure remains trapped in theblocked accumulator circuit. Since there is no way to discharge the accumulator, thisstored energy can throw parts with enough force to cause injury or spray oil asfittings are loosened.

Figure 4-3. Bleed-down circuit using a needle valve.

All industrial accumulator circuits have a positive discharge allowing you todepressurize, or bleed-down the circuit after the power unit is turned off. This isusually a needle valve connected directly into the pressure line near theaccumulator, as Figure 4-3 shows. Closing the needle valve blocks the flow through

Accumulators

4-7

the valve and allows the accumulator to charge when the power unit is turned on.When the power unit is turned off, the accumulator can be discharged safely byopening the needle valve.

REFERENCE MATERIAL

For additional information on accumulators, refer to the chapter entitled CheckValves, Accumulators and Cylinders in the Parker-Hannifin’s manual IndustrialHydraulic Technology.

Procedure summary

In the first part of the exercise, you will measure the storage capacity of a spring-loaded accumulator at several different pressures. To do so, you will fill theaccumulator with oil to a specified pressure and then discharge and measure the oilinto a flask to determine the volume of stored oil.

In the second part of the exercise, you will use a spring-loaded accumulator as anemergency device to actuate a cylinder when the pump is turned off.

In the third part of the exercise, you will use the spring-loaded accumulator as anauxiliary power source to increase the cycling speed of a cylinder.

EQUIPMENT REQUIRED


PROCEDURE

Measuring the storage capacity of a spring-loaded accumulator

G 1. Remove the 3.81-cm (1.5-in) bore cylinder from its adapter by unscrewingits retaining ring. Make sure the cylinder tip (bullet) is removed from thecylinder rod end. Screw the cylinder into the spring Loading Device.

Note: If the rod of the 3.81-cm (1.5-in) bore cylinder is not fullyretracted, do not try to screw it into the spring load device. Insteadretract the rod of the cylinder hydraulically, using the cylinderactuation circuit shown in Figure 2-10. Disconnect the circuit. Nowscrew the cylinder into the spring load device.

G 2. Connect the circuit shown in Figures 4-4 and 4-5. This circuit uses a spring-loaded accumulator to store hydraulic pressure. The directional valve servesas an on-off control to control the discharge of oil into the plastic flask. Thecheck valve inside the Flow Control Valve prevents the accumulator fromdischarging through the Power Unit pressure line port when the Power Unitis turned off.

Accumulators

4-8

Note: Make sure the plastic flask is empty before connecting it tothe directional valve.

Figure 4-4. Schematic diagram of circuit used to accumulate energy.

Accumulators

4-9

Figure 4-5. Connection diagram of circuit used to accumulate energy.



position.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve completely (turn knob fully counterclockwise).g. Make sure the Loading Device is securely mounted to the work

surface.

G 4. Close the Flow Control Valve completely (turn knob fully clockwise).


Accumulators

4-10

G 6. With the oil flow blocked at the directional valve, the oil from the pump isnow pushing on the accumulator piston, which compresses the accumulatorspring and charges the accumulator. Slowly turn the Relief Valve adjustmentknob clockwise until the system pressure at gauge A equals 1400 kPa(200 psi).

Note: When using industrial gas-loaded accumulators, it is veryimportant to open the relief valve completely before turning on thepower unit, and then to increase the relief valve pressure settinggradually to protect against accumulator overpressure.

G 7. Turn off the Power Unit. The pressure reading at gauge A should drop, whilethe pressure reading at gauge B should remain near 1400 kPa (200 psi)even though the Power Unit is turned off. Record the pressure readings atgauges A and B in Table 4-1.

Note: The pressure reading at gauge B may drift down after thePower Unit is turned off due to internal leakage in the directionalvalve. Take your reading immediately after the Power Unit isturned off.

SYSTEM PRESSURE GAUGE A GAUGE BDISCHARGE

VOLUME

1400 kPa (200 psi)

2800 kPa (400 psi)

Table 4-1. Storage capacity versus system pressure.

G 8. Firmly hold the plastic flask upright with one hand, and move the lever of thedirectional valve toward the valve body to discharge the oil into the flask.Keep the lever in this position until Pressure Gauge B reads approximately0 kPa (0 psi).

G 9. Unscrew and remove the plastic flask cap, then empty the collected oil intoa graduated beaker. Measure and record the approximate volume ofcollected oil in Table 4-1 under "DISCHARGE”. Then, replace the plasticflask cap.

Note: The trainer beaker is graduated in milliliters. Milliliter is ametric unit of measurement for volume (liquid capacity). 1 milliliterequals 0.001 liter.

If you are working with English units, multiply the measuredvolume in milliliters by 0.000264 to obtain the equivalent volumein US gallons.

Accumulators

4-11

G 10. Repeat steps 5 to 9 for the other system pressure given in Table 4-1.Record your data in Table 4-1. It is not necessary to empty the graduatedbeaker before measuring the new volume of collected oil. Instead keep trackof the accumulated oil volume at 1400 kPa (200 psi) and subtract to find theincrease in volume.

G 11. Open the Flow Control Valve completely (turn knob fully counterclockwise)to discharge the accumulator. Make sure gauge B reads approximately0 kPa (0 psi) before disconnecting any components from the accumulatorline.

CAUTION!

Before disconnecting any hose or component from theaccumulator line, the accumulator must be dischargedcompletely. Otherwise, oil under pressure will be trapped inthe accumulator circuit, making hose and componentdisconnection difficult or impossible.

G 12. Pour the collected oil into a container (capped plastic jugs, topped bottles,milk cartons, etc.) for transport to a disposal site. Oil recycling centers willnormally accept the oil, which can be refined and used again. Do not emptythe oil back into the pump reservoir, since it could have been contaminatedby dirt particles. Dirty oil can be very harmful to the hydraulic systembecause it causes flow paths to become clogged, valves to stick, and pumpto overheat.

G 13. According to Table 4-1, does the volume of oil stored in the accumulatorincrease when the Relief Valve setting is increased? Explain.

G 14. When the Power Unit was turned off in step 7 of the procedure, why did thepressure drop at gauge A and remain high at gauge B?

G 15. How did the spring-loaded accumulator absorb some of the shock producedwhen the pump flow was suddenly stopped at port P of the directional valveafter power up?

Accumulators

4-12

Using an accumulator as an emergency source of power


Figure 4-6. Accumulator used as an emergency source of power.

Note: If the rod of the cylinders are not fully retracted, do notconnect the circuit of Figure 4-6. Instead retract the rod of eachcylinder hydraulically, using the cylinder actuation circuit ofFigure 2-10. Once the cylinder rods are fully retracted, connectthe circuit of Figure 4-6.

G 17. Open the Relief Valve completely (turn knob fully counterclockwise), thenclose it 1 turn.


Accumulators

4-13

G 19. Turn on the Power Unit and watch the two Pressure Gauges. Why do thePressure Gauge readings increase a little after the Power Unit is turned on?

G 20. Turn off the Power Unit. Move the lever of the directional valve toward thevalve body to extend the rod of the 2.54-cm (1-in) bore cylinder. Why doesthe cylinder rod move even though the pump is not running?

G 21. Turn on the Power Unit. Move the lever of the directional valve outward fromthe valve body to retract the rod completely, then release the valve lever.

G 22. Adjust the Relief Valve adjustment knob clockwise until the system pressureat gauge A is 1400 kPa (200 psi).


G 24. Move the lever of the directional valve toward the valve body to extend therod. Does the rod extend over its full stroke?

G Yes G No

G 25. Turn on the Power Unit. Move the lever of the directional valve outward fromthe valve body to retract the rod completely, then release the valve lever.



G 28. Move the lever of the directional valve toward the valve body to extend therod. Does the rod extend over its full stroke?

G Yes G No

G 29. Does a slight pressure level remain at the piston of the accumulator cylinder(gauge B) when the cylinder rod is fully extended?

G Yes G No

Accumulators

4-14

G 30. Discharge the accumulator. To do so, open the Flow Control Valvecompletely (turn knob fully counterclockwise). The accumulator iscompletely discharged when gauge B reads approximately 0 kPa (0 psi).

G 31. Given that the volume of a cylinder is equal to the full piston area, Af,multiplied by the length of its stroke, L, calculate the theoretical volume ofoil, V, required to fully extend the 2.54-cm (1-in) bore x 1.59-cm(0.625-in) rod x 10.16-cm (4-in) stroke cylinder.

G 32. Based on the volume calculated in step 31, what is the pressure required atthe piston of the accumulator cylinder to store enough oil to allow completeextension of the 2.54-cm (1-in) bore cylinder when the Power Unit is turnedoff ?

HINT: The volume of oil stored in the accumulator is equal to the full pistonarea, Af, of the accumulator cylinder [2.54-cm (1.5-in) bore] multiplied by thedistance, D, the spring is compressed. Assume the spring rate, K, tobe 728 N/cm (416 lb/in).

G 33. Based on the calculations you made in steps 31 and 32, explain why the rodof the 2.54-cm (1-in) bore cylinder did not extend over its full stroke whenthe system pressure was set to 200 psi (13.8 bar).

G 34. In step 29 of the experiment, why did a slight pressure level remain at thepiston of the accumulator cylinder after extension of the 2.54-cm (1-in) borecylinder rod?

G 35. Keep your circuit connected since you will use it to perform the next part ofthe exercise.

Accumulators

4-15

Using an accumulator as an auxiliary source of power

G 36. Switch the two hoses connected to the ports of the Flow Control Valve witheach other in order to obtain the circuit shown in Figure 4-7. In this circuit,the Flow Control Valve is used to restrict the oil flow into the circuit tosimulate a small capacity pump.

Figure 4-7. Accumulator used as an auxiliary source of power.

G 37. Close the Flow Control Valve completely (turn knob fully clockwise), thenopen it ½ turn.

G 38. Open the Relief Valve completely (turn knob fully counterclockwise).

G 39. Remove the accumulator from the circuit. To do so, disconnect both endsof the hose connecting the cap end of the 3.81-cm (1.5-in) bore cylinder tothe 4-port manifold installed on port P of the directional valve.


Accumulators

4-16

G 41. Turn the Relief Valve adjustment knob clockwise until the system pressureat gauge A is 2100 kPa (300 psi). Move the lever of the directional valveoutward from the valve body to retract the 2.54-cm (1-in) bore cylinder rodcompletely.

G 42. Move the lever of the directional valve toward the valve body to extend the2.54-cm (1-in) bore cylinder rod, and measure the extension time. Recordthis value in the “NOT IN CIRCUIT” row of Table 4-2.

ACCUMULATOR EXTENSION TIME RETRACTION TIME

NOT IN CIRCUIT

IN CIRCUIT

Table 4-2. Cylinder cycling times with and without the accumulator.

G 43. Retract the cylinder rod and measure the retraction time. Record this valuein the “NOT IN CIRCUIT” row of Table 4-2.


G 45. Place the accumulator in the circuit by connecting the cap end of the3.81-cm (1.5-in) bore cylinder to the 4-port manifold on the directional valve.

G 46. Turn on the Power Unit. The accumulator spring should slowly compress asthe restricted oil flow fills the accumulator. During this time, the pressurereading at gauge B should slowly increase.

G 47. When the pressure reading at gauge B has stopped increasing, extend thecylinder rod and measure the extension time. Record this value in the “INCIRCUIT” row of Table 4-2.

G 48. Wait until the accumulator has fully recharged, then retract the cylinder rodand measure the retraction time. Record this value in Table 4-2.

G 49. Turn off the Power Unit. Open the Relief Valve completely (turn knob fullycounterclockwise). Notice that the accumulator will automatically dischargeafter the Power Unit is turned off because the check valve inside the FlowControl Valve now allows the oil to move freely to the reservoir through thePower Unit pressure line port.

Accumulators

4-17

G 50. According to Table 4-2, is the cylinder cycling speed higher when theaccumulator is connected to the circuit? Why?




CONCLUSION

In this exercise, you learned that pressurized oil can be stored by means of springcompression, gas pressure, and weights.

The volume you measured in the first part of the exercise was the amount of oilavailable to drive an actuator if the Power Unit were turned off. The higher thesystem pressure, the greater the volume of stored oil in the accumulator. The checkvalve you put in the circuit kept the accumulator from forcing oil back through thePower Unit pressure line port when the Power Unit is turned off.

The second test circuit showed you how an accumulator can provide emergencypower. You used pressurized oil stored in the accumulator to operate a cylinder withthe Power Unit turned off. You saw that the cylinder extends more or less, dependingon the amount of stored oil. Accumulators used in industrial circuits, however, areusually large enough to allow cycling a cylinder several times after the power unit isturned off.

You observed that the system pressure did not rise quickly when you turned on thePower Unit. The accumulator seemed to cushion sudden pressure changes. Thiscan be an advantage by eliminating shock to a system due to pressure surges.

You learned that another reason for using accumulators is to compensate for a pumpwhich is too small. You imitated a small capacity pump by controlling the flow into thecircuit. In this application, you could see how the energy stored in the accumulatorhelped your “low-volume pump” by acting as an auxiliary source of power. Theenergy is not free – the system pump still has to work to charge the accumulator.

Another reason for using accumulators is to apply pressure to hold a load in place.In this application, the accumulator compensates for the leakage through the cylinderand the directional valve and still keeps pressure to hold the load, thus taking a greatburden off the power unit. Constant start-stop operation is hard on the equipment.The pump operates long enough to charge the accumulator and stops for a whilerather than continuously turning on and off to hold the load.

Accumulators

4-18

Finally, you learned that an important component in an accumulator circuit is apositive means to discharge (bleed down) the accumulator after the power unit isturned off. In some cases, a directional control valve can be used. In other cases, aseparated bleed-down line incorporating a solenoid-operated 2-way valve or needlevalve is needed.

REVIEW QUESTIONS

1. In the circuit of Figure 4-4, why is a check valve (making part of the flow controlvalve) used between the accumulator and the power unit pressure line port?

2. Why should the accumulator and accumulator line be emptied beforedisconnecting any hose or component from the accumulator circuit?

3. How can an accumulator be used to reduce the cycle time of a cylinder?

4. What happens when a spring-loaded accumulator is used to suppress shocksin a circuit?

5. How does a gas-loaded accumulator work?

4-19

Exercise 4-2

Hydraulic Motor Circuits

EXERCISE OBJECTIVE

C To describe the design and operation of a hydraulic motor;C To calculate the torque and speed of a hydraulic motor;C To determine the effect a change in flow rate or pressure has on motor operation.

DISCUSSION

Hydraulic motors convert hydraulic energy into mechanical rotary energy. Thisenergy is used to turn a resisting object by means of a shaft. Hydraulic motors haveseveral advantages over electric motors. They can be uni- or bi-directional. They areinstantly reversible and they can absorb severe shock loading without damage to themotor. They are smaller than electric motors and respond faster. Hydraulic motorscan be overloaded and stalled without damage because excess pressure is divertedover the relief valve.

As Figure 4-8 shows, a hydraulic motor consists of the following elements:

C An inlet port supplying pressurized oil.

C A housing containing a rotating mechanism, as drive gears, exposed to systempressure. The rotating mechanism is connected to the motor shaft. Aspressurized oil enters the motor inlet, pressure on the rotating mechanism causesthe motor shaft to rotate.

C An outlet port through which oil exits the motor.


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Figure 4-8. Hydraulic motor.

Types of hydraulic motors

There are three basic types of hydraulic motors, named after the type of rotatingmechanism used inside the motor. These are gear, vane, and piston motors, asFigure 4-9 shows.

C The gear motor is the type of motor supplied with your Hydraulics Trainer. Itconsists of two intermeshing gears enclosed in the motor housing. One gear isattached to the motor shaft. When pressurized oil is pumped into the inlet port,the pressure acts upon the surface area of the gear teeth and the gears areforced to rotate. Gear motors are usually less expensive than the other types ofmotors. They are commonly used in agriculture and mining applications.


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Figure 4-9. Types of hydraulic motors.

C The vane motor consists of a slotted rotor connected to a shaft. The slots containvanes which are free to slide in and out. When pressurized oil is pumped into theinlet port, the pressure acts on the vanes and the rotor is forced to rotate. Vanemotors are often used on industrial equipments.

C The piston motor consists of several pistons fitted into a rotating barrel. The barrelis connected to a shaft. When pressurized oil is pumped into the inlet port, thepressure acts on the pistons and the cylinder barrel is forced to rotate. Pistonmotors are used in applications requiring precise rotation speeds.

Motor displacement

Displacement of a hydraulic motor is the volume of oil required for the motor shaftto turn one complete revolution. It is expressed in cubic centimeters per revolution(cm3/r) in S.I. units, or in cubic inches per revolution (in3/r) in English units.


4-22

Motor speed and volumetric efficiency

The speed at which a hydraulic motor turns is determined by how fast it is filled withoil. This speed, then, is directly proportional to the oil flow rate through the motor,and inversely proportional to the motor displacement. The formula for calculating thetheoretical speed of a motor is:

S.I. units:

English units:

Due to internal leakage, the actual motor speed will be less than the theoreticalspeed given by the above formula. The actual speed depends on the volumetricefficiency of the motor. Volumetric efficiency is the ratio of the actual motor speedto the theoretical speed, expressed as a percentage:

Since motor leakage is lower at higher speeds, volumetric efficiency tends toincrease as the speed of the motor is increased. In industry, calculation of thevolumetric efficiency at several different speeds is useful to determine if the motorshould be replaced.

Motor torque

Torque is the turning effort or rotary force generated at the motor shaft. Forexample, the force applied to the end of a wrench to tighten a bolt is called torque.In hydraulic systems, torque is often expressed in Newtons-meters (N@m) in S.I.units, or in pounds-inches (lb@in) in English units. A rotary force of 20 N (4.5 lb)applied to a shaft 5 cm (2 in) from the center of the shaft would be expressed as atorque of 1 N@m (9 lb@in).

The amount of torque generated at the shaft of a hydraulic motor is directlyproportional to the motor displacement and system pressure at the motor inlet. Theformula for calculating the theoretical torque output of a motor is:

S.I. units:

English units:


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The resistance of the load connected to the motor shaft determines the amount ofsystem pressure developed at the motor inlet, and, therefore, the amount of torquegenerated at the motor shaft. No torque will be generated if there is no load on theshaft.

Motor output power

The amount of power generated by a hydraulic motor is equal to the torquedeveloped at the motor shaft multiplied by the speed of the shaft. In equation form:

S.I. units:

English units:

The actual amount of power generated by a hydraulic motor will be less than thetheoretical value given by the above formula because of the losses, mechanicalfriction and flow friction suffered inside the motor.

Hydraulic transmissions

A hydraulic motor can be teamed up with a variable-displacement pump to form ahydraulic transmission. Hydraulic transmissions are circuits that match the torqueand speed of a hydraulic pump to the torque and speed requirements of a hydraulicmotor driving a load.

Hydraulic transmissions are extensively used in mobile, marine, and aircraftapplications. They can be either “closed loop” or “open loop”, as Figure 4-10shows.


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Figure 4-10. Hydraulic closed- and open-loop transmission circuits.

In a closed-loop transmission, the pump outlet is connected to the motor inlet, andthe motor outlet is connected to the pump inlet, as Figure 4-10 (a) shows. A smallpump, called replenishing pump, compensates for leakage by keeping the loop full


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of oil. The motor turns in one direction only. Instead of being returned to thereservoir, the oil discharged from the motor is re-circulated back to the pump inlet atlow pressure. The advantage of this type of transmission is that it is highlycontrollable and compact in size. Closed-loop transmissions are also calledhydrostatic transmissions.

In an open-loop transmission, the pump outlet is connected to the motor inlet, andthe motor outlet is connected to the reservoir, as Figure 4-10 (b) shows. The motorcan turn in either direction. The direction of rotation is controlled through the use ofa directional control valve. This type of transmission is less expensive tomanufacture than closed loop transmissions, but it is less efficient and less easy tocontrol.

Hydraulic cylinders compared to hydraulic motors

Hydraulic cylinders are linear actuators, whereas hydraulic motors are rotaryactuators. As with hydraulic cylinders, however, the speed of a hydraulic motor is afunction of flow rate, while its output force, or torque, is a function of pressure.Therefore, increasing the flow rate through a motor increases the rotation speedwithout affecting the torque output capability. Conversely, increasing the systempressure available to a motor increases the torque output capability without affectingthe rotation speed.

Conversion factors

Table 4-3 shows the conversion factors used to convert measurements ofdisplacement and torque from S.I. units to English units, and vice versa.

Displacement

Cubic centimetersper revolution (cm3/r)

x 0.061 = Cubic inches perrevolution (in3/r)

x 16.387 = Cubic centimeters perrevolution (cm3/r)

Torque

Newton-meters (N@m) x 8.85 = Pounds-inches(lb@in; lbf@in)

x 0.113 = Newton-meters (N@m)


REFERENCE MATERIAL

For additional information on hydraulic motors, refer to the chapter entitled HydraulicMotors in the Parker-Hannifin’s manual Industrial Hydraulic Technology.


4-26

Procedure summary

In the first part of the exercise, you will determine the effect a change in flow rate hason the speed of a hydraulic motor. To do so, you will measure the speed of the motorat several different flow rates, using a tachometer. You will then compare theobtained speeds with the theoretical values, demonstrating that the motor volumetricefficiency increases slightly due to a decrease in internal leakage as the speedincreases.

In the second part of the exercise, you will determine the effect a change in themotor load has on the torque output of the motor. To do so, you will vary the load onthe motor, which will vary the system pressure at the motor inlet. For each loadcondition, you will evaluate the torque generated at the motor shaft by stopping themotor flywheel with your hand. The resistance of the motor to be stopped willindicate whether the generated torque is high or low.

EQUIPMENT REQUIRED


PROCEDURE

Speed and flow relationship in a hydraulic motor

G 1. Connect the circuit shown in Figures 4-11 and 4-12. In this part of theexercise, you will determine the effect a change in flow rate has on thespeed of a hydraulic motor. The Flow Control Valve will be used to vary theflow rate through the motor.

IMPORTANT!

The trainer hydraulic motor comes with alightweight flywheel and a heavyweight flywheel.For this exercise, install the heavyweight flywheelon the shaft of the motor, if not already installed.Make absolutely certain the flywheel is firmlysecured to the motor shaft. To do so, verify that theset screw in the flywheel hub is completely screwedinto its hole. The head of the set screw must beflush with the flywheel hub.


4-27

Figure 4-11. Schematic diagram of the circuit used to test motor operation.




G 3. Make sure the hydraulic motor is securely fixed to the work surface. Clearaway hoses, tools, and other objects from the motor.





4-28

Figure 4-12. Connection diagram of the circuit used to test motor operation.

G 7. Slowly move the lever of the directional valve toward the valve body to routeoil out of port A, which will make the motor turn. Looking at the motor shaftfrom the front of the motor, in which direction does the motor turn?

G 8. What happens to the motor operation when the directional valve is releasedin its center position? Why?


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G 9. Move the lever of the directional valve toward the valve body and hold it sothat the motor runs. Slowly turn the Flow Control Valve adjustment knobclockwise until the valve is completely closed. As you do this, observe themotor shaft. What happens to the motor operation?

G 10. While keeping the lever of the directional valve in the inward position, slowlyincrease the Flow Control Valve opening by turning its adjustment knobcounterclockwise. According to the pitch of the sound made by the motor,what happens to the speed of the motor when the oil flow rate through themotor increases?

G 11. As the motor is turning, adjust the Flow Control Valve so that the Flowmeterreads 2.0 l/min [0.53 gal(US)/min].

G 12. As the motor is turning, place a tachometer on the motor shaft and measurethe motor speed. Record this speed in Table 4-4 under "ACTUAL".

FLOW RATE ACTUAL SPEEDTHEORETICAL

SPEEDVOLUMETRICEFFICIENCY

2.0 l/min[0.53 gal(US)/min]



Table 4-4. Motor speed and efficiency versus flow rate.


G 14. Turn off the Power Unit. Open the Relief Valve completely (turn knob fullycounterclockwise). Do not disconnect your circuit since you will use it toperform the next part of the exercise.

G 15. According to the speeds registered in the "ACTUAL" column of Table 4-4,how does the flow rate affect the speed of a motor?


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G 16. Given that the theoretical displacement of your motor is 1.77 cm3/r(0.108 in3/r), calculate the theoretical motor speed for each of the flow rateslisted in Table 4-4. Record your results in the “THEORETICAL” column ofTable 4-4.

G 17. Based on the actual and theoretical speeds registered in Table 4-4,calculate the volumetric efficiency of the motor for each of the flow rateslisted in Table 4-4. Record your results in Table 4-4 under “VOLUMETRICEFFICIENCY”.

G 18. Does volumetric efficiency increase as the speed of the motor increases?Why?

G 19. What flow rate would be required to make the motor turn at 2000 r/min?

Pressure and torque relationship in a hydraulic motor

Note: Wear protective gloves when conducting this part of theexercise.

G 20. Make sure your circuit is connected as shown in Figure 4-11. In this part ofthe exercise, you will demonstrate that the pressure at the motor inlet willincrease if the mechanical load applied to the motor increases.

G 21. Examine the motor flywheel to make sure it is smooth and free of burrs orsharp edges.

G 22. Make sure the Relief Valve is open completely (turn knob fullycounterclockwise).



G 25. Increase the Relief Valve pressure setting until the circuit pressure atgauge A is 2800 kPa (400 psi).


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G 26. Move the lever of the directional valve toward the valve body to make themotor turn. Since there is no load on the motor, no torque is beinggenerated at the motor shaft. While keeping the valve lever in the inwardposition, note and record the pressure readings at gauges A and B in the“NO LOAD” row of Table 4-5.

LOAD CONDITIONMOTOR INLET

PRESSURE (GAUGE A)MOTOR OUTLET

PRESSURE (GAUGE B)

NO LOAD

STALLED

Table 4-5. Motor pressure and torque versus load condition.

G 27. Put on protective gloves. Move the lever of the directional valve toward thevalve body to make the motor turn. As the motor is turning, hold the motorflywheel with your hand to increase the mechanical load on the motor. Whiledoing this, observe the motor inlet pressure at gauge A. What happens tothis pressure as the mechanical load on the motor is increased?

G 28. Move the lever of the directional valve toward the valve body to make themotor turn. Hold the motor flywheel with your hand to keep the motor fromturning. Note and record in Table 4-5 the motor inlet and outlet pressuresat gauges A and B when the motor is stalled.

G 29. Where does the oil from the pump go when the motor is stalled?


G 31. Calculate the amount of theoretical torque being generated at the motorshaft when the motor stalls, based on the motor inlet pressure registered inthe “STALLED” row of Table 4-5.


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G 32. What is the mechanical power output of the motor in question 31?




CONCLUSION

In the first part of the exercise, you operated a hydraulic motor at several differentspeeds and demonstrated that motor speed is directly proportional to the flow rate.You saw how speed can be varied greatly in a hydraulic motor which leads to oneof the many advantages of using a hydraulic motor as a mechanical drive. You havealso noticed that when running this type of motor at a low speed, there is muchinternal leakage, and therefore, the speed is not as great as it should be. As the flowrate increases, the speed increases and the volumetric efficiency gets better.

In the second part of the exercise, you operated the hydraulic motor under differentload conditions and demonstrated that motor torque is directly proportional to systempressure at the motor inlet. As the load on the motor is increased, the systempressure at the motor inlet increases, and, therefore, the output torque increases.

As with hydraulic cylinders, the speed of a hydraulic motor is a function of flow rate,while its output force, or torque, is a function of pressure. Therefore, increasing thesystem flow rate increases the motor speed without affecting the motor torque.Increasing the system pressure increases the motor torque without affecting themotor speed.

REVIEW QUESTIONS

1. What is the function of a hydraulic motor in a hydraulic circuit?


4-33

2. Name the three most common types of hydraulic motors.

3. What is a hydraulic transmission?

4. What happens to the speed and torque output capability of a hydraulic motorwhen the displacement of the motor is doubled?

5. State two methods of increasing the speed of a hydraulic motor.

6. A hydraulic motor is required to drive a load at a speed of 2000 r/min with atorque of 4.52 NAm (40 lbAin). Calculate the minimum motor power outputrequired to drive the load.

4-34

4-35

Exercise 4-3

Pressure Reducing Valves

EXERCISE OBJECTIVE

C To describe the design and operation of a pressure reducing valve;C To test the operation of a clamp and bend circuit using a pressure reducing valve.

DISCUSSION

Pressure limitation in a branch circuit

The pressure control valve used in many exercises so far is the relief valve. Thepressure setting of this valve limits the maximum circuit pressure, and when youchange the setting, you affect pressure in all branches of the circuit.

In some hydraulic circuits, however, it is necessary to operate two cylinders inseparate branches at different pressures. A typical example is a clamp and bendcircuit requiring that the clamp cylinder apply a lesser force than the bend cylinderto prevent distortion or damage to the workpieces. A pressure reducing valve mightbe used to limit the pressure in the branch of the clamp cylinder.

Pressure reducing valves

The pressure reducing valve is another member of the pressure control valve family.This type of valve limits, or regulates, the pressure in a circuit branch to a level lessthan the system (relief valve) pressure by closing partially to restrict the oil flow intothe branch. It compensates for pressure changes in the system by adjusting thepressure drop across its inlet and outlet ports to maintain the pressure in the branchat the desired level.

Figure 4-13 shows the Pressure Reducing Valve supplied with your HydraulicsTrainer. The valve body has three ports: an inlet pressure (P) port, a regulatedpressure (R) port, and a tank (T) port. Unlike the Relief Valve, the PressureReducing Valve is normally open and it senses the pressure downstream, asindicated by the valve symbol in Figure 4-13.


4-36

Figure 4-13. The trainer Pressure Reducing Valve.


4-37

An internal spool controls the oil flow through the valve by acting on a large spring.The pressure level where the spool begins to close to restrict the oil flow through thevalve is called cracking pressure. The pressure level where the spool is completelyclosed and no oil flows through the valve is called operating pressure. Theoperating pressure can be adjusted by using the adjustment knob on the valve body.Turning the knob counterclockwise decreases the valve operating pressure, whichreduces the level of allowable downstream pressure. Once the operating pressureis adjusted, tightening a locking nut on the adjustment screw will prevent vibrationsand shocks from modifying the adjustment.

As long as the pressure demanded by the load connected downstream from thevalve is lower than the valve cracking pressure, the spool will remain wide open(lowest position) and oil will flow freely through the valve. If the pressure demandbecomes higher than the valve cracking pressure, the spool will move up to someintermediate position to stop the rise in pressure. The higher the pressure demand,the closer the spool will move toward its completely closed position.

If the oil flow becomes blocked downstream from the valve, as for cylinder stall orcylinder at full stroke, the valve will close completely and allow the excess oil at theregulated pressure port to dump back to the reservoir through its tank port, therebymaintaining the downstream pressure at its operating pressure setting.

If the pressure downstream drops off, the valve will reopen and allow the pressureto build up to the operating pressure again.

Pressure reducing valves will not allow the oil to flow very well in the reversedirection, because they will try to close, resulting in a restricted flow rate. Whenreverse flow is required, as in a circuit containing an extending and retractingcylinder, a pressure reducing valve with a built-in check valve may be used, or anexternal check valve may be connected across the valve inlet pressure (P) andregulated pressure (R) ports. The Pressure Reducing Valve supplied with yourHydraulics Trainer has an internal built-in check-valve.

In order for a pressure reducing valve to operate properly, the tank port mustimperatively be connected to the reservoir. If this connection is missing or blocked,the spool will not be able to move to a throttling position to adjust the pressure dropacross its inlet and outlet ports, which will cause the pressure downstream from thevalve to climb to the maximum system (relief valve) pressure. The valve will remainfully open, preventing any control of the downstream pressure.

Clamp and work application

The most common application of pressure reducing valves is for “clamp and work”circuits. The clamp operation, which consists in holding a workpiece in a fixture, isperformed by a small bore cylinder requiring less than full system pressure to limitthe force applied to the workpiece. The work operation, which consists in doing aparticular task on the clamped workpiece, such as bending, pressing, drilling, cutting,or grinding, is performed by a cylinder of larger bore requiring the full systempressure.


4-38

Figure 4-14. Clamp and bend circuit.

Figure 4-14 shows a clamp and bend circuit used to form metal workpieces. Thecircuit is made up of two cylinder branches connected in parallel. The clampingbranch consists of a small bore cylinder requiring a maximum pressure of 1200 kPa(175 psi) to clamp the workpieces with limited force. A pressure reducing valveconnected upstream limits the pressure to this cylinder to 1200 kPa (175 psi). Thebending branch consists of a large bore cylinder requiring the maximum systempressure [3500 kPa (500 psi)] to bend the workpieces with full force. A flow controlvalve connected downstream limits the speed of this cylinder. This valve is set sothat it creates an additional resistance of 1400 kPa (200 psi) downstream from thebend cylinder. Limiting the cylinder speed minimizes the impact produced when thecylinder rod hits the workpiece and ensures that the workpiece is clamped withsufficient force at the moment when the bend cylinder starts to bend the workpiece.


4-39

When the directional valve is shifted as shown in Figure 4-15, the pumped oil isdirected to the cap side of both cylinders at the same time. The clamp cylinderbegins to extend first because it requires a very low pressure to counteract theresistance of the oil flowing back to the reservoir, while the bend cylinder requiresa 1400-kPa (200-psi) pressure to counteract the resistance of the flow control valve.Since the clamp cylinder extends under no load, it requires a pressure [550 kPa(80 psi)] lower than the 1200-kPa (175-psi) pressure setting of the pressure reducingvalve, so the valve stays fully open.

Figure 4-15. Clamp cylinder extends.

When the clamp cylinder stalls against the workpiece (see Figure 4-16), the oil flowbecomes blocked, causing the system pressure to rise quickly. When the pressuredownstream from the pressure reducing valve reaches 1200 kPa (175 psi), thepressure reducing valve closes completely to limit the clamping force. The systempressure then rises to 1400 kPa (200 psi), causing the bend cylinder to startextending. When this cylinder contacts the workpiece, the system pressure rises tothe 3500-kPa (500-psi) setting of the relief valve to bend the workpiece with fullforce.


4-40

Figure 4-16. Workpiece bent with full force while clamped with limited force.

When the workpiece is bent, the directional valve is shifted to retract the cylinders(see Figure 4-17). The oil from the cap end of the clamp cylinder returns to thereservoir through the bypass check valve, while the oil from the cap end of the bendcylinder returns freely to the reservoir.


4-41

Figure 4-17. Bend and clamp cylinder retract.

REFERENCE MATERIAL

For additional information on pressure reducing valves, refer to the chapter entitledPressure Control Valves in the Parker-Hannifin’s manual Industrial HydraulicTechnology.

Procedure summary

In the first part of the exercise, you will test the operation of a pressure reducingvalve. A flow control valve will be used to vary the load (pressure demand)downstream from the pressure reducing valve.


4-42

In the second part of the exercise, you will connect and test the operation of theclamp and bend circuit described in the DISCUSSION section of the exercise.

EQUIPMENT REQUIRED


PROCEDURE

Operation of a Pressure Reducing Valve

G 1. Connect the circuit shown in Figures 4-18 and 4-19. In this circuit, the FlowControl Valve will be used to simulate a varying load downstream from thePressure Reducing Valve.

Figure 4-18. Schematic diagram of the circuit used to test the operation of a pressure reducingvalve.


4-43

Figure 4-19. Connection diagram of the circuit used to test the operation of a pressure reducingvalve.




G 3. Close the Flow Control Valve completely by turning its adjustment knob fullyclockwise. This will block the oil flow downstream from the PressureReducing Valve.

G 4. Open the Pressure Reducing Valve completely. To do so, first loosen thelocking nut on the adjustment screw by turning this nut fullycounterclockwise. Then, turn the valve adjustment knob fully clockwise. Thevalve is now wide open and the operating pressure is set at the highestpossible pressure.


4-44



G 7. According to the pressure reading at gauge B, is the pressure downstreamfrom the Pressure Reducing Valve approximately equal to the circuit (ReliefValve) pressure at gauge A? Why?

G 8. Decrease the reducing valve operating pressure by turning its adjustmentknob counterclockwise and observe the pressure reading at gauge B. Whathappens to the pressure downstream from the reducing valve (gauge B) asthe valve operating pressure is decreased?

Note: The reducing valve adjustment knob can be turned overapproximately six turns. You may have to turn the valve knob fouror five turns counterclockwise before the downstream pressure atgauge B starts to change.

G 9. Close the reducing valve completely by turning its adjustment knob fullycounterclockwise.

G 10. Observe the pressure reading at gauge B. This is the minimum pressurelevel allowable downstream from the reducing valve. Record this pressurebelow.

Minimum downstream pressure: kPa or psi

G 11. Set the operating pressure of the reducing valve to 1400 kPa (200 psi). Todo so, turn the valve adjustment knob clockwise until the pressure leveldownstream from the valve (gauge B) is 1400 kPa (200 psi).


4-45

G 12. Increase the circuit pressure by slowly turning the Relief Valve adjustmentknob clockwise until gauge A reads 3500 kPa (500 psi). While doing this,observe the pressure level downstream from the reducing valve (gauge B).Does increasing the circuit pressure increases the pressure downstreamfrom the valve? Explain.

G 13. Decrease the load (pressure demand) downstream from the reducing valveby opening the Flow Control Valve 1 turn counterclockwise. Oil is nowflowing downstream from the reducing valve, as indicated by the Flowmeter.Is the pressure level downstream from the reducing valve (gauge B) still1400 kPa (200 psi)?

G Yes G No

G 14. Further decrease the load (pressure demand) downstream from thereducing valve by turning the Flow Control Valve adjustment knob fullycounterclockwise. While doing this, observe the Flowmeter reading and thepressure reading at gauge B. What happens to the oil flow rate and to thepressure level downstream from the reducing valve (gauge B) as thepressure demand is decreased?

G 15. Increase the load (pressure demand) downstream from the reducing valve.To do so, reduce the Flow Control Valve opening by slowly turning itsadjustment knob clockwise. While doing this, observe the Flowmeterreading and the pressure level at gauge B. What happens to the oil flow rateand to the pressure level downstream from the reducing valve (gauge B) asthe pressure demand is increased? Explain why.

G 16. Close the Flow Control Valve completely (turn knob fully clockwise) to blockthe oil flow downstream from the reducing valve. The circuit pressure nowat gauge A is the Relief Valve pressure setting [3500 kPa (500 psi)], whilethe downstream pressure at gauge B is the reducing valve operatingpressure [1400 kPa (200 psi)].


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G 17. Turn off the power. Do not modify the setting of the Relief Valve andreducing valve.

G 18. Now test the effect on the reducing valve operation from removing its tankconnection. Disconnect both ends of the hose connecting the PressureReducing Valve tank (T) port to the return manifold. Then, turn on the PowerUnit.

G 19. According to the pressure reading at gauge B, is the pressure leveldownstream from the reducing valve still limited to the 1400-kPa (200-psi)setting of this valve?

G Yes G No

G 20. Try to decrease the level of the pressure downstream from the reducingvalve (gauge B) by closing this valve completely (turn knob fullycounterclockwise). Did the downstream pressure decrease? Why?

Note: The reducing valve adjustment knob may be hard to turn inthis situation.


Clamp and bend application

G 22. Connect the clamp and bend circuit shown in Figure 4-20. In this circuit, the2.54-cm (1-in) bore cylinder will simulate a clamp cylinder, while the 3.81-cm(1.5-in) bore cylinder will simulate a bend cylinder. The circuit (relief valve)pressure will be set to 3500 kPa (500 psi).

The reducing valve will limit the pressure to the clamp cylinder to 1200 kPa(175 psi) when this cylinder becomes fully extended.

The Flow Control Valve will be adjusted so that the circuit pressure atgauge A must build up to 1400 kPa (200 psi) before the bend cylinder canextend.


4-47

Figure 4-20. Circuit for testing the operation of a clamp and bend circuit.

G 23. Open the reducing valve completely by turning its adjustment knob fullyclockwise.

G 24. Open the Flow Control Valve completely by turning its adjustment knob fullycounterclockwise.



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Preliminary settings

G 26. Move the lever of the directional valve toward the valve body to extend therods of the clamp and bend cylinders. With the piston rods extended, all theoil from the pump is now being forced through the relief valve and gauge Aindicates the minimum pressure setting of this valve.

While keeping the directional valve lever in the inward position, turn therelief valve adjustment knob clockwise until the circuit pressure at gauge Ais 3500 kPa (500 psi).

Then, limit the pressure at the piston of the clamp cylinder to 1200 kPa(175 psi). While keeping directional valve lever in the inward position, turnthe reducing valve adjustment knob counterclockwise until the pressure atgauge B is 1200 kPa (175 psi).

G 27. Move the lever of the directional valve outward from the valve body toretract both cylinder rods completely.

G 28. Turn the Flow Control Valve adjustment knob 3 turns clockwise.

G 29. Extend the cylinder rods and readjust the Flow Control Valve so thatgauge A reads 1400 kPa (200 psi) as the bend cylinder [3.81-cm (1.5-in)bore] extends. Accurate adjustment may require that the cylinders beextended and retracted several times.

G 30. Move the lever of the directional valve outward from the valve body toretract both cylinder rods completely.

Testing circuit operation

G 31. Now test the operation of the clamp and bend circuit. Move the lever of thedirectional valve toward the valve body to extend the cylinders. Whichcylinder begins to extend first? Why?

G 32. Retract both cylinders completely.


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G 33. Extend the cylinders while observing the pressure readings at gauges A andB. Does the gauge B pressure climb to the 3500-kPa (500-psi) pressuresetting of the Relief Valve when the clamp cylinder [2.54-cm (1-in) bore]becomes fully extended? Why?

G 34. Is the force available from the clamp cylinder limited by the reducing valve?

G Yes G No

G 35. Retract both cylinders completely.

G 36. Extend the cylinders while observing the pressure reading at gauge A. Whatis the gauge A pressure when the bend cylinder [3.81-cm (1.5 in) bore] isextending? When this cylinder becomes fully extended?

G 37. Is the force available from the bend cylinder limited by the reducing valve?

G Yes G No

G 38. Make sure both cylinders are completely retracted, then turn off the PowerUnit. Open the Relief Valve completely (turn knob fully counterclockwise).





4-50

CONCLUSION

In the first part of the exercise, you tested the operation of a reducing valve. You sawthat this type of valve compensates for pressure changes in the system by adjustingthe pressure drop across its inlet and outlet ports to maintain the pressuredownstream at the desired level. As you changed the relief valve and flow controlsettings, you saw that the reducing valve was able to hold a fixed pressure as longas the pressure demand downstream was higher than its operating pressure.

In the second part of the exercise, you tested a clamp and bend circuit using areducing valve. The pressure at the bend cylinder was able to rise up to the system(relief valve) pressure, while the pressure at the clamp cylinder was limited to a levelless than the system pressure. This showed you that the pressure in some part ofthe circuit can be controlled so it is not affected by the load on the cylinder. Withoutsuch a control, excessive clamping force would cause damage.

REVIEW QUESTIONS

1. Why are pressure reducing valves used?

2. Is the pressure reducing valve normally open or normally closed?

3. How is the pressure reducing valve different from the relief valve?


4-51

4. Why must the pressure reducing valve be externally drained (connected totank)?

4-52

4-53

Exercise 4-4

Remotely Controlled Pressure Relief Valves

EXERCISE OBJECTIVE

C To explain how a pressure relief valve can be controlled remotely;C To control the tonnage of a press cylinder remotely.

DISCUSSION

Remote control of a relief valve

Remote control of a relief valve consists in overriding the pressure setting of a reliefvalve from a remote control location. Remote control is particularly convenient whenseveral levels of system pressure are required and the relief valve is located at adistant point from the operator control panel.

Remote control is achieved by connecting the vent (V) port of the relief valve to theinlet of another pressure control valve, usually a second relief valve of smaller size.As an example, Figure 4-21 shows remote control of a main relief valve with asecondary relief valve. The secondary relief valve is installed in the vent line of themain relief valve. The highest desired system pressure is adjusted on the main reliefvalve itself, lower pressures being obtained with the secondary relief valve. Circuitoperation is as follows:

C When the pressure setting of the secondary relief valve is higher than thepressure setting of the main relief valve, the secondary relief valve stays fullyclosed and all the pumped oil discharges to the reservoir through the main reliefvalve at whatever pressure this valve is set for.

C When the pressure setting of the secondary relief valve is decreased below thepressure setting of the main relief valve, a small amount of oil passes out the ventport of the main relief valve and discharges to the reservoir through the secondaryrelief valve. This prevents the pressure in the spring chamber of the main reliefvalve from building up, causing the main poppet inside this valve to open. As aresult, most of the pumped oil discharges to the reservoir through the main reliefvalve at whatever pressure the secondary relief valve is set for.

As you can see, decreasing the pressure setting of the secondary relief valve has thesame effect on the circuit pressure as decreasing the pressure setting on the knobof the main relief valve. However, it is important to understand that the secondaryrelief valve can reduce the pressure setting of the main relief valve but can neverraise it.


4-54

Figure 4-21. Overriding the pressure level set on the knob of a main relief valve from a remotelocation.

The advantage to using a remotely controlled main relief valve comes from the factthat only a small amount of oil is required for opening the secondary relief valve andventing the main relief valve, the remainder of the oil being discharged to thereservoir through the main relief valve. This allows large-size relief valves to beadjusted remotely using very small relief valves and small-diameter tubing lines.

Applications

Remotely controlled relief valves can be used in applications where the process mayrequire one or more cylinders to operate at pressures lower than the main systempressure. An example is a hydraulic press. Since different materials will require anadjustment of the maximum force, or tonnage, available from the hydraulic press, asmall relief valve can be mounted on the operator panel and connected to the ventport of the main relief valve through a small-diameter tubing line. The maximumrequired tonnage is to be set on the main relief valve, lower tonnages being obtainedwith the secondary relief valve.

Procedure summary

In the first part of the exercise, you will demonstrate the operation of a remotelycontrolled relief valve.

In the second part of the exercise, you will control the tonnage of a press cylinderremotely.

EQUIPMENT REQUIRED



4-55

PROCEDURE

Remote control of a relief valve

G 1. Connect the circuit shown in Figures 4-22 and 4-23. Since your HydraulicsTrainer comes with only one Relief Valve, you will use a Flow Control Valveas the secondary relief valve to vary the pressure level at the vent port ofthe Relief Valve.

Figure 4-22. Schematic diagram of the circuit used to test the operation of a remotely controlledrelief valve.

G 2. Close the Flow Control Valve completely by turning its adjustment knob fullyclockwise. This will block the oil flow downstream of the relief valve ventport, simulating a secondary relief valve in the fully-closed condition.


a. Make sure the hoses are firmly connected.b. Check the level of the oil in the reservoir. Add oil if required.c. Put on safety glasses.d. Make sure the power switch on the Power Unit is set to the OFF pos-

ition.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve completely (turn knob fully counterclockwise).


4-56

Figure 4-23. Connection diagram of the circuit used to test the operation of a remotely controlledrelief valve.



G 6. Very slowly open the Flow Control Valve (turn knob counterclockwise)½ turn to decrease the pressure level at the vent port of the Relief Valve(gauge B). While doing this, observe what happens to the circuit pressureat gauge A. Record your observations below.


4-57

G 7. Slowly close the Flow Control Valve (turn knob fully clockwise) to increasethe pressure level at the vent port of the Relief Valve (gauge B). While doingthis, observe what happens to the circuit pressure at gauge A. Record yourobservations below.

G 8. According to the pressure reading at gauge A, can the Flow Control Valvebe used to raise the pressure setting (2800 kPa/400 psi) of the Relief Valve?

G Yes G No

G 9. Open the Flow Control Valve completely (turn knob fully counterclockwise).According to the pressure reading at gauge A, can the venting action reducethe circuit pressure to zero? Why?

G 10. According to the Flowmeter reading, does most of the pumped oil dischargeto the reservoir through the Relief Valve or through the Flow Control Valve?Explain.


Remote control of the tonnage of a hydraulic press

G 12. Get the 3.81-cm (1.5-in) bore cylinder and Loading Device from yourstorage location. Remove the cylinder from its adapter by unscrewing itsretaining ring. Make sure the cylinder tip (bullet) is removed from thecylinder rod end. Screw the cylinder into the Loading Device.

Note: If the rod of the 3.81-cm (1.5-in) bore cylinder is not fullyretracted, do not try to screw it into the Loading Device. Insteadretract the cylinder rod hydraulically, using the cylinder actuationcircuit shown in Figure 2-10. Once the cylinder rod is retracted,screw the cylinder into the Loading Device.

G 13. Connect the circuit shown in Figure 4-24. In this circuit, the 3.81-cm (1.5-in)bore cylinder simulates a press cylinder. The maximum force (tonnage)available from the press cylinder is remotely controlled through the use ofa Flow Control Valve connected in the vent line of a Relief Valve.


4-58

Figure 4-24. Remote control of the maximum force (tonnage) available from a press cylinder.

G 14. Clip the NEWTON/LBF-graduated ruler to the Loading Device, aligning the“0” mark with the colored line on the load piston.



G 17. The circuit pressure should now be controlled by the Relief Valve becausethe Flow Control Valve is fully closed. Turn the Relief Valve adjustment knobclockwise until the circuit pressure at gauge A is 2800 kPa (400 psi).

G 18. Move the lever of the directional valve toward the valve body to compressthe spring. Note and record the tonnage of the press cylinder as indicatedon the ruler of the Loading Device. This is the maximum rated tonnage asset on the Relief Valve.

Maximum rated tonnage: N or lb



4-59

G 20. Open the Flow Control Valve (turn knob counterclockwise) until the circuitpressure at gauge A is 1400 kPa (200 psi).

G 21. Move the lever of the directional valve toward the valve body to compressthe spring. Note and record the new tonnage of the press cylinder asindicated on the ruler of the Loading Device.

New tonnage: N or lb

G 22. From the observations you made, can the tonnage of a press cylinder beremotely reduced using a remotely controlled Relief Valve?

G Yes G No





G 27. Clean up any hydraulic oil from the floor and the trainer. Properly disposeof any towels and rags used to clean up oil.

CONCLUSION

In the first part of the exercise, you demonstrated the operation of a remotelycontrolled Relief Valve. As you decreased the pressure downstream of the reliefvalve vent port, you noticed that the circuit pressure decreased as if the pressuresetting of the Relief Valve were decreased on the valve adjustment knob. As youincreased the pressure downstream of the relief valve vent port, you noticed that thecircuit pressure could not be raised beyond the 2800-kPa (400-psi) pressure settingof the Relief Valve.

In the second part of the exercise, you used a remotely controlled Relief Valve toadjust the maximum force, or tonnage, available from a press cylinder. With the FlowControl Valve fully closed, the force you measured on the Loading Device was themaximum rated tonnage as set on the remotely controlled Relief Valve. With theFlow Control Valve partially open, the force you measured was the reduced tonnagelevel as set on the Flow Control Valve.


4-60

Remote control is a fact of life today. Almost anywhere you look, complicatedhydraulic systems can be controlled through a single central site for efficiency andsafety. Small valves located on a control panel operate massive valves and let theoperator control large hydraulic equipment from points of safety and comfort.

REVIEW QUESTIONS

1. What is meant by “remote controlling a relief valve”?

2. How can a relief valve be remotely controlled?

3. Name an advantage to using a remotely controlled relief valve.

4. Can the pressure setting of a remotely controlled relief valve be raised bymodifying the adjustment of a secondary relief valve?

5. On a hydraulic press, which valve (main/secondary relief valve) should be usedto adjust the highest desired tonnage?

5-1

Unit 5

Troubleshooting

UNIT OBJECTIVE

When you have completed this unit, you will be able to test the main components ofa hydraulic system based on the manufacturer specifications and on the firstprinciples of hydraulics. You will also be able to explain how temperature affects theoperating characteristics of a hydraulic system.


Troubleshooting is the activity or process of diagnosing and locating the cause ofmalfunction in a hydraulic system. Troubleshooting hydraulic equipment is basicallythe same as troubleshooting any mechanical or electrical device. Individual initiativeand imagination, coupled with effective and efficient techniques, are importantelements in successful troubleshooting. Good troubleshooting techniques dependon a sound knowledge of the system or device and the way it normally operates, aswell as a procedure that limits the number of verification steps.

Troubleshooting can be structured according to four levels of activity designed toidentify, locate, and correct a problem. Each level brings us closer to the problemsource. The levels of activity, listed in order, are:

1. System function;2. Location of the defective part of system;3. Component checking;4. Substitution or replacement.

In this unit, you will concentrate on the third level of activity, with special emphasisplaced on the verification of the pump, directional valve, and flowmeter.

To successfully perform any troubleshooting activity, it is necessary to understandthe equipment and its operation. The first step to do when testing a hydraulic systemis to locate system specifications. Pump delivery, relief valve setting, and cylindercycle times must be known if thorough testing is to be done. This kind of information,along with hydraulic schematics and functional block diagrams, can be useful andtime saving. Estimate specifications only if manufacturer’s specifications cannot befound.

In all cases, troubleshooting should be approached using two fundamental rules asa guide. First, observe the symptoms of the problem, and second, relate the problemto a specific part of the system. This will usually lead you to the device which isdefective.

5-2

5-3

Exercise 5-1

Hydraulic Pumps

EXERCISE OBJECTIVE

C To describe the basic operation of a hydraulic pump;C To be able to use manufacturer pump specifications to test a pump in a hydraulic

system;C To explain how oil temperature affects flow rate and volumetric efficiency.

DISCUSSION

Pump design and operation

The overall construction of a hydraulic pump is very similar to a hydraulic motor.Pump operation is basically the reverse of a motor. While hydraulic motors converthydraulic energy into mechanical energy, hydraulic pumps convert mechanicalenergy into hydraulic energy.

Most hydraulic pumps, including the pump in your Power Unit, are positive-displacement pumps. These pumps deliver a fixed volume of oil for each revolutionregardless of pressure in the system.

As Figure 5-1 shows, a hydraulic pump typically consists of the following elements:

C An inlet port supplying oil from the reservoir.

C An outlet port connected to the pressure line port.

C A housing containing a rotating mechanism connected to a drive shaft. The shaftis usually turned by an electric motor.

Hydraulic Pumps

5-4

Figure 5-1. Hydraulic pump.

Hydraulic pumps all operate on the same principle:

C A vacuum is created at the pump inlet by increasing the volume within thepump. The pressure difference between the vacuum condition in the pump andatmospheric pressure in the reservoir causes the oil from the reservoir to flow tothe pump inlet through the suction line.

C Oil is then expelled out of the pump by decreasing the volume within the pump.

Types of positive-displacement pumps

As with hydraulic motors, there are three basic types of hydraulic pumps, namedafter the type of rotating mechanism used inside the pump. These are gear, vane,and piston pumps.

Hydraulic Pumps

5-5

Gear pumps

The gear pump, shown in Figure 5-2, is the type of pump used in your Power Unit.It consists of two intermeshing gears that rotate within the pump housing. One gear,called the drive gear, is turned by the electric motor, and the other gear, called theidler gear, is turned by the drive gear. As the gear teeth separate on the inlet side,volume is increased and a vacuum is formed, causing oil to be drawn into thehousing. As the gears turn, oil is trapped between the teeth and the housing, andcarried to the outlet port, where the meshing teeth decrease the volume and forcethe oil into the system. Gear pumps are usually less expensive than the other typesof pump. They are popular in mobile equipment.

Figure 5-2. Gear pump.

Vane pumps

The vane pump consists of a slotted rotor connected to a drive shaft. These slotscontain vanes which are thrust outward by centrifugal force when the rotor turns. Theedges of the vanes seal against the housing walls, transporting oil in much the samemanner as the gears in the gear pump. Vane pumps are used extensively onindustrial manufacturing equipment.

Piston pumps

The piston pump consists of several pistons fitted into a rotating cylinder barrel. Asthe cylinder barrel is rotated, the pistons move in and out in the barrel, creatingincreasing and decreasing volumes within the pump. Piston pumps are commonlyused in high pressure applications.

Hydraulic Pumps

5-6

Pump slippage

Slippage may be defined as the internal leakage of oil in a pump. Some slippage isnecessary into all pumps to lubricate the various internal parts of the pump. In a gearpump, for example, slippage results from the required clearances between the gearteeth and between the sides of gears and the housing.

Excessive slippage, however, reduces the efficiency of the pump. Slippage increaseswhen the pump clearances increase from wear. More oil will flow through a largeopening than through a small clearance.

Hydraulic pump ratings

The hydraulic pump is the most important component in any hydraulic system. Thepump operating characteristics influence the performance, efficiency, and operatingcost of the entire system.

The first thing to do when testing a hydraulic pump is to get manufacturer’sspecifications. Important specifications to know are displacement, nominal flowrate, volumetric efficiency, and overall efficiency.

C Displacement is the volume of oil discharged by the pump in a single revolution.It is expressed in cubic centimeters per revolution (cm3/r) in S.I. units, or in cubicinches per revolution (in3/r) in English units.

C Nominal flow rate is the theoretical amount of oil supplied by the pump at agiven rotation speed. It is equal to the pump displacement multiplied by therotation speed. In equation form:

S.I. units:

English units:

The nominal flow rate is usually specified for a pump pressure around 0 kPa(0 psi). Due to internal leakage, however, the actual amount of oil supplied by thepump will decrease as the pump pressure increases.

For this reason, manufacturers often specify a slip value for their pump. The slipvalue indicates the amount of oil that does not reach the pump output for a givenincrease in pump pressure. If, for example, a pump has a nominal flow rate of10 l/min and a slip value of 0.1 l/min per 1000 kPa, this means that when thepressure demanded at the pump output is 5000 kPa, 0.5 l/min will be lost due tointernal leakage, and the pump flow rate will be reduced to 9.5 l/min.

Hydraulic Pumps

5-7

The nominal flow rate is also specified for an operating temperature around 49°C(120°F). At higher temperatures, the actual flow rate may be less than thespecified value because pump leakage increases as the oil temperatureincreases.

C Volumetric efficiency is the ratio of actual to nominal flow rate, expressed as apercentage:

Pump manufacturers often provide a graph showing pump volumetric efficiencyversus pump pressure. Figure 5-3 shows the volumetric efficiency curve specifiedfor the hydraulic pump in your Power Unit. As the pump pressure increases,volumetric efficiency decreases because the increasing amount of internalleakage causes the output flow to decrease.

Figure 5-3. Pump volumetric efficiency versus pressure.

When the pump pressure is limited by a relief valve, the discharge flow begins todecrease as soon as the valve cracking pressure is reached, which affects thevolumetric efficiency curve. If, for example, the valve pressure setting is 6200 kPa(900 psi), volumetric efficiency will start to decrease at a faster rate in the regionof 5500 kPa (800 psi), and it will decrease more and more as the pump pressureapproaches 6200 kPa (900 psi), as Figure 5-3 shows.

Hydraulic Pumps

5-8

The condition of a pump can be evaluated by measuring its volumetric efficiencyat several different pressures and comparing the obtained curve with thatprovided by the manufacturer.

C Pump overall efficiency is the ratio of the pump output power to the pump inputpower, expressed as a percentage. It is equal to pump volumetric efficiencymultiplied by pump mechanical efficiency, as shown below:

Notice that the output and input power values for this equation must be stated inthe same kind of units.

Pump manufacturers often provide a graph showing pump overall efficiencyversus pressure. Figure 5-4 shows an example. Overall efficiency is low below1700 kPa (250 psi) because industrial pumps are usually designed to operateabove this pressure.

The overall efficiency graph can be used to calculate the amount of mechanicalpower required at the shaft of a hydraulic pump to obtain a certain amount ofpower at the pump output, based on the following formula:

S.I. units:

English units:

Hydraulic Pumps

5-9

Figure 5-4. Overall efficiency versus pressure.

Cavitation

Cavitation is the formation and collapse of gaseous cavities in the hydraulic oil.Gaseous cavities are in the form of air dissolved in the oil. These cavities causedamage to the metal parts of the pump and also shortens the service life of the oil.

Air can enter a pump through loose pipe joints and fittings. Air can also enter thepump if the oil level in the reservoir is allowed to go below its minimum level, or if thepump is run at an excessive speed so that insufficient oil is drawn from the reservoir.

Cavitation of a pump is characterized by a decrease in the pump flow rate, andsystem pressure becomes erratic. This is often accompanied by severe vibration inthe hydraulic system and by a rattling sound coming from the pump. Cavitation alsoproduces excessive heat that dries out the pump bearings resulting in completefailure of the pump.

What can be done to prevent cavitation? The best safeguard is to use oil that isspecifically designed for use in hydraulic systems. In addition, all pipe joints andfittings must be tight to prevent air from entering the system, and the oil level mustbe kept above the minimum.

Hydraulic Pumps

5-10

Conversion factors

Table 5-1 shows the conversion factors used to convert measurements ofdisplacement from S.I. units to English units, and vice versa.

Displacement

Cubic centimetersper revolution (cm3/r)

x 0.061 = Cubic inches perrevolution (in3/r)

x 16.387 = Cubic centimetersper revolution (cm3/r)


REFERENCE MATERIAL

For detailed information on hydraulic pumps, refer to the chapter entitled HydraulicPumps in the Parker-Hannifin’s manual Industrial Hydraulic Technology.

Procedure summary

In the first part of the exercise, you will measure the pump maximum flow rate whenthe system pressure is nearly zero. You will compare the measured flow rate with thenominal flow rate specified by the pump manufacturer.

In the second part of the exercise, you will measure the pump flow rate at severaldifferent pressures. You will use the registered data to plot the pump volumetricefficiency against pressure on a chart.

EQUIPMENT REQUIRED


PROCEDURE

Measuring the pump maximum flow rate


Hydraulic Pumps

5-11

Figure 5-5. Measuring volumetric efficiency versus pump pressure.



position.e. Plug the Power Unit line cord into an ac outlet.



G 5. With the Flow Control Valve fully open, the system pressure at gauge A isnearly zero. At this very low pressure, the effects of internal leakage arenegligible, so the Flowmeter indicates the pump maximum flow rate. Recordbelow the Flowmeter reading.

Maximum flow rate: l/min or gal(US)/min

Note: The trainer Flowmeter is graduated in liters per minute only. If you are working with English units, multiply the measuredflow rate in liters per minute by 0.264 for determining theequivalent flow rate in US gallons per minute.

Hydraulic Pumps

5-12

G 6. Record below the approximate temperature of the oil as indicated by thetemperature/oil level indicator on the Power Unit.

Oil temperature: °C or °F

G 7. Turn off the Power Unit. Do not disconnect your circuit since you will use itin the next part of the exercise.

G 8. The nominal flow rate of your pump, as specified by the manufacturer, is3.1 l/min [0.82 gal(US)/min].

Compare this nominal flow rate to the actual flow rate recorded in step 5. Arethese values within 15% of each other?

G Yes G No

Note: A flow rate of less than 85% of the rated pump flow ratecould indicate pump failure due to worn. As the pump wears, theclearances become greater and slippage increases. A pump thathas been in service for extended periods will typically operate atlower than rated flow rate. A flow rate lower than the rated flowcould also indicate a failure of the Power Unit relief valve or animproper pressure setting of this valve.

On the other hand, a flow rate higher than the rated pump flowrate can be measured when the temperature of the oil is below38°C (100°F). This is because the trainer Flowmeter is designedto accurately read the flow rate at 38°C (100°F). Below thistemperature, the oil is thicker, which places extra pressure on theinternal parts of the Flowmeter and causes the Flowmeter readingto be slightly higher than the actual flow rate.

G 9. Calculate the theoretical pump displacement, based on a nominal flow rateof 3.1 l/min [0.82 gal(US)/min] and on a motor speed of 1725 r/min.

Measuring volumetric efficiency versus pressure


G 11. Turn the Flow Control Valve adjustment knob clockwise until the systempressure at gauge A is 1400 kPa (200 psi). Since the relief valve inside thePower Unit is set to 6200 kPa (900 psi), the relief valve is closed and theFlowmeter now reads the full pump flow at 1400 kPa (200 psi). Record theFlowmeter reading in Table 5-2 under “ACTUAL FLOW RATE”.

Hydraulic Pumps

5-13

PRESSURE ACTUAL FLOW RATE VOLUMETRIC EFFICIENCY

1400 kPa (200 psi)

2800 kPa (400 psi)

4100 kPa (600 psi)

5500 kPa (800 psi)

Table 5-2. Flow rate and volumetric efficiency versus pressure.

G 12. Repeat step 11 for the other pressures listed in Table 5-2. Read theFlowmeter as accurately as possible.


G 14. According to Table 5-2, does the pump flow rate decrease as the systempressure increases? Why?

G 15. Based on the actual flow values registered in Table 5-2, calculate the pumpvolumetric efficiency at each of the pressures listed in this table. Recordyour calculations in Table 5-2 under “VOLUMETRIC EFFICIENCY”. Use theflow value registered in step 5 as the nominal flow rate.

Hydraulic Pumps

5-14

Figure 5-6. Pump volumetric efficiency versus pressure curve.

G 16. In Figure 5-6, plot the pump volumetric efficiency versus pressure curvebased on the data registered in Table 5-2.

G 17. From the curve plotted in Figure 5-6, what effect does system pressure haveon the pump volumetric efficiency? Why?

G 18. Evaluate the condition of your pump. To do so, compare the curve plottedin Figure 5-6 with the 0-4100 kPa (0-600 psi) portion of the manufacturercurve in Figure 5-3. Is your pump in satisfactory condition?

G Yes G No

G 19. According to Figure 5-6, what is the volumetric efficiency of your pump at4100 kPa (600 psi)?

Hydraulic Pumps

5-15

G 20. What is the pump overall efficiency at 4100 kPa (600 psi), if the mechanicalefficiency is 90%?

G 21. Calculate the pump output power at 4100 kPa (600 psi), based on the actualflow rate registered in Table 5-2.

G 22. Calculate the electrical power required on the shaft of your pump to operatethe system at 4100 kPa (600 psi), based on the overall efficiency calculatedin step 20, and on the pump output power calculated in step 21.

G 23. Based on your answers in steps 21 and 22, is the mechanical powerrequired at the shaft of the pump to operate the system at a given pressuregreater than the actual power generated at the pump output? Why?




Hydraulic Pumps

5-16

CONCLUSION

This exercise has exposed you to pumps and how they are used in a hydraulicsystem. You learned some terms related to pumps, such as displacement, nominalflow rate, volumetric efficiency, and overall efficiency.

You determined the relationship between pump flow rate and pressure by measuringthe pump flow rate at several different pressures. This test showed that you could geta high flow rate or a high pressure but not both at the same time.

You also learned about volumetric efficiency, which is the ratio of the actual pumpflow rate to the theoretical, or nominal, flow rate. As a result you are aware that acertain amount of oil leaks through clearances inside the pump and does not makeit into the circuit. The condition of a pump can be evaluated by measuring the pumpflow rate versus pressure and comparing the obtained curve with that provided bythe manufacturer.

REVIEW QUESTIONS

1. What happens inside the pump that causes oil to enter the pump and then leavethe pump?

2. What are the three design classifications of positive-displacement pumps?

3. Why does the pump flow rate decrease as the system pressure increases?

4. Does the pump flow rate decrease as the oil temperature increases? Why?

Hydraulic Pumps

5-17

5. Define the term volumetric efficiency. What relationship exists between pumpslippage and volumetric efficiency?

5-18

5-19

Exercise 5-2

Directional Valve Testing

EXERCISE OBJECTIVE

C To show normal leakage of a directional valve;C To evaluate the condition of a directional valve according to the amount of

leakage flow.

DISCUSSION

Directional valve internal leakage

Most directional valves consist of a spool moving within the body of the valve tocover and uncover ports or pairs of ports. The spool is carefully machined to providea tight seal between the spool lands and the valve body.

However, microscopic clearances are intentionally left between the spool lands andthe valve body through which a small amount of oil is continuously flowing andlubricating, as Figure 5-7 shows. Oil leaking through the valve clearances is not lost.It returns to the reservoir through the return lines.

Figure 5-7. Directional valve internal leakage.

Leakage will usually increase gradually over a long period of time as the valve spoolgets wear. It is rare that a valve spool becomes so worn that it will bypass all thepump oil to the reservoir. Excessive leakage, however, tends to reduce cylinderspeed and becomes a problem when a load is required to be stopped in someposition for a long period of time. Excessive leakage is also a source of wastedenergy.


5-20

Figure 5-8 shows the effect of excessive leakage on cylinder speed. When the valveis shifted to extend the cylinder, some oil from the P port leaks across the spool landedges into the T port, so there is less oil available to extend the cylinder. Theextension speed is reduced.

Figure 5-8. Excessive leakage reduces the extension speed.

Figure 5-9 shows the effect of leakage on a load required to be suspended in somemid-point position when the power unit is off and the valve is in the center position.Pressurized oil leaks from the B cylinder line across the spool land edges into the Pand T ports, causing the load to drift down rapidly.


5-21

Figure 5-9. Leakage causes the load to drift down.

Figure 5-10 shows the effect of leakage on a load required to be stopped in somemid-point position when the power unit is on and the valve is in the center position.Pressurized oil leaks from the P port across the spool land edges into the A andB cylinder lines, acting on both ends of the cylinder. Since the cap end of the pistonhas a larger surface area exposed to pressure than the rod end, a greater force isexerted on the full piston area, which tends to extend the rod. If the cylinder has alight load attached to its rod, it will tend to drift out.

Figure 5-10. Leakage causes the cylinder rod to drift out.


5-22

Testing 4-way directional valves

Industrial 4-way directional valves are usually tested by measuring the amount ofleakage from the P port into the T port according to the following procedure:

1. Turn off the power unit and disconnect the plumbing at ports A, B, and T of thedirectional valve, as Figure 5-11 shows.

2. Block ports A and B of the directional valve with caps. This will deadhead thepump flow into the directional valve.

3. Turn on the power unit.

4. Shift the valve spool back and forth. Oil coming out of port T is spool leakage.Turn off the power unit.

Figure 5-11. Testing a 4-way directional valve for leakage.

When a directional valve is delivering oil at the maximum rated pressure, up to10% loss of flow rate from the P port into the T port is usually tolerated, if the valveis used only intermittently. However, industrial valves may require repair orreplacement if there is a 1% loss of flow rate between ports P and T.

REFERENCE MATERIAL

For detailed information on directional control valves, refer to the chapter entitledDirectional Control Valves in the Parker-Hannifin’s manual Industrial HydraulicTechnology.


5-23

Procedure summary

In this exercise, you will measure the leakage of a directional valve from the P portinto the T, A, and B ports. The directional valve will be held in the center position withpressure applied to the P port. The plastic flask will be connected to the T, A, andB ports in turn. The amount of collected oil, as well as the velocity of oil in the plastichose, will be measured using the graduated beaker.

Since the amount of leakage is very small, the amount of leakage per minute, orleakage rate, will be calculated by multiplying the velocity of the oil in the plastic hoseby the cross-sectional area of this hose.

The leakage rate will then be compared to the oil flow rate supplied by the pump todetermine if the tested valve is in satisfactory condition.

EQUIPMENT REQUIRED


PROCEDURE

Testing a 4-way directional valve for leakage


Figure 5-12. Measuring leakage from the P port into the A, B, and T ports.


5-24

Note: Do not connect the plastic flask to the directional valve yet.This will be done later in the exercise.






G 5. Record below the Flowmeter reading:

Pump flow rate: l/min or gal(US)/min

Note: The trainer Flowmeter is graduated in liters per minute. Ifyou are working with English units, multiply the measured flowrate in liters per minute by 0.264 for determining the equivalentflow rate in US gallons per minute.


G 7. Make sure the graduated flask and its plastic hose are empty. Connect theplastic flask to port T of the directional valve.

G 8. Turn on the Power Unit and let it run until some oil appears at the valve endof the clear plastic hose.

CAUTION!

Do not shift the directional valve lever during this exercise.Shifting the lever will result in dumping oil into the plasticflask.


G 10. Mark the oil level in the clear plastic hose with a piece of sticky tape.


5-25

G 11. Turn on the Power Unit and let it run for exactly 1 minute, then turn it off.

G 12. Measure the distance the oil level has risen beyond the tape mark todetermine the speed at which oil leaked from port T, in cm/min (in/min).Record this speed in Table 5-3 under “SPEED”.

PORT SPEED LEAKAGE RATE LOSS OF FLOW (%)

T

A

B

Table 5-3. Leakage from the P port into the T, A, and B ports.

G 13. Disconnect the plastic flask from port T of the directional valve, then connectit to port A.

G 14. Turn on the Power Unit and let it run for approximately 30 seconds toeliminate air bubbles in the oil inside the plastic hose.


G 16. Place a second piece of tape on the plastic hose to indicate the oil level (topof the oil column).


G 18. Measure the distance the oil level has risen beyond the tape mark todetermine the speed at which oil leaked from port A, in cm/min (or in/min).Record this speed in Table 5-3.

G 19. Disconnect the plastic flask from port A of the directional valve, then connectit to port B.

G 20. Turn on the Power Unit and let it run for approximately 30 seconds toeliminate air bubbles in the oil inside the plastic hose.



5-26

G 22. Place a third piece on the plastic hose to indicate the oil level (top of the oilcolumn).


G 24. Open the Relief Valve completely (turn knob fully counterclockwise).

G 25. Measure the distance the oil level has risen beyond the tape mark todetermine the speed at which oil leaked from port B, in cm/min (in/min).Record this speed in Table 5-3.

G 26. Measure and record the inside diameter of the clear plastic hose.

Hose diameter: cm or in

G 27. Using this diameter and the formula given below, calculate the leakage rateat ports T, A, and B in l/min [or gal(US)/min]. Record your calculated valuesin Table 5-3 under “LEAKAGE RATE”.

Metric units:

English units:

G 28. Using the formula below, calculate the percentage of flow loss caused byleakage from ports T, A, and B, based on the pump flow rate registered instep 5. Record your calculated values in Table 5-3 under “LOSS OF FLOW”.

G 29. Was the valve tested in this exercise in satisfactory condition, if the toleratedpercentage of flow loss from port T is 5%?

G Yes G No

G 30. Empty the collected oil into a container (capped plastic jugs, topped bottles,milk cartons, etc.) for transport to a disposal site. Oil recycling centers willnormally accept the oil which can be refined and used again. Do not emptythe oil back into the pump reservoir, since it could have been contaminatedby dirt particles. Dirty oil can be very harmful to the hydraulic system


5-27

because it causes flow paths to become clogged, valves to stick, and pumpto overheat.




CONCLUSION

In this exercise, you observed that a small amount of oil leaks from the P port to theother ports of a directional control valve when the valve is “closed” (centered). As aresult, you have an idea of how much oil actually might pass through a valve whichis closed. If you have records showing results from previous exercises, you may beable to determine the increase in leakage due to wear.

Clearances are intentionally left between the spool lands and the valve body forlubrication purposes. The size of the clearances is a compromise between sealingand lubrication. With loose clearances, the valve will leak more so the system willwaste energy. On the other hand, clearances that are too tight mean that the valvewill be insufficiently lubricated and that moving parts will quickly wear out.

Cheaper valves have looser clearances, so they leak more. As a result the systemwastes energy. You can either spend money on better components or spend moneypumping flow through leaks. The same holds for periodic maintenance. You canspend money to fix valves or spend it pumping oil through worn parts.

Try to imagine how you might use the cost of power and maintenance to set aleakage limitation (like the 1% of full flow figure in the DISCUSSION reading) beyondwhich the system inefficiency would cost more than fixing the leak...

REVIEW QUESTIONS

1. Why are small clearances intentionally left between the spool lands and the bodyof a directional valve?


5-28

2. In the circuit of Figure 5-13, predict what will happen to the suspended load andto the reading of the pressure gauge if the directional valve remains centered forseveral hours. Explain.


3. In the circuit of Figure 5-14, how much flow must be supplied by the pump toextend the cylinder rod in 2 seconds, if the percentage of flow loss from theP port into the T port is 10%?

Assume the leakage through the cylinder seals to be negligible.


5-29


5-30

5-31

Exercise 5-3

Flowmeter Accuracy

EXERCISE OBJECTIVE

C To verify the accuracy of a flowmeter;C To determine the effect of temperature on flowmeter accuracy.

DISCUSSION

Flowmeter construction and operation

Figure 5-15 shows the Flowmeter supplied with your Hydraulics Trainer. ThisFlowmeter consists of a graduated transparent tube and an indicating ring. The ringslides along the tube, indicating the amount of flow.

Inside the Flowmeter body is a spring-loaded piston sliding over a tapered meteringcone. As the flow increases, the piston slides down the metering cone until the forceon the piston is equal to the force of the spring.

The movement of the piston is proportional to the oil flow rate. The force on a ringmagnet on the piston controls the position of the indicating ring.

The Flowmeter will work in any position because the piston is spring-loaded. Gravityhas little or no effect on operation. However, the Flowmeter will operate in onedirection only. An internal check valve will bypass the oil flow through the Flowmeter,should the Flowmeter be connected in a line where flow direction is reversed.

There are other flowmeters in which the flow of the oil is opposed by gravity actingon a ball or a rotor. These flowmeters must be in the upright position to operateproperly.

Flowmeter Accuracy

5-32

Figure 5-15. Pictorial and cutaway views of the trainer Flowmeter.

Oil viscosity and specific gravity

Two very important characteristics of any hydraulic oil are viscosity and specificgravity. These characteristics affect the accuracy of flow measurement.

C Viscosity is a measure of the resistance of a fluid to flow. Water is a low viscosityfluid because it is “thin” and flows easily. Molasses is a high viscosity fluidbecause it is “thick” and shows more resistance to flow.

In hydraulics, the most common system used to rate the viscosity of oil is theSaybolt Universal Seconds (SSU or SUS) system. An oil is rated 1 SSU when60 milliliters (0.0158 US gallon) of this oil at 38°C (100°F) takes 1 second to flowthrough a standard orifice of 0.1765-cm (0.0695-in) diameter. For example, if ittakes 150 seconds for 60 milliliters of another type of oil at 38°C (100°F) to flowthrough the standard orifice, this oil will be rated at 150 SSU.

Oil viscosity is affected by temperature, as Figure 5-16 shows. As temperatureincreases, viscosity decreases and the oil gets thinner. For this reason, viscosityis usually associated with a temperature. Industrial hydraulic systems normallyuse oils that are rated between 150 and 250 SSU @ 38°C (100°F).

Flowmeter Accuracy

5-33

Figure 5-16. Oil viscosity versus temperature.

C Specific gravity is the quotient of the weight of a volume of fluid divided by to theweight of an equal volume of water. The specific gravity of pure water is 1.00.

Figure 5-17 shows an example. If a given volume of water weighs 100 kg (221 lb)and an equal volume of oil weighs 84 kg (185 lb), the specific gravity of this oil is0.84. This means that the weight of oil is 84% that of the water.

Flowmeter Accuracy

5-34

Figure 5-17. Determining the specific gravity of a certain type of oil.

Specific gravity is affected by temperature. As temperature increases, oil expandsand a given volume weighs less, so specific gravity decreases.

Flowmeter accuracy

Flowmeters are designed to accurately read the flow rate based on using a hydraulicoil which has a certain viscosity and specific gravity. The Flowmeter supplied withyour Hydraulics Trainer, for example, is calibrated for hydraulic oil with a viscosity of150 SSU and a specific gravity of 0.876. The oils commonly used in industrialhydraulic systems usually correspond to these ratings at a temperature around 38°C(100°F).

Accuracy becomes increasingly difficult to achieve as oil temperature cools below38°C (100°F), because the thicker oil places extra pressure on the internal parts ofthe flowmeter, giving a reading slightly higher than the actual flow rate.

The accuracy of a flowmeter can be verified by measuring the actual volume of oilpassing through it and comparing that volume to the indicated volume. This test canbe made at different temperatures to determine if the flowmeter is relatively immuneto temperature changes.

REFERENCE MATERIAL

For detailed information on oil viscosity and specific gravity, refer to the chapterentitled Petroleum Base Hydraulic Fluid in the Parker-Hannifin’s manual IndustrialHydraulic Technology.

Flowmeter Accuracy

5-35

Procedure summary

In this exercise, you will determine the accuracy of a flowmeter by measuring theactual volume of oil passed through it and comparing this volume to the indicatedvolume. This test will be made at two different temperatures to determine if theflowmeter is sensitive to temperature changes.

EQUIPMENT REQUIRED


PROCEDURE

Measuring flowmeter accuracy with cold and warm oil

G 1. Connect the circuit shown in Figure 5-18. Make sure the plastic flask isempty before connecting it to port B of the directional valve.





G 4. Note the oil temperature indicated by the temperature/oil level indicator onthe Power Unit. Record this temperature in Table 5-4 under “COLD OIL”.

Flowmeter Accuracy

5-36

Figure 5-18. Measuring the Flowmeter accuracy.



G 7. Move the lever of the directional valve toward the valve body to connect portP to port A and adjust the Flow Control Valve until the Flowmeter reads2.0 l/min [0.53 gal(US)/min]. Release the valve lever.

CAUTION!

Make sure not to move the directional valve lever outwardfrom the valve body, or oil will be dumped into the plasticflask.

G 8. Firmly hold the plastic flask upright with one hand. Move the lever of thedirectional valve outward from the valve body and keep it shifted duringexactly 10 seconds to dump the pump flow into the plastic flask, thenrelease the lever.

G 9. Turn off the Power Unit. Do not modify the relief valve pressure setting.

Flowmeter Accuracy

5-37

G 10. Disconnect the plastic flask from port B of the directional valve, then emptythe collected oil into the graduated beaker. Measure and record the volumeof collected oil below.

Volume of collected oil : ml

G 11. If you are working with S.I. units, multiply the volume registered in step 10by 0.001 to obtain the equivalent volume in liters. Record your calculatedvalue in Table 5-4 under “VOLUME”.

If you are working with English units, multiply the volume registered instep 10 by 0.000264 to obtain the equivalent volume in US gallons. Recordyour calculated value in Table 5-4 under “VOLUME”.

TEMPERATUREFLOWMETER

READING

VOLUMEDELIVERED IN

10 s

ACTUAL FLOWRATE

FLOWMETERERROR %

COLD OIL( _____°C or _____ °F)

2.0 l/min(0.53 gal(US)/min)

WARM OIL( _____°C or _____°F)

2.0 l/min(0.53 gal(US)/min)

Table 5-4. Flowmeter accuracy versus oil temperature.

G 12. Empty the collected oil into a container (capped plastic jugs, topped bottles,milk cartons, etc.) for transport to a recycling center. Do not empty the oilback into the pump reservoir since it could have been contaminated by dirtparticles. Dirty oil can be very harmful to the hydraulic system because itcauses flow paths to become clogged, valves to stick, and pump tooverheat.

G 13. Turn on the Power Unit and let it run for about 30 minutes. Then note the oiltemperature indicated by the temperature/oil level indicator on the PowerUnit. Record this temperature in Table 5-4 under “WARM OIL”.

G 14. Turn off the Power Unit. Reconnect the plastic flask to port B of thedirectional valve, then turn on the Power Unit.

G 15. Move the lever of the directional valve toward the valve body to connect portP to port A and adjust the Flow Control Valve until the Flowmeter reads0.53 gal(US)/min (2.0 l/min). Release the valve lever.

G 16. Firmly hold the plastic flask upright with one hand. Move the lever of thedirectional valve outward from the valve body and keep it shifted duringexactly 10 seconds to dump the pump flow into the plastic flask, thenrelease the lever.

Flowmeter Accuracy

5-38


G 18. Disconnect the plastic flask from port B of the directional valve, then emptythe collected oil into the graduated beaker. Measure and record the volumeof collected oil below.

Volume of collected oil : ml

G 19. If you are working with S.I. units, multiply the volume registered in step 18by 0.001 to obtain the equivalent volume in liters. Record your calculatedvalue in Table 5-4 under “VOLUME”.

If you are working with English units, multiply the volume registered instep 18 by 0.000264 to obtain the equivalent volume in US gallons. Recordyour calculated value in Table 5-4 under “VOLUME”.

G 20. Based on the volumes registered in Table 5-4, calculate the actual flow ratedelivered by the pump at both oil temperatures using the formula givenbelow. Record your calculated values in Table 5-4 under “ACTUAL”.

Actual flow rate = Volume delivered in 10 s x 6

G 21. Calculate the error in the Flowmeter reading at both oil temperatures, usingthe formula given below. Record your calculated values in Table 5-4 under“FLOWMETER ERROR %”.

G 22. According to your calculations in Table 5-4, does temperature affectFlowmeter accuracy?

G Yes G No

G 23. Is the Flowmeter error higher at the lower temperature? Explain why.

G 24. Empty the collected oil into a container (capped plastic jugs, topped bottles,milk cartons, etc.) for transport to a recycling center.

Flowmeter Accuracy

5-39




CONCLUSION

In this exercise, you found that the trainer Flowmeter is temperature sensitive andthat it reads more accurately at higher temperatures in the normal operating range.This is because temperature affects the viscosity and the specific gravity of fluids.Flowmeters are designed to accurately read the flow rate based on using a hydraulicoil which has a certain viscosity and specific gravity. Industrial flowmeters usuallyhave an accuracy within ± 5% of full scale under almost any field conditionsencountered.

REVIEW QUESTIONS

1. In the trainer Flowmeter, to what is piston movement proportional?

2. In what position will the trainer Flowmeter work?

3. Will flowmeters work with the flow in either direction?

4. How does temperature affect oil viscosity?

5. How does temperature affect oil specific gravity?

5-40

5-41

Exercise 5-4

Effects of Temperature on System Operation

EXERCISE OBJECTIVE

C To explain the effect of temperature changes on the viscosity of oil;C To demonstrate the effect of temperature changes on pressure drop and circuit

flow rate.

DISCUSSION

Oil viscosity and viscosity index

Viscosity reflects the resistance of an oil to flow. A low viscosity oil tends to be thinand flows easily while a high viscosity oil is thick and shows more resistance to flow.A low viscosity oil allows the oil to be pumped through the lines more easily. Goodlubrication, however, requires a reasonable degree of viscosity.

Viscosity index (VI) determines how an oil resists viscosity changes due totemperature. The viscosity of an oil with a high viscosity index will change little as thetemperature of the oil changes. A low viscosity index will allow greater change duringsimilar temperature changes. This property is important in a system subjected to awide range of temperatures, as a uniform oil viscosity is desirable in any system.

Typical viscosity indexes for petroleum oils range from 90 to 105; for polyglycols from160 to 200. Figure 5-19 shows the relationship between temperature and viscosityin 50 VI and 90 VI oils. With both types of oil, the oil viscosity decreases as the oiltemperature increases. However, the 90 VI oil has a lower slope and therefore hasa higher viscosity index than the 50 VI oil, which means that its viscosity varies lesswith temperature changes.

The pour point is the lowest temperature at which an oil will flow. At or below pourpoint, an oil will not flow into the pump intake port, possibly causing damage to thepump through cavitation. As a general rule, an oil should have a pour point lowerthan the lowest operating temperature of the system.


5-42

Figure 5-19. Relationship between temperature and viscosity in 50 VI and 90 VI oils.

Effect of temperature on pressure drop

Pressure drop ()P) occurs in a circuit when the oil flow is restricted. The amount ofpressure drop across a component depends on the component size and shape, andon the circuit flow rate and oil viscosity.

Pressure drop is affected by temperature changes. The higher the temperature, thelower the pressure drop across a component. This is because the oil viscositydecreases as the temperature increases. Reduced viscosity allows the oil to bepumped through a component more easily, because there is less friction, orresistance to flow.

Oil temperature is a very important point to consider while testing a hydraulic circuit.A test run at a temperature below the normal system operating temperature wouldbe inaccurate because of high resistance to flow resulting in higher than normalpressure readings.


5-43

Effect of temperature on flow setting

In a system where circuit flow rate is rather critical, as when precise cylinder speedis necessary, the system should be allowed to warm up before the circuit flow rateis adjusted. If a non-compensated flow control valve were adjusted to provide10 l/min [2.64 gal(US)/min] while at room temperature, the circuit flow rate might beonly 9.5 l/min [2.5 gal(US)/min] or less once the oil heated because of the increasedinternal leakage of the pump. Therefore, oil temperature is an important point toconsider when adjusting the circuit flow rate.

REFERENCE MATERIAL

For detailed information on the effect of temperature on oil viscosity and systemoperation, refer to the chapter entitled Petroleum Base Hydraulic Fluid in the Parker-Hannifin’s manual Industrial Hydraulic Technology.

Procedure summary

In this exercise, you will determine the effect a change in temperature has on thepressure drop across a flow control valve and on the circuit flow rate. To do so, youwill measure the pressure drop and circuit flow rate when the oil is cold. You will heatthe oil by allowing the Power Unit to run for 20 minutes, then you will measure thenew pressure drop and circuit flow rate. You will heat the oil another 20 minutes andagain measure the pressure drop and circuit flow rate. Finally, you will compare theresults obtained at the different operating temperatures.

EQUIPMENT REQUIRED


PROCEDURE

Effect of temperature changes on pressure drop and flow rate



5-44

Figure 5-20. Effect of temperature changes on pressure drop and flow rate.





G 4. Note and record the oil temperature indicated by the temperature/oil levelindicator on the Power Unit.



G 6. With the oil flow blocked at the Flow Control Valve, all the oil from the pumpis now being forced through the Relief Valve and gauge A indicates theminimum pressure setting of this valve. Adjust the Relief Valve so thatgauge A reads 4100 kPa (600 psi).


5-45

G 7. Partially open the Flow Control Valve by turning its adjustment knobcounterclockwise until the pressure reading at gauge A is 3500 kPa(500 psi).

G 8. The frictional resistance of the Flow Control Valve causes a pressuredifference between gauges A and B. Record below the pressure readingsof gauges A and B. Then, calculate the pressure drop, )P.

Gauge A: kPa or psi

Gauge B: kPa or psi

)P = Gauge A - Gauge B = kPa or psi

G 9. Record below the Flowmeter reading.

Flow rate: l/min or gal(US)/min

Note: The trainer Flowmeter is graduated in liters per minute only. If you are working with English units, multiply the measuredflow rate in liters per minute by 0.264 for determining theequivalent flow rate in US gallons per minute.

G 10. Let the Power Unit run for about 20 minutes, then check the temperature/oillevel indicator again. What is the temperature of the oil now?


G 11. Is the temperature recorded in step 10 different from the temperaturerecorded in step 4? If so, what caused the temperature change?

G 12. Record below the new pressure readings of gauges A and B. Then,calculate the pressure drop, )P.

Gauge A: kPa or psi

Gauge B: kPa or psi


G 13. Record below the new Flowmeter reading.



5-46

G 14. Again let the Power Unit run for 20 minutes, then check the temperature/oillevel indicator again. What is the temperature of the oil now?


G 15. Record below the new pressure readings of gauges A and B. Then,calculate the pressure drop, )P.

Gauge A: kPa or psi

Gauge B: kPa or psi


G 16. Record below the new Flowmeter reading.



G 18. Compare the pressure drops ()P) recorded in steps 12 and 15 with thepressure drop recorded in step 8. As the temperature increased, whathappened to the pressure drop? Why?

G 19. Compare the flow rates recorded in steps 13 and 16 with the flow raterecorded in step 9. As the temperature increased, what happened to theflow rate? Why?

G 20. How would the results of this exercise be affected by the use of an oil witha higher viscosity index?



5-47



CONCLUSION

In this exercise, you demonstrated the effect of temperature changes on pressuredrop and circuit flow rate. As the temperature increased, the pressure dropdecreased due to reduced oil viscosity. The flow rate also decreased because of theincreased internal leakage of the pump.

The operating temperature of industrial hydraulic systems change over the courseof the day, ranging from 27°C (80°F) in the morning to 60°C (140°F) in the afternoon.If the system pressure and flow rate settings must be precise throughout theworkday, an oil with a high viscosity index must be used.

REVIEW QUESTIONS

1. Why does pressure decrease as the oil temperature increases?

2. Why does flow rate decrease as the oil temperature increases?

3. What is meant by "viscosity index"?

5-48

A-1

Appendix A

Equipment Utilization Chart

The following Lab-Volt equipment is required to perform the exercises in this manual.

EQUIPMENT EXERCISE

MODEL DESCRIPTION 1-1 1-2 2-1 2-2 2-3 2-4 3-1 3-2 3-3 3-4

6310 Power Unit 1 1 1 1 1 1 1 1 1 1

6320 Directional Valve, Lever-Operated 1 1 1 1 1 1 1 1 1 1

6321 Flow Control Valve 1 1 1 1 1 1 1 1

6322 Relief Valve 1 1 1 1 1 1 1 1 1 1

6323 Pressure Reducing Valve 1

6340 Double-Acting Cylinder, 2.5-cm Bore 1 1 1 1 1 1 1 1

6341 Double-Acting Cylinder, 3.8-cm Bore 1 1 1 1 1 1 1

6342 Bidirectional Motor and Flywheels

6350 Pressure Gauge 2 1 2 2 2 1 2 1 2

6351 Flowmeter 1 1 1 1 1

6380 Loading Device 1 1 1

6390 Manifold, 5 Ports, Fixed 2 2 2 2 2 2 2 2 2 2

6391 Manifold, 4 Ports, Mobile 2 1 1 2 1

6392 Hose Set 1 1 1 1 1 1 1 1 1 1

(continued on next page)

Equipment Utilization Chart

A-2

EQUIPMENT EXERCISE

MODEL DESCRIPTION 4-1 4-2 4-3 4-4 5-1 5-2 5-3 5-4

6310 Power Unit 1 1 1 1 1 1 1 1

6320 Directional Valve, Lever-Operated 1 1 1 1 1 1

6321 Flow Control Valve 1 1 1 1 1 1 1

6322 Relief Valve 1 1 1 1 1 1 1

6323 Pressure Reducing Valve 1

6340 Double-Acting Cylinder, 2.5-cm Bore 1 1

6341 Double-Acting Cylinder, 3.8-cm Bore 1 1 1

6342 Bidirectional Motor and Flywheels 1

6350 Pressure Gauge 2 2 2 2 1 1 1 1

6351 Flowmeter 1 1 1 1 1 1 1

6380 Loading Device 1 1

6390 Manifold, 5 Ports, Fixed 2 2 2 2 2 2 2 2

6391 Manifold, 4 Ports, Mobile 1 2

6392 Hose Set 1 1 1* 1 1 1 1 1

* If additional hoses are required to connect the circuits in this exercise, hoses may be taken froma second Hydraulics Trainer by having the students from two workstations working together at asingle workstation.

ADDITIONAL COMPONENTS

Stopwatch, tachometer 0-2000 r/min, chemical-resistant gloves.

B-1

Appendix B

Care of the Hydraulics Trainer

General rules of good maintenance

a. Keep all components and work area in a clean, dirt-free condition.

b. Spilled or drained hydraulic oil should NOT be re-used. If re-use is imperative,the oil must be stored in a clean container. It should be carefully strained offiltered as it is returned to the Power Unit reservoir.

c. Use only a clean, lint-free cloth to wipe or dry component parts or to clean dustand dirt from the outside of the system.

d. Clean quick disconnects carefully before each re-assembly.

e. Flush out old oil and replace with clean oil at least once per year.

Oil and filter change

Regular oil changes are the most important preventive maintenance procedures thatcan be done. As hydraulic oil ages, it becomes diluted and contaminated, whichleads to premature pump wear. A new filter should also be installed every time theoil is changed.

Oil change

To change oil, perform the following steps:

1. If the Power Unit is running, turn it off. Allow the system oil to drain back into thereservoir for 5 minutes.

2. A drain pan large enough (about 20 liters /5 US gallons) to hold all of the oil inthe Power Unit reservoir is required to drain the Power Unit. Place such a panunder the reservoir drain, as shown in Figure B-1. Remove the drain cap with awrench and allow the reservoir to drain completely.


B-2

Figure B-1. Changing the Power Unit oil.

3. Replace the drain cap and remove the drain pan after the reservoir is completelyemptied. Use teflon tape or pipe joint compound to seal the drain cap threads.

4. Open the reservoir breather/filler cap, as shown in Figure B-1. Fill the reservoirup to the black line on the thermometer/oil level indicator. Use one of the fluidslisted on the Power Unit information decal on the front of the reservoir.

5. Replace the reservoir breather/filler cap.

6. Empty the drained oil into a container (capped plastic jugs, topped bottles, milkcartons, etc.) for transport to a disposal site. Oil recycling centers will normallyaccept the oil, which can be refined and used again.

Filter change

To change the filter, perform the following steps:

1. Turn off the Power Unit if it is running. Allow oil to drain out of the filter and intothe reservoir for 5 minutes.

2. When the filter has drained or if the Power Unit is cold, completely unscrew thefilter, as Figure B-2 shows. Be careful; its full of oil. Empty the oil inside the filterinto the drain pan.

3. Compare the old filter with the new one to make sure they are the same typeand micron rating (10-microns or less). Lubricate the gasket of the new filter witha few drops of oil and screw the new filter onto the Power Unit assembly. Thefilter should be hand-tightened only.


B-3

Figure B-2. Changing the oil filter.

4. Empty the drained oil into a container (capped plastic jugs, topped bottles, milkcartons, etc.) for transport to a disposal site. Oil recycling centers will normallyaccept the oil, which can be refined and used again.

B-4

C-1

Appendix C

Conversion Factors

Use the following conversion factors to convert S.I. or metric measurements toEnglish measurements and vice versa.

Length(distance)

Centimeters(cm)

x 0.394 = Inches (in) x 2.54 = Centimeters(cm)

Meters (m) x 3.281 = Feet (ft) x 0.305 = Meters (m)

Volume(capacity)

Cubiccentimeters(cc; cm3)

x 0.061 = Cubic inches(in3)

x 16.387 = Cubiccentimeters(cc; cm3)

Liters (l) x 0.264 = US gallons(US gal)

x 3.785 = Liters (l)

Mass(weight)

Kilograms (kg) x 2.205 = Pounds (lb) x 0.454 = Kilograms(kg)

Force

Newtons (N) x 0.225 = Pounds-force (lb; lbf)

x 4.448 = Newtons (N)

Pressure

Bars (bar) x 14.5 = Pounds-force persquare inch(psi; lb/in2)

x 0.069 = Bars (bar)

Kilopascals(kPa)

x 0.145 = Pounds-force persquare inch(psi; lb/in2)

x 6.895 = Kilopascals(kPa)

Conversion Factors

C-2

Area

Squarecentimeters(cm2)

x 0.155 = Squareinches (in2)

x 6.45 = Squarecentimeters(cm2)

Flow rate

Liters perminute (l/min)

x 0.264 = Gallons (US)per minute [gal(US)/min]

x 3.79 = Liters perminute (l/min)

Velocity

Centimetersper minute(cm/min)

x 0.394 = Inches perminute(in/min)

x 2.54 = Centimetersper minute(cm/min)

Work

Joules (J) x 0.738 = Foot-pounds(ft@lb)

x 1.355 = Joules (J)

Power

Watts (W) x 0.0013 = Horsepower(hp)

x 745.7 = Watts (W)

D-1

Appendix DHydraulics and Pneumatics Graphic Symbols

Figure D-1.

Hydraulics and Pneumatics Graphic Symbols

D-2

Figure D-2.

Bibliography

Bohn, Ralph C. and MacDonald, Angus J., Power: Mechanics of Energy Control,Second Edition, Bloomington, Illinois: McKnight Publishing Company, 1970.

ISBN 87345-256-9

Hedges, Charles S., Industrial Fluid Power, Volume 1, Third Edition, Dallas, Texas:Womack Educational Publications, Department of Womack Machine SupplyCompany, 1984.

ISBN 0-9605644-5-4

Hedges, Charles S., Industrial Fluid Power, Volume 2, Fourth Edition, Dallas, Texas:Womack Educational Publications, Department of Womack Machine SupplyCompany, 1988.

ISBN 0-943719-01-1

Corporate authors, Industrial Hydraulic Technology, Bulletin 0232-B1, ParkerHannifin Corporation, Cleveland, Ohio, 1991.

ISBN 1-55769-025-6

Corporate authors, Mobile Hydraulics Manual, Second Edition, Vickers Incorporated,Michigan, 1979.

We Value Your Opinion!

Your comments and suggestions help us produce better manuals and developinnovative systems to meet the needs of our users. Please contact us by e-mail at:

[email protected]

For further information, visit our website at www.labvolt.com.

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