James a. Sullivan Fluid Power Theory and Applications

FLUID POWER Theory and Applications ------

Third Edition

James A. Sullivan Southern Illinois University Carbondale, Ill ino is

A RESTON BOOK PRENTICE HALL, Englewood Cliffs, New Jersey 07632

Library of Congress Cata log ing - In -Pub lIcation Data

Sullivan, James A. Flui d power : th e or y and a pplicati ons I James A. Sull ivan . p. cm. - A Reston book. Includ es b ib l iograp h ica l ref er ences and In de x. IS BN 0- 13-32308 0-5 1. F lu id power technology. 1. Tit l e. TJB40 . S882 1989 620.1' 06- -d c 19 88 - 16883 CIP

Editori al/production supervision and interior design: Carol L. A tkins Cover design: Wanda Lub elska Manufacturing buyer: Robert A nderson

© 1989, 1982, 1975 by Prentice-Hall, Inc. A Division of Simon & Schuster

Engle wood Cliffs, New Jer sey 07632

All rights reserved . No part of this book may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.

Printed in the United State s of America 10 9 8 7 6 5 4 3 2 I

ISBN

0-13-323080-5

Prentice-Hall Internation al (UK) Limited , London Prent ice-Hall of Australia Pty. Limited , Syndey Pren tice-H all Canada Inc., Toronto Prentice-Hall Hispanoamericana , S.A ., Mexico Prentice-Hall of India Private Limited , N ew Delhi Prentice-Hall of Japan , Inc ., Tokyo Simon & Schu ster Asia Pte . Ltd ., Singap ore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro

CONTENTS

FOREWORD

xi

PREFACE CHAPTER 1

I-I

1-2 1-3 1-4 1-5 1-6 1-7 CHAPTER 2

xiii

INTRODUCTION

Introduction I Historical Perspective Capabilities 2 Careers in Fluid Power 8 The Use of Units II How Flu id Power Works II Summary 14 Study Questions and Problems

1

15

BASICS OF HYDRAULICS

2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9 2-10

17

Introduction 17 How Oil Transmits Power 17 Work Performed by the System 18 Multiplication of Force 19 Power and Horsepower 21 Flow Rate in a System 23 Fluid Horsepower 24 Torque and Torque Horsepower 26 Torque Horsepower and Fluid Horsepower Relationships 30 Summary 32 Study Questions and Problems 32 v

CHAPTER 3

3-1 3-2 3-3 3-4 3-5 3-6 3-7 CHAPTER 4

4-7

56

Introduction 56 56 Applications of the Displacement Formula 58 Applications of the Continuity Equation Viscosity 60 Non compressible Flow in Pipes 65 Reynolds Number 66 Summary 70 Study Questions and Problems 70

Introduction 72 Darcy-Weisbach and Hagen-Poiseuille Formulas f-Factor for Turbulent Flow 75 C-Coefficients 79 K-Values 81 Equivalent Length 86 Circuit Calculations 87 Summary 91 Study Questions and Problems 92

72

73

94

Introduction 94 Types of Fluids 95 Fluid Applications 97 Service Related Properties 100 Fluid Storage and Handling 108 Filtration 110 Summary 119 Study Questions and Problems 120 122

PUMPS

7-1 7-2 7-3 vi

53

HYDRAULIC FLUIDS

6-1 6-2 6-3 6-4 6-5 6-6 6-7 CHAPTER 7

42

FRICTION LOSSES IN HYDRAULIC SYSTEMS

5-1 5-2 5-3 5-4 5-5 5-6 5-7 5-8 CHAPTER 6

Introduction 36 Pressure and Head 37 Continuity Equation 40 Potential and Kinetic Energy Bernoulli's Equation 48 Torricelli's Theorem 52 Summary 53 Study Questions and Problems

HOW FLUIDS FLOW

4-1 4-2 4-3 4-4 4-5 4-6

CHAPTER 5

36

ENERGY IN HYDRAULIC SYSTEMS

Introduction 122 123 Pumping Theory Pump Types 128 Contents

15-6 15-7 CHAPTER 16

406

SIZING PNEUMATIC SYSTEMS

16-1 16-2 16-3 16-4 16-5 16-6 16-7 16-8 16-9 16-10 16-1I CHAPTER 17

Conditioning Air 401 Summary 405 Study Questions and Problems

Introduction 408 Pneumatic Distribution Systems 409 Compressors 413 Sizing Compressors 419 Receivers 421 Compressor Controls 423 Pneumatic Valves 425 Sizing Pneumatic Valves 431 Cylinders 434 Rotary Pneumatic Activators 438 Summary 443 Study Questions and Problems 444

PNEUMATIC CIRCUIT APPLICATIONS AND CONTROLS

17-1 17-2 17-3 17-4 17-5 17-6 17-7 17-8 17-9 17-10 17-11 17-12 17-13

408

446

Introduction 446 Basic Symbols 447 Basic Circuit Designs 451 Single-Acting Circuits 452 Double-Acting Circuits 455 Sequence Circuits 459 Interlock Circuits 466 Pneumatic Logic Control 470 Logic Control Concepts 471 Basic Air Logic Circuit Design 475 Miniature Pneumatics 476 Moving Part Logic and Modular Air Components 476 Summary 478 Study Questions and Problems 478

APPENDIX A

LETTER SYMBOLS AND ABBREVIATIONS

482

APPENDIX B

CONVERSION FACTORS

486

APPENDIXC

APPROXIMATE VISCOSITY CONVERSIONS

487

APPENDIX D

STANDARD GRAPHIC SYMBOLS FOR FLUID POWER DIAGRAMS

489

APPENDIX E

OFFSET BEND CALCULATIONS FOR STEEL TUBING

510

APPENDIX F

STANDARD TEST PROCEDURES

513

INDEX

519

Contents

ix

7-4 7-5 7-6 7-7 CHAPTER 8

HYDRAULIC CYLINDERS AND CUSHIONING DEVICES

8-1 8-2 8-3 8-4 CHAPTER 9

Content s

181

Introduction 181 Torque, Speed, Power, and Efficiency 182 Limited Rotation Act uators 186 Continuous Rotation Actuators 188 High Torque-Low-Speed Motors 193 Wheel Motors 196 Electrohydraulic Pulse Motors (EHPM) 200 Output Performance and Testing 202 Summary 205 Study Que stions and Problems 206 207

Introduction 207 Pres sure Control Valves 208 Flow Control Valves 21 5 Directional Control Valves 221 Proportional Control Valves 234 Servo Valves 245 Summary 249 Study Questions and Problems 250

SEALS AND PACKINGS

11-1 11 -2 11 -3 11-4 11-5 11 -6 11-7

160

179

VALVES

10-1 10-2 10-3 10-4 10-5 10-6 10-7 CHAPTER 11

Introduction 160 Cylinders 161 Cushioning Devices 170 Summary 179 Study Questions and Problems

HYDRAULIC MOTORS

9-1 9-2 9-3 9-4 9-5 9-6 9-7 9-8 9-9 CHAPTER 10

Pump Performance 150 Pump Selection Factors 153 Other Pump Characteristics 154 Summary 156 Study Questions and Problems 157

252

Introduction 252 Static Seal s 254 Dynamic Seals 260 Materials and Compounds 262 Seal Configurations 268 Elastomer Seal Testing 275 Summary 275 Study Que stion s and Prob lems 276 vii

CHAPTER 12

SYSTEMS COMPONENTS

12-1 12-2 12-3 12-4 12-5 12-6 12-7 12-8 12-9 12-10 CHAPTER 13

13-13

viii

320

Introduction 320 Sizing Actuators from Output Objectives 322 Constant Flow Systems (C-Q) 328 Constant Pressure Systems (C-P) 330 Constant Horsepower Systems (C-Hp) 331 Load Sensing Systems 333 Basic Circuits 336 Open Center Versus Closed Center Circuits 344 Pump Selection 347 Matching Components 349 Performance Testing 350 Compliance With Safety Recommendations and Standards 352 Summary 353 Study Questions and Problems 354

TROUBLESHOOTING HYDRAULIC SYSTEMS

14-1 14-2 14-3 14-4 14-5 14-6 14-7 CHAPTER 15

Introduction 278 Reservoirs and Heat Exchangers 278 Accumulators 284 291 Intensifiers Conductors and Connectors 293 Instrumentation 303 Hydraulic Power Units 309 Hydrostatic Transmissions 310 Electrical Controls 312 Summary 317 Study Questions and Problems 317

BASIC CIRCUITS AND SIZING HYDRAULICS COMPONENTS

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

CHAPTER 14

278

362

Introduction 362 Writing Problem Statements 366 Problem Statements for the System 366 Problem Statements for Components 370 Basic Calculations 374 Tee Test for Hydraulic Circuits 386 Summary 388 Study Questions and Problems 388

BASICS OF PNEUMATICS 15-1 Introduction 391 15-2 Standard Air 392 15-3 The Gas Laws 393 15-4 Application of the Gas Laws 397 15-5 Moisture and Dew Point 399

391

Contents

FOREWORD

Fluid power (hydraulics and pneumatics), mechanical , and electrical power are the three basic means for transmitting power. Although fluid power is a very old tech nology, it was not until the 20th century that machinery and equipment designers discovered the many instances where the ease of co ntrol, compactness , and versatility of fluid power made it the ideal replacement for mechanical and electrical transmissions , and enabled machinery and equipment manufacturers to accomplish functions previously impossible within the limits of mechanical and electrical technology. Some common examples are articulated man-lifts, hydraulic pre sses, 747 flight controls, and even the actuators that blink the eye s and move the fingers on the almost-human mannequins at Disney World . Once the advantages of fluid power became known, machinery and equipment manufacturers were quick to incorporate them into their products . The industry had an almost explo sive growth to its present $6 billion level. The rapid growth of the new technology generated an unprecedented demand for educational support. Professor James A. Sullivan, author of this text, is highly qualified in fluid power education and instruction. He is one of the dedicated cadre of teachers who were introduced to fluid power during the 1965-66 Institutes on Fluid Power Education . Since then he has enriched his knowledge, both theoretical and practical. He is currently working on validation of the occupations in fluid power, identification of job competencies for the Fluid Power Educational Foundation, and certification of Fluid Power Specialists and Mechanics . I heartily recommend this text' s use by the instructor and the student. James I. Morgan, CAE President National Fluid Power Association

xi

PREFACE

This third edition contains seventeen chapters, five more than in the previous edition. Much of the material has been reorganized and improved with new illustrations of components and circuits. Three new chapters have been added to reflect the increased interest in troubleshooting and pneumatics . Additionally, the problems sections at the end of each chapter have been expanded. As with previous editions, the present text relates theory to approved practice. When theory is explained, a concerted attempt is made to identify and solve related problems-and then to identify appropriate laboratory work, suggest how it is performed , and cite sources of related information from which other learning activities can be organized. ASTM standards and conventional tests, for example, are explained and suggested as appropriate exercises for the fluid power technician. And several maintenance procedures are laid out for use by the fluid power mechanic. The author wishes to gratefully acknowledge the contributions of many companies and individuals who made this work possible. Included in this list are professional and standards organizations, publishers, schools and educational equipment companies, petroleum companies, original equipment manufacturers , and personal associates. Credit lines are given for these contributions whenever possible throughout the text. James A. Sullivan Southern Illinois University Carbondale, Illinois

xiii

_ _ _ _ _ _ _ _~CHAPTER

1

INTRODUCTION

1-1 INTRODUCTION

Fluid power means using pre ssurized fluids in a confined sys tem to acco mplish work. Both liquid s and gase s are fluids. Most hydraulic sys tems use petroleum oils, but often synthetic oils and water base fluids are used for safety rea son s. Pneumatic systems use air which is exhausted to the atmosphere after doing the work. The oil used in hydraulic systems exhibits the characteristic s of a solid and provides a rigid med ium to transfer power through the system . Conversely, the air used in pneumatic systems is spongy and provi sion must be made in its control to effect smooth operation of actuators . Air and oil are frequently combined in one system to provide the advantages of both in accomplishing the work specified . A fluid power system accomplishes two main objectives . First, it provides substantial fluid force to move actuators in locations awa y from the power source where the two are connected by pipe s, tubes, or hoses. A power source, for example a gasoline or diesel engine coupled to a hydraulic pump, can be housed in one area to power a cy linder or hydra ulic motor one hundred feet or more away in another location. This is a decided advantage over sy stems using a mechanical drive train as the location of the output becomes less accessible . Second, fluid power sy stems accomplish highly accurate and precise movement of the actuator with relative ease . This is partic ular ly important in such applications as the machine tool industry where tolerances are often specified to one ten thousandth of an inch and must be repeatable during several million cycles . .....

1-2 HISTORICAL PERSPECTIVE

The modern era in fluid power began around the turn of the century. Hydraulic application s were made to such installations as the main armament system of the USS Virgi nia in 1906. In this app lication , a variable speed hydrostatic transmis-

sion was installed to drive the main guns (see Fig. 1-1). Since that time , the marine industry has applied fluid power to cargo handling and winch systems, controllable-p itch propellors , submarine control system s, shipboard aircraft elevators , aircraft and missile launch systems, and radar-sonar drive s.

Figure I-I

USS Virginia (courtesy of Sperry Vickers) .

1-3 CAPABILITIES

Hydra ulics and pneumatics have almost unlimited applications in the production of goods and services in nearly all sectors of the country. Several industries are dependent on the capabilities that fluid power affords . Among these are agriculture, aerospace and av iation , construction, defense, manufacturing and machine tool, marine , material handling, mining, transportation, undersea technology , and public utilities, including communications transmission systems . The world ' s need for food and fiber production has caused unprecedented leadership in agricultural equipment development , and particularly in applying hydraulics to solve a variety of problem s. Figure 1-2 illustrates a modern four wheel drive tractor that feature s exten sive use of hydraulic power. A 26.5 gal/min eight cylinder variable volume radial piston pump supplies fluid to a closed-center load-sensitive circuit for instant response . The pump unload s to a minimum standby pressure to reduce power con sumption when demand is low. Hydraulics power the rear and front main drive and power take-off (pto) clutches, wet disc brakes, remote valves , implement hitch, draft sensing, power shift transmission , differential lock , and hydrostatic steering. A closed accumulator-style reservoir is used to supplement the flow from the charge pump during maximum dem and from large bore cylinders. Other applications to agriculture include combine s, forage harvesters, backhoes, chemical sprayers , and organic fertilizer spreaders. Extensive design and 2

Introduction

Chap. 1

Figure 1-2

Modem wheel tractor (courtesy of John Deere Company).

application are also being made to wheel motors driven in remote locations to assist in marginal tractive conditions. Wheel motors consist of a hydraulic motor mounted integrally with a wheel and tire assembly . Braking is usually designed into the system. All that is required to make the wheel motor functional is mounting and hydraulic line connections back to the main control valve and power source. Auxiliary power drive wheel motors have several advantages to the agricultural industry. For example, they can be used to maintain and improve the turning ability of regular row crop tractors. They allow for high design which maintains under-axle crop clearances. They can be used to retain front axle adjustability and to provide on-the-go engagement and disengagement in forward as well as reverse. Figure 1-3 illustrates the application of hydraulic power to the aviation and aerospace industries. Another secto r of our economy that has benefited from the brute power of hydraulics and pneumatics is the construction industry. Crawler tractors, road graders, bucket loaders , trenchers, backhoes, hydraulic shovels and pan scrapers are just a few of the many applications. Figure 1-4 shows a large (235 ton payload) off-the-road truck with many hydraulic and pneumatic components. Notice the large telescope hydraulic dump cylinders. Sec. 1-3

Capabilities

3

,J>.

Auxiliar y power uni t

Main syste m pumps

Figure 1-3 Aerospace a nd aviation a pplica tio ns of fluid power (courtesy of Fluid Po wer Education Fo undation).

Gu n dri ve motor

Figure 1·4 Off-the-road truck (court esy of WABCO Constru ction and Minin g Equipment Group , An American Standard Company ).

The manufacturing and machine tool industry is very dependent on hydraulic power to provide the power and close tolerance necessary in controlled production. Figure 1-5 illustrates a 2,000 ton forging press, one of the largest selfcontained presses in existence. Hydraulic fluid is supplied by 20 pumps , each of which has a capacity of 35 gal/min. At an operating pre ssure of 2,000 lbf/in? the press consumes over 800 hp. The pre ss is driven by fluid supplied from a central hydraulic pumping system consisting often double-end electric motors driving the 20 hydraulic pumps. Undersea applications of hydraulics are receiving widespread attention in research, development, and utilization. The Pisces series of submersibles , developed by a Vancouver, B.C . company, can work at depths to 6,500 ft and are engaged in a variety of work ranging from rescue to salvage operations. Figure 1-6 illustrates a two-man operation inside the sub in the control of the steering and stabilizing system s, bow thrusters , grappling hooks , net winches , mooring winches , and anchor windlasses. Also shown is the DSRV-I developed by Lockheed for the U.S. Navy undersea research effort. Material handling conveys products from the point of production to the consumer. Several systems are in use and nearly all of them depend on hydraulics Sec. '-3

Capabilities

5

Figure 1-5 2,OOO-ton self-contained forging press (courtesy ofAb ex Corporation . Deni son Division ).

and pneumatics in the transport of goods. These include garbage trucks , conveyors, hydraulic leveling dock systems , hoist s, truck and trailer tailgate systems , and even industrial robots . Fluid power has wide application in continuous coal mining. Machines equipped with several hydraulic cylinders, jacks, and motors dig and load coal at the rate of two tons per minute. Conventional cutting and blasting are not required. Other mining applications utilizing fluid power pneumatics include rock drills, hydraulic track laying machines, shuttle cars , roof bolting machines, and conventional hydraulic jacks. Transportation systems provide examples of the most varied uses of fluid power. Road and off-the-road vehicles are typically suspended by pneumatic tires. Air assisted brakes (both high pressure and vacuum assisted) are standard equipment. Power steering systems, either of the hydraulic assist or full power steering type, are also standard and relieve the operator of the fatigue commonly associated with physical exertion over long periods of time. Hydrostatic transmissions are very common on all types of vehicles. Suspension systems are 6

Introduction

Chap. 1

Figure 1-6 (a) U.S . Navy Undersea Research Vehicle (court esy of Parker-Hannifin); (b) Intern ational operatio n of Undersea Research Vehicle (courtesy ofFluid Power Education Founda tion) .

dampened with hydr aulic shock absorbers , and some combine pneumatics by using the compressible nature of gases as the basis for air-oil suspension sys tem s. Hydr aulic wheel drive motors are a recent addition to the tran sportation line. They give almo st unlimited flexibility to the design which can mount the power plant in a convenient location to power individu ally driven wheel s as they are required to support the load and pro vide traction sufficient to propel it in almo st any direction with a variety of stepless speeds . Utilities and communication s are two industries where the use of fluid power is vital. Line utility vehicles are now a must to support per sonn el working above the ground. Similar vehicles are used in applic ation s that vary throughout indu str y from fruit tree mainten anc e and picking to jumbo jet service. Almo st unlimited movement is ava ilable to the operator with these systems. The y can be given direction from the bucket , from the side of the truck vehicle, from the end of an umbilical cord , a nd even by remot e signals from an electronic control box. Oth er applications to the communication and utilitie s indu strie s include hydraulic trenchers, cable boring machines, earth augers, pipe laying machin es, tampers, and man y others. Th ree applications given recent attention in the fluid power indu stry are Sec. 1-3

Capabilities

7

miniature pneumatics, moving part logic, and fluidics. Miniature pneumatics makes use of small air-powered com ponents such as cylinders and valves to carry out small assig ned tasks as well as to co ntrol ot her large components . Moving part logic components and fluidics make use of logic elements with functions similar to several electronics counterparts, such as capacitors , resistors , and amplifiers to control other hydraulic and pneumatic systems . 1-4 CAREERS IN FLUID POWER

Career opportunities in the fluid power industry look promising. Workers function primari ly at three levels : (I) mec hanics and maintenance men, (2) technicians , and (3) engineers. Work at each of the three levels is diverse . Through the efforts of the National Fluid Power Association , I the Fluid Power Education Foundation-, and Fluid Power Society", however, substa ntial progress has been made developing new job de scriptions for fluid power mechanic , fluid power technician, and fluid power engineer for inclusion in the Dictionary of Occupational Titles, the nat ion 's data base for occupational information . Two of the job title s and definitions formulated and which have been accepted by the U .S . Department of Labor are: Fluid Power Mechanic (DOT 600.281) and Fluid Power Technician (DOT 007.161). In general , mechanics and maintenance men work with fluid power components in all indu stries and make up one of the faste st growing occupational groups in a nation ' s labor force . In the U.S. , with a labor force of 110 million workers , employment of mechanics and repairmen total s nearly three million. More than 800 thousand of these are automobile mechanics. Recent trend s in licensure of this group require training in fluid power related accessories and components, including braking systems , hydromatic transmissions , power steering, air emission control sys tems, air conditioning, and stabilizing systems . Indu stry employs about 300 thousand industrial machinery repairmen , another 100 thousand millright s, 75 thou sand farm eq uipment mechanics , and other miscellaneous mec hanic s. All told , nearly 30% of the mechanics and repairmen employed are in the manufacturing industrie s , and the majority of these are employed in plants that produce durable good s such as tran sportation equipment machinery, primar y metals , and fabricated met al products. Another 20% are employed in shops that specialize in serv icing such equipment. Most of the remaining mechanics and repairmen are employed in the transportation, construction, and public utilitie s industries, and by the government at all level s. Most employment opportunities for mechanics and repa irmen occur in the more populous and indu strialized areas . Many mechanics and repairmen learn their skills on the job or through t.2 Na tio nal Fluid Power Assoc iatio n and Fluid Power Educa tion Foundatio n are located at 3333 N . Mayfair Road , Milwauk ee , Wiscon sin, 53222 USA. J At the time of this writing , the Fluid Power Societ y was located at 2900 N . I 17th Street , Milwaukee , Wiscon sin 53226.

8

Introduction

Chap. 1

apprenticeship training . They are union affiliated and earn high wages. Some apprenticeship training programs require and give credit for related courses from high schools , trade and technical schools, and private schools. Training and experience in the armed services also help young men and women prepare for occupations such as aircraft hydraulic mechanic and others. Some employers con sider a formal apprentice training program to be the best way to learn skilled maintenance and repair work. An apprenticeship consists of three to five year s of paid on-the-job training, supplemented each year by at least 144 hours of related classroom instruction . Formal apprenticeship agreements are registered with the state apprenticeship agency and the U.S . Department of Labor' s Bureau of Apprenticeship and Training. Emplo yers look for applicants who have mechanical aptitude and manual dexterity . Man y emplo yer s prefer people whose hobbies or interests include automobile repair and model building. A high school education or the equivalent is a must. Also nece ssary prior to training, or as a part of training , are courses in mathematics, physics, blueprint reading, and machine shop. Many manufacturers of fluid power systems offer specialized training courses for mechanics and maintenance per sonnel through private schools. Workers in most maintenance and repair occupations have several avenues of advancement. Some move into supervisory positions, such as foreman , maintenance manager, or service manager. Specialized training prepares others to advance to sale s, technical writing, and technician jobs. A substa ntial number of servicemen have also opened their own businesses. Fluid power technicians assist engineering team s with such work as sales, applied research, planning production , maintenance , testing , and installation. Education includes two years of post high school training containing several courses in phy sics, mathematics, fluid power, mechani sms, and communication s skills, and culminates in an associate degree. The technician can describe idea s about machinery to superiors, client s, and peer workers, and write technical reports. Work is usually performed at a drafting desk station, in specialized laboratories , and in the shop or plant. Figure 1-7 shows an engineering technician at work in a modern fluid power design facility. Joint effort s between the Fluid Power Society (FPS) and indu str y over the last several year s have resulted in certification of fluid power speciali sts". The se efforts grew out of the realization that certification is closely tied to specialized technologies. Membership in cert ification is open to men and women who function in sale s, technical aspects, and other specialties in the fluid power industry. Engineers compri se the second largest profe ssional occupation in this country . Of the more than 1.1 million engineers in the labor force , nearl y 600 thou sand are in the manufacturing industrie s. Engineers who work with fluid power compo4 Certific ation of techni cian s is also done by two other group s: Institu te for Th e Certification of Engineeri ng Technician s, spo nso red by the Nat iona l Society of Professional Enginee rs , 2029 K Stree t, N .W.. Wash ington , D.C. 20006., and America n Society of Certifie d Enginee ring Tec hnicians, Na tional Office , 2029 K Stree t, N .W., Washington , D.C. 20006.

Sec. 1-4

Careers in Fluid Power

9

Figure 1-7 Engineering technician in a modern fluid power design facilit y (courtesy of John De ere Company).

nents and systems are found primarily in this manufacturing group. The background of most engineers who work primarily with fluid power is rooted in a mechanical engineering degree from one of the many four-year institutions located throughout the states. Another area producing several engineers for fluid power is industrial engineering. Whereas mechanical engineers are concerned primarily with the production , transmission , and use of power, industrial engineers determine the most effective methods of using the basic factors of production, manpower, and materials. They are concerned with people and things in contrast to engineers in other specialties who generally are concerned more with developmental work in subject fields such as power and mechanics . Industrial engineering is an excellent background for sales related work and customer relations in the fluid power industry .

10

Introduction

Chap. 1

1-5 THE USE OF UNITS

English and SI metric units are used throughout this book. The basic units used from the English system are length, force, and time. Derived units include area, pressure, velocity, volume, and flow rate. The units of length are the foot and inch (ft, in.). The units of force are the pound and ounce (Ibf, oz). These same units are used for weight. The units for time are the minute and second (min, sec). Derived units are made up of basic units. Area is derived in square feet and square inches (ft ' , in."), pressure is in pounds per square inch (lbf/in."), velocity is in feet per minute or feet per second (ft/min , ft/sec), and volume is in cubic feet or cubic inches (ft ', in."). One gallon equals 231 in". The units for distance in the SI metric system most often used are the meter and centimeter (m, cm). The unit offorce is the Newton (N), and the unit of time is the second (sec) . Area is derived in square meters and square centimeters (rn-, cm-), The unit of pressure is derived in Pascals in honor of the French physicist. A Pascal equals one Newton per square meter (Pa, N/m 2 ) . This is somewhat confusing because the Newton also is an honorary name given after the English mathematician, Isaac Newton . Because the Pascal is small, the kilo Pascal (kPa), which is 1000 Pascals, and mega Pascal (MPa), which is one million Pascals, are used often in hydraulics. Velocity in the SI metric system is derived in meters per second (m/sec). Volume is derived in cubic meters or cubic centimeters (m', ern"). Flow rate is a derived unit, but it is not the same in hydraulics as in pneumatics. In hydraulics, flow rate is a liquid measure. In the English system the units are gallons per minute (gal/min), and in SI metrics liters per minute (l/min). In pneumatics, flow is a volumetric measure: cubic feet per minute or cubic feet per second in the English system (ft3/min, ft 3/sec), and cubic meters per minute or cubic meters per second (m3/min, m3/sec) in the SI metric system. In most cases the units should be included with the numbers when calculations are made. This is because without the units , the answer would be incomplete. And unless the units are included with each number in the calculation, there would be no easy way to check through the problem to see if errors have been made. The use of units also is important when conversions are made from one system to another, from liquid to volumetric measures, or from larger to smaller units. Conversion factors for derived units are difficult to remember so if one can work through the conversion with the units themselves, it usually is easy to find the conversion between base units like inches, centimeters , pounds, and Newtons.

1-6 HOW FLUID POWER WORKS

Fluid power works in accordance with the laws governing the behavior of the fluid itself. The two most common fluids used are air in pneumatic systems and oil in hydraulic systems . The beginning of an understanding of how fluid power works

Sec. 1-6

How Fluid Power Works

11

is attributed to Blaise Pascal (1650), who disco vered that the pressure exerted by a confined fluid acts undiminished the same in all dir ect ion s at right angles to th e inside wall of the container. Thi s is known as Pascal' s Law and is often thought of as the foundation of the discipline . The law can be ex te nded to include tr an smi ssion and multiplication of force , as show n in Fig. 1-8. Fluid po wer works by applying a for ce agai ns t a movable area. In a ty pica l fluid power system, a fluid power pump or co mpressor del iver s fluid th rou gh th e control va lve to a syste m actuator, suc h as a cy linde r, th rou gh lines at high pressure. U nuse d fluid in hydraulic syste ms is ret urne d to a reservo ir at low pressure for cooling, storage, and later use . Figure 1-9 illustrat es a simu lated

•

•

l------- --J ~r=r=At_-_t-l0

1 sq . in.

1.

A n input for ce of 10 pounds (44.8N) on a one square inch (6.45 cm 2 piston) . . .

2.

Develops a pressure of 10 pounds per square inch (I b f/ in~ or 68.95 kN /m 2 ) throughout the container.

3.

Th is pressure w ill suppo rt a 100·p ou nd (444 .8 N) weight if t his is a 10·square-inch piston

4.

Th e forces are proportional to th e piston areas.

sq. in.

Figure 1·8 Transmi ssion and multiplication of fluid power force.

Oil reservoir

Cyl inder

_~r ""

Figure 1·9

12

How fluid power works.

Introduction

Chap. 1

hydraulic system used for lowering and raising the nose wheel of an aircraft landing gear. Movement of the control valve to lower the wheel causes fluid to be delivered at high pressure from the pump to the blank end of the cylinder. To raise the wheel, the control valve directs fluid to the rod end of the cylinder. In both cases fluid power works by applying a force against the area of each side of the movable piston in the cylinder. Examining the definition and the example closer Force

= Pressure x Area

(I-I)

All we need to remember is that the units of force, pressure, and area must be consistent. This means that in the English system of measurement the weight or force is in pounds (lbf'), the area is in square inches (in."), and the pressure is in lbf/irr'. In SI metric units, the force is in Newtons (N), the area is in square meters (m-), and the pressure is in N/m 2 , which is given the name Pascals (Pa) after that famous physicist. Conversions among the various units used for pressure around the world' are given in Table 1-1. Only the lbf/in. ? and Pa will be used here. The Pa is the recognized international unit of pressure, but the kg/ern? and bar are still used in some countries in the world . TABLE 1-1

UNITS OF PRESSURE AND CONVERSION FACTORS

:;:

lbf/in. ?

Convert from

lbf/in. ? kPa

kPa

kgf /crn ?

bars

0.07

0.069

0.0101

0.010

I

0.98

Multiply by I

6.895

0.145

I

kgf/cm?

14.22

bars

14.5

98.07 100

1.02

I

pascal (Pa) = I N/m 2 Conversions are made by multiplying the units in the left margin by the conversion factors in the boxe s to arrive at the unit s across the top . Example: 1000 lbf/in. ? x 6.895 = 6895 kPa

When designing or applying fluid power to a system, the force required at the output is used to determine the pressure of the system and the cross section area of the cylinder. System pressures are also determined from the strength of components, cost factors, and safety precautions. When the system pressure is l Conversions are made this way l-lbf/in. ? = (I-lbf)(4.448 222-N/tbf)(I /in. 2 )(I550-in.2 /m l )

lbf/in .? x 6895

Sec. 1-6

= 6895 N/m 2 or

= N/m l = 6895 Pa = 6.895 kPa

How Fluid Power Works

13

known and the force that the sys tem must appl y is specified, the area of the cylinder then can be computed by solving the formula for this value. A rea

=

Force Pres sure

Exa mple 1

F=p XA

F A

A=

p= -

t

p

Figure 1·10 Force , pressure, and area relationships.

Pressure = 1500 lbf/in .?

Area = (0.7854)(4 in.' ) = 12.57 in2 Force

=p

x A

= (1500 1bf/in .2 )(l 2.57 in. ' ) =

18,855 lbf

To simplify calculations using thi s formula , the desired value can be determined using Fig . 1-10. By covering the desired value , the relationship between the other two is given in the proper order. That is, force equal s pre ssure time s area, pressure equals force divided by area, and area equal s force divided by pre ssure. Look at the example that follow s the figure , which as ks for the solution of the force that would be applied to the cylinder rod if the cylinder has a bore of 4 in. and the pre ssur e in the cylinder is 1500 lbf/in-. Man y helpful devices such as thi s have been developed to assist the fluid power mechanic and technician to make calculations. 1-7 SUMMARY

Fluid power is used throughout industry to accomplish work and de liver power from the source to some other location. Hydraulic and pne umatic systems have been put to a variety of uses , from the more common plant air systems and hydraulic production systems to more sophisticated systems used in airplanes, aerospace vehicles, and industrial robots. Miniature pneumatics and fluidic logic systems provide tremendous possibilities for fluid power system control. The modern era in fluid power was heralded by the use of applications made to suc h installations as the main armament system on the USS Virginia in 1906. Fluid power is now con sidered to be a support system in all indu stries, including 14

Introduction

Chap. 1

agricultu re, ae rospace, aviation, co nstruction , manufacturing, materials hand ling, and man y othe rs . Fluid power is a multi-billion dollar industry ann ually . While several factors account for the growth of the industry, the primary one is the need for more flexible power for a variety of uses in the durable goods production, agricultural, construction, and services industries. Career op portunities are pro mising for fluid power mec hanics , maintenance men , tec hnicians, and engineers. Certification and extensive training are now avai lable. This is in response to the demand for qualified personnel and the efforts of such groups as the National Fluid Power Association , Fluid Power Education Foundation, and Fluid Power Society over the last several years to give recognition to this important and growing segment of highly skilled professionals. STUDY QUESTIONS AND PROBLEMS 1. 2. 3. 4.

5.

6. 7. 8. 9.

10. 11. U.

List ten hydraulic applications and te n pneumatic applications. List the components on one hydraulic application and give their specifications . List the components of one pn eum atic application and give their specifications. Con sult the Fluid Power Society to determine the current requirements to become a certified fluid power specialist. List five schools in your state which offer fluid power course s or a program in fluid power leading to a degree . Compute the are a and volume for a cylinder with a 2-in. bore and a 12-in. stro ke. Compute the bore of a cylinder which must exert 10,000 Ibf with a pre ssure limitation of 1200 lbf/in' . Compute the force available from a hydraulic cylinder with a bore of 75 mm under a pressure of 10 MPa. What bore would be necessary on a cylinder operating at 15 MPa to exert a force of 65 kN ? A log splitter is to exert a 10,000 Ibf force using a 3-in. bore cylinder. At full capacity, what will be the system pre ssure? How much fluid is needed to stroke a cylinder with a 100 mm bore and 0.50 m stroke? Co nstruct a table of values which lists the force in Ibf available from cylinders with bore diameters of2 , 4,6,8 and 10 in., and pre ssure s of 400,800, 1200, 1600, and 2000 lbf/in", Use a table like Table 1-2 to record the value s. TABLE 1-2

CAPACITY OF HYDRAULIC CYLINDERS IN POUNDS FORCE

Pressure (lbf/in ,")

400

800

Diam et er (in.) 2 in. 4 in. 6 in. 8 in. 10 in.

Chap. 1

Study Questions and Problems

1,200

1,600

2,000

----- --

--- - 15

13. Construct a table using the same values as in Problem 12, but this time convert the diameter of the cylinder to mm, the pressure to MPa and the force to MN . 14. If a hydraulic cylinder has a rod diameter half the size of the bore, what will be the difference in force available if the pressure applied alternately to each end remains constant? (Clue: Try two convenient values such as 4 in. for the piston and 2 in. for the rod diameter.) 15. In Problem 14, what would happen if the pressure were applied to both sides of the piston at the same time? (Make a drawing to prove your answer.)

16

Introduction

Chap. 1

---

CHAPTfR

2

BASICS OF HYDRAULICS

2-1 INTRODUCTION

Ther e are a number of basics that explain how a fluid power system work s. One of these has to do with raising its energy level by incre asing the pre ssure within the fluid. This is done by a pump moving fluid against the load resistanc e . Anoth er explains how a small force by a per son operating the hydraulic system, for example on a hydraulic jack handle, can produce a very large force on the object being lifted . Still other prin ciple s have to do with the flow of fluid in the sys tem as it tran sfer s fluid power from one place to another in a sys tem. The concepts of work, power , and horsepower show what can be accomplished with a fluid power system. The output of fluid power system s is measured in fluid horsepower and this puts it all together. Finally the measurements of torque and speed show how fluid hor sepo wer can be broken down to size components for the sys tem. 2-2 HOW OIL TRANSMITS POWER

The two cylinder s in Fig. 2- 1 are conn ected by a piece of tubing so that when one of them retracts, the other extends. This is how fluid power is transmitted by the pre ssurized fluid moving from one place to another. If both cylinders have the same bore, the force at the output cylinder as well as the speed are the same as the forc e and speed applied to the input cylinder. There are some friction losse s, of course, but in simplified term s, what enters at the input will exit at the output. This is really an application of the law of co nservation of energy whic h simply mean s that energy can neither be created nor de stro yed. And while you don 't get something out of the sys tem for nothing , you do get out everything that is put in , minus a small amount which is accounted for by friction and slippage in the parts. Thi s loss usually is no more than 5-10%. Keep in mind that during the time that 17

t

Distance = L

~ Extend

Retract Fo rce = F

/

O utput

Input Work = F X L

Figure 2-1 Transmission of force .

the system is transmitting fluid power from one cylinder to the other the pressure remains the same everywhere . It is the same in both cylinders and in the line between the cylinders. This is really just another demonstration of Pascal's Law that says the pressure in the system is the same everywhere . There may be a slight drop in pressure when the fluid flows from one cylinder to the other, if the line is very long, but again, this can be accounted for by flow friction losses.

2.3 WORK PERFORMED BY THE SYSTEM Notice that work was accomplished by the output cylinder in Fig. 2-1. By this we mean that the cylinder rod moved a force F through a distance L. This is the definition of work, and the units are always in ft-lbf or m-N. The formu la used is Work (W)

= Force (F) x Distance (L)

or simply (2-1)

W=F xL

Example 1 Let's say that the weight (N) represented 1000 Ibf (4448 N) and the distance was 15 in . (38 ern) . Then the work from the system would be

W

=

(1000 Ibf)(l .25 ft)

=

1250 ft-Ibf

In SI units the solution is the same but with different units : W

18

=

(4448 N)(0.381 m)

=

1695 N -m

Basics of Hydr auli cs

Chap . 2

2-4 M ULTIPLICATION OF FORCE

Now what happens is the input piston , called the pump , and the output piston , called the linear actuator, have different are as? Well, as might be expected , if the pump piston is smaller than the actuator piston , then the force is greater at the output. Thi s multiplication offorce isn't free , of course , and what happens is that when the force increases, the distance that the output piston rod travels is less. Since the pressure in the system and the work accomplished rem ain the same , there must also be a balance between the input and output force s and distances traveled . This says simply that the work input equals the work output. Figure 2-2 shows how the force is multiplied as fluid moves from the pump cy linde r to the ram. We can call this a ram bec au se the cylinder rod is at least half as large as the cylinder bore . The balance is given by the formula (2-2) When three of the variables in the formula are known, the fourth can be determined simply by substitution. In practical examples, usually one of the ratios is known or three of the four quantities are given by the de sign requirements of the system. Let' s look at an example.

Figure 2-2

Multiplication of force .

Example 2 An operator ca n exert 100-lbf (444.8-N ) on the pump piston . If the ram must lift 2000 Ibf (8896-N) a distance of 12 in. (0.3048 m), how far would the pump piston have to travel? Look at the hydraulic jack sys tem in Fig. 2-3. Th is allows the operator to move the handl e up and do wn to lift the weight. When the handl e is lifted , fluid from the reservoir is drawn into the pumping ch amb er at atmos pheric pre ssure . When the handle is shov ed do wn , the check valve to the right opens and the fluid ente rs the ram at high pre ssure . You ca n see that the left che ck valve ope ns on the up stro ke, and the right one op en s on the do wn stro ke when the left one closes. Now to solve the problem, what must be don e is to solve the formula for L) . L ) = L2

X

F2 (2000 Ibf) . • F ) = (12 in.) (l OO lbf) = 240 In .

In SI unit s (8896 N) L) = (0.3048 m) (4448 N) = 6.096 m

Sec. 2-4

Multiplication of Force

19

Vent

.J!'----~

r..

Reservoir

~2/

t

t

t

Check :.. valve

Figure 2·3 Single-acting hydraulic system.

Now all that would have to be done would be to set the stroke of the pump to some convenient unit, for example 1 in., and you would know that it would take 240 strokes to lift the weight. That's really a small price to pay for the tremendous force that can be exerted by the ram. There is another interesting way to look at the balance between the pump and the ram, and that is to begin with the pressure balance. That is, the pressure in the system is the same everywhere. This is taken from the Bernoulli principle and applied to both the pump and the ram cylinder chambers which hold the fluid . This can be shown with a simple formula: (2-3) Example 3 Look again at the system shown in Fig. 2-3. If the pump piston had a diameter of 1/4 in. (6.35 mm) and the ram a diameter of 1 in . (25.4 mm), how much weight could be lifted by a downward force on the pump piston of 150 lbf (667.2 N)? Here the problem is almost the same as the previous example, except that we 're solving for F 2 • 2

A2 (15 Ibf (0.785 in. ) 31bf . F2 = F I A = 0 ) (0.049 in. 2) = 240 ,Just over a ton J

In SI units F2

(506 mm 2)

= (667.2 N) (31.7

mm-)

.

= 10659 N, wntten as 10.7 KN

This makes it relatively easy to see why hydraulics is the natural choice when it comes to multiplying a small force from a person to provide a large force to lift or move a heavy object. The largest hydraulic jack in Fig. 2-4, for example , has a capacity of 100 tons , yet its lightweight alloy construction permits an aver20

Basics of Hy d raulics

Chap . 2

Figure 2-4

Hydraulic jacks (courtesy of Charles S. Madan Co.• Ltd.).

age workman to carry it from place to place. It is also common with these large capacity jacks to have a progressive pump. This means that when the load is light, the larger pump piston can be selected, and as the load gets progressively heavier, the smaller pump piston is used. This saves the operator a great deal of time because when the smallest pump piston is used, the movement of the ram piston is very slow . 2-5 POWER AND HORSEPOWER

Work and power are different. Where work means to move an object from one place to another, power introduces the element of time. It is not enough to just move an object with a force through some distance. It must be done within a specific time, for example, one minute, two minutes, or an hour. This is what power means , the rate at which work is done , and the formula is F « t

t.

W t

P=-- =-

(2-4)

When the power is known it gives a convenient way to describe the output of many machines , but it doesn 't give us a convenient wa y to compare one machine with another, for example, two hydraulic motors of the same size. James Watt had this same problem when he was trying to compare his steam engines to horses used in the mines back in the late 1800s. To solve the problem, he tested the capacity of draft hor ses to lift weight s that were hung over pulleys. The actual layout was similar to Fig. 2-5. After several tests he estimated that a strong horse could raise 150 Ibf about 37,13 ft/sec which works out to be about 550 ft -lbf/sec. While the value of a horsepower turned out to be high for most horses , this has been the standard for rotating machinery since that time. The formula is F x L

llP Sec. 2-5

W

= t x 550 = t x 550

Power and Horsepower

P 550

(2-5)

21

-1 ~

l I

,"1

~- ..J.,

> __

\

:_W')

w, )

l::L_ _----f::.i.---=-- _ _

F :1501bf

Figure2-5 Horsepower.

and since the top and bottom have the same units, horsepower is dimensionless. Its simply called horsepower, without any units. The Sl metric equivalent of power is measured in watts, and the formula for horsepower is P

HP

(2-6)

= 746

It is also dimensionless. All that really must be remembered is that there are 550 ft-Ibf/sec in an English horsepower, and 746 watts in an Sl metric horsepower. Let's look at a couple of examples to see how the formula works. Example 4 What horsepower output would result if the horse in Fig. 2-5 lifts 800 Ibs 20 ft in 20 seconds when pulling at full capacity. That sounds like a strong horse, but let 's put the numbers in the formula to see (800 Ibf)(20 ft) That is a very strong horse. Notice in the formula that all the units cancel. This is important because if they don't, it indicates there's an error somewhere in the solution. Example 5 An electric motor is to be coupled to a 3 hp hydraulic pump (Fig. 2-6) . If the motor will draw 2000 watts at its rated output, will it do the job? First of all it has to be assumed the motor will operate at 100% efficiency . which it won't. but that gives us a place to start. Using the SI metric horsepower formula for the motor HP

22

= (2000 watts) = 268 h (746 watt s)

.

p

Basics of Hydraulics

Chap. 2

Motor

Tank

Inlet

Figure 2-6 Motor -pump assembly.

No , the motor won't do the job, even at 100% efficiency. What would likely happen is that it would overload and overheat when the pump was at full capacity.

2-6 FLOW RATE IN A SYSTEM In mechanical systems, horsepower is defined as the product of force and distance, divided by the product of time and a constant, 550 ft-lbf', because these are common units. In hydraulics, however, the common units are pressure and flow rate, and these are used to compute fluid horsepower. In hydraulics, the flow rate Q is measured in gal/min and liter/min. It is a measure of the displacement volume of fluid V, divided by the time t. The general formula for flow rate is Q

=

Displ~cement time

=~ (

(2-7)

For hydraulic cylinders the displacement equals the product of the bore area and stroke.

V=A xL where V A

L

= displacement in in.Vstroke = cross section of the bore in = cylinder stroke in in.

in

2

Pumps and motors with vanes or gears instead of pistons have a different internal geometry, but the displacement is still measured in in.Vrev. In the metric system, displacement is measured in m 3/rev or mrn-/rev , depending on the size of the unit. When the displacement of cylinders, pumps , or motors is used to compute the flow rate, the displacement in in.Vstroke or in.Vrev must be converted to gal/ min. Thus the general formula for flow rate becomes Q

=

V(in .3/rev) x l(gal) (min) x 231(in. 3)

Sec. 2-6 ' Flow Rate in a System

=

V ( x 231

--=-=.-7

(2-8)

23

For hydraulic pumps and motors, the general formula for computing the displacement is (2-9)

And if the pump or motor has cylinders, the flow rate equals (A x L x n) x N

Q= where Q

=

(2- 10)

231

flow rate in gal/min

L

= stroke in in.

n

=

no. of cylinders

N

=

speed in rev /min

Example 2 Compute the displacement of a 5-cylinder piston pump having a II/S in . bore a nd a I in. stroke . What would be the flow rat e if the pump turns at 1200 rev /min ?

V - 3. 14 -

X

(l Vs in.}2 x I in. x 5 cyls/rev _ 4 97· 3/ 4 - . m. re v

Fin ally , computing th e flow rate

Q

=

V x N 23 1

= (4. 97 in.3/ rev)

x (1200 re v/min) 231 in.Vga l

= 258

1/ . . ga min

2-7 FLUID HORSEPOWER

When a cylinder must tran smit power in two directions, a system like that shown in Fig. 2-7 can be used . The components of the system are the reservoir, gear pump, pre ssure relief valve to pre vent overloading, the four-wa y control valve , double-acting cylinder, and the connecting tubing. When the control valve is in the neutral position , the oil circulates through the pump, center spool, and back to the reservoir and the system is idling. When the hand-operated control valve is shifted, the system begins to perform useful work. Shifting the operator handle to the left directs oil from the pump at high pres sure to the cap end of the cylinder and moves the load to the right. At the same time , oil leaving the rod end of the cylinder returns to the reservoir through the control valve at low pre ssure. Shifting the operator handle to the right directs oil to the rod end of the cylinder which returns the load to its original position and again the oil from the blank end of the cylinder returns to the reservoir at low pre ssure through the control valve . What happens when the cylinder reaches the end of its stroke , or encounters an immovable object? When the se situations occur, the cylinder stalls and the relief valve opens to allow the oil from the gear pump to return to the reser voir without damaging any of the components in the system . This is another great advantage of fluid power systems . 24

Basics of Hydraulics

Chap. 2

Vent

- - -

Reservoir

Cylinder

Figure 2-7 Double-acting hydraul ic sys tem.

The fluid power that a sys tem delivers equals the pressure multiplied by the flow rate. In English unit s the formula is FHP

=P

x Q

=

(Ibflin . 2)(gal/min)

1714

1714

(2-11)

In SI metric units, the formula is FHP

=P

x Q

=

(kPa)(Iiters/min)

44 760

44 760

(2-12)

Notice that the formulas are followed by the unit s that mu st be used to make the answers correct. Example 6

A simple ex ample would ask a question like what is the fluid horsepower of a system operating at 2000 lbf/in .? delivering 5 gal/min ; and the an swer would be 2

FHP = (2000 Ibf/in. )(5 gal/min) = 583 h 1714 . P

Sec. 2-7

Fluid Horsepower

25

Co mpleti ng the sa me exa mple agai n in SI metric unit s, not ice how they are co nverted from the E nglish unit s

FHP = (2000

Ibf/i n. 2)(4.4 48

Nllbf)(I 550 in. 2 / m2)(5 gal/ min)(3.785 l/gal)(O.oo l m3 /1) (44 760)(m . N/m in)

and

FHP = 5.83 hp Or , sta ted in SI metr ic units, what is the horsepower tran smitted by a system deli vering 23 liters/m in at a pressur e of 13 793 kPa. Sub stituting in the SI metric formul a

FHP

= (13 793)( 18.9 liter s/min) = 583 (44 760)

.

h P

which is another way of say ing the sa me thing. For con ven ience , Table s 2-1 and 2-2 are used to simplify computing the FHP when the pressure and flow rat e are known. Perh ap s th e easiest way to co nvert p x Q in English unit s to SI metri c unit s is to multipl y by 26.1. Example 7 Con verting a flow rat e of 5 gal/ min at 2000 lbf/in.? to a flow rate in liter s/m in at a certain pressur e in kPa (5 gal/ min)(2000 Ibf/in . 2)(26.1)(I /gal)(N /lbf)(in. 2fm 2 )(kN /N ) = 261 000 The co nversion of the unit s see ms overpo we ring so j ust to keep it simple rem ember to use the co nve rsion fac tor of 26. 1 and refe r to this example onl y when a refres her abo ut the units is necessary.

2-8 TORQUE AND TORQUE r ORSEPOWER Torque

Pumps receive a twisting effort called torque to make them turn. Motors deliver a torque at their output shaft. Torque simply means to make an effort to twist or turn and is measured as a force multiplied by the distance along the radiu s of the shaft. Figure 2-8 illustrates a physical example of torque where a per son exerts a 50 lbf pulling force on a pipe wrench 12 in. from the center of the pipe. Thu s the pipe receives a torque of 50 Ibf-ft. You may have noticed that torque is measured in the same unit s as work , but conceptually the y are different. Work requires that the force move an object some distance. Thu s, to perform work , movement is required. In Fig. 2-8 when the pipe becomes tight the wrench will stop even though it will continue to exert a turning force on the pipe. The same is true with fluid power applications. A gear motor may stall, for example , and continue to deliver a torque at its output shaft. In fact , the torque will typicall y increa se at stall because the pre ssure will increase to the relief valve sett ing. This is another ad vantage of fluid power. Stalling does 26

Basics of Hydraul ics

Chap. 2

TABLE 2 -1

FLUID HORSEPOWER (ENGLISH UNI TS )

Q From the formula FHP = PI;14

N

.....

Gallon s/ minute

100 Ibf/in.?

200 lbf/in .?

300 Ibf/in . 2

400 lbf/in .?

500 Ibf/in .2

750 lbf/in. ?

1000 lbf/in .?

1500 lbf/in. ?

2000 Ibf/in . 2

2500 Ibf/in. 2

3000 lbf/in .?

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 5 6 7 8 9 10

0.029 0.058 0.088 0.1 17 0.146 0.175 0.204 0.233 0.292 0.350 0.408 0.467 0.525 0.583

0.058 0.117 0. 175 0.233 0.292 0.350 0.408 0.467 0.583 0.700 0.817 0.933 1.050 1.167

0.088 0.175 0.263 0.350 0.438 0.525 0.613 0.700 0.875 1.050 1.225 1.400 1.575 1.750

0.117 0.233 0.350 0.467 0.583 0.700 0.8 17 0.933 1.167 1.400 1.634 1.867 2.100 2.333

0.146 0.292 0.438 0.583 0.729 0.875 1.021 1.167 1.459 1.750 2.042 2.333 2.625 2.917

0.219 0.438 0.656 0.875 1.094 1.313 1.532 1.750 2.188 2.625 3.063 3.501 3.938 4.376

0.292 0.583 0.875 1.167 1.459 1.750 2.042 2.333 2.917 3.50 1 4.084 4.667 5.251 5.834

0.438 0.875 1.313 1.750 2.188 2.625 3.063 3.501 4.376 5.251 6.126 7.001 7.876 8.751

0.583 1.167 1.750 2.334 2.917 3.500 4.084 4.667 5.834 7.001 8.168 9.335 10.502 11 .669

0.729 1.459 2.188 2.917 3.646 4.376 5.105 5.834 7.293 8.751 10.210 11.669 13.127 14.586

0.875 1.750 2.625 3.501 4.376 5.251 6.126 7.001 8.751 10.502 12.252 14.002 15.753 17.503

N 00

TABLE 2-2

FLUID HORSEPOWER (SI M ETRIC UN ITS)

p XQ From the formul a F H P = 44 760

L iters/ minute

500 kPa

1000 kPa

1500 kPa

2500 kPa

3500 kPa

5000 kPa

7500 kP a

10 000 kPa

12 500 kP a

15 000 kPa

17500 kP a

I 2 3 4 5 10 15 20 25 30 35 40 45 50

0.011 0.022 0.034 0.045 0.056 0.112 0.168 0.223 0.279 0.335 0.391 0.447 0.503 0.559

0.022 0.045 0.067 0.089 0.1 12 0.223 0.335 0.447 0.559 0.670 0.782 0.894 1.005 1.117

0.034 0.067 0.101 0. 134 0. 168 0.335 0.503 0.670 0.838 1.005 1.172 1.340 1.508 1.676

0.056 0. 112 0. 168 0.223 0.279 0.558 0.838 1.117 1.396 1.676 1.955 2.234 2.5 13 2.793

0.078 0. 156 0.235 0.3 13 0.391 0.782 1.173 1.564 1.955 2.346 2.737 3.128 3.5 19 3.910

0.112 0.223 0.335 0.447 0.559 1.1 17 1.676 2.234 2.793 3.351 3.910 4.462 5.027 5.585

0.168 0.335 0.501 0.670 0.838 1.676 2.513 3.351 4. 189 5.027 5.865 6.702 7.540 8.378

0.224 0.447 0.667 0.893 1.117 2.235 3.351 4.468 5.585 6.702 7.819 8.937 10.054 11.17 1

0.280 0.558 0.835 1.117 1.397 2.793 4.188 5.585 6.982 8.378 9.774 11. 171 12.567 13.963

0.336 0.670 1.002 1.340 1.676 3.352 5.026 6.702 8.378 10.054 11.729 13.404 15.080 16.756

0.392 0.782 1.169 1.563 1.955 3.91 1 5.864 7 .819 9. 774 11.730 13.684 15.639 17.593 19.549

~ ?? 50 Ibf pull

Tor que = 50 1bf X 1 ft = 50 1bf - ft

Figure 2-8 Torque.

not damage the sys tem. Look at Fig. 2-9 . The sprocket receives a torque from a hydraulic motor to give the ch ain a linear motion . There is a torque at the sprocket whether the chain is moving or not. What is important to understanding the concept of torque is that the force is delivered at right angles to the radius at some distance from the center of the sprocket. The formula for torque is T= F x D

Notice that the torque is mea sured in Ibf-ft or N-m, where work is measured in ft-Ibf or m - N . Thi s is done by general agreement to keep the two separate whe n calculations are made . Torque Horsepower

Knowing the torque gives us a va luable tool to compute the horsepower of turning machiner y such as pumps and motors. There are two formulas that are used the

Sprocket

Hydraulic motor

5 gal/min at 1000 Ibf/in?

Figure 2-9 Torque delivered to a sprocket by a hydraulic motor.

Sec. 2-8

Torque and Torque Horsepower

29

most. In Engli sh unit s , 21TN T 60 THP = ----s5()

21TN T

=

33 ,000

=

N x T

(2 - 13)

5252

Where N is re v/m in and T is in lbf-ft. In SI metric unit s THP

=

21TN T 60 ~

=

21TN T 44 760

N x T

(2 - 14)

= 712l

Where N is rev/min and T is mea sured in N . m. Example 8

Now let us look aga in at a motor receiving th e delivery of 5 gal /min at a pre ssu re of 2000 lbf/in .? from the last example, and mak e a computation for the torque th at might be delivered at the output shaft turning at 500 rev /min . Thi s would ass ume 100% efficiency , of course , but that matter will be de alt with later. All th at has to be done is to solve the torque hor sepower formul a for torque T. Here is the computation . T

= ( TH P)(3 3,000) = (5.83)(33,000 ft-Ib s/min) = 61 27 Ibf-ft 27TN

(2)(3. 14)(500 rev /min)

.

.

Notice th at torque is measured in Ibf-ft , wher e wo rk is measured in ft-Ibf.

2-9 TORQU E HORSEPOWER AND FLUID HORSEPOWER RELATIONSHIPS

The que stion about what the difference is bet ween fluid horsepower and torque horsepower may have occurred to you during the pre viou s discu ssion. Well , there reall y isn ' t any, a hor sepower is a hor sep ower ; ju st the variables in the formula are different. For that rea son , the two ca n be set equal to each other and used to solve for any one of the variables if the other three are known , and frequently the y are. Setting the two equal in English unit s p XQ

FHP

N xT

=

THP

(2-15)

= 7T22 =

THP

(2-16)

= I7T4 =

525 2

The SI metric equi valent is FH P

p XQ 760

= 44

N xT

The formulas can be used to size a fluid power component when the pre ssure and flow rate are known , and the desired speed or torque are specified by the application . Let's tr y two applications. Example 9

Th e pre ssure of the co mpo nents in a fluid power sys te m have a rating of 2000 lbf/in.? a nd we wish to use a hydraulic motor which co uld turn a drum hoi st similar to tha t in Fig . 2-10 at 20 rpm with a torque of 2000 Ibf-ft. Wh at would be the flow rat e and horsepower of th e motor? First , we should se t the two horsepower formulas equal and solve for th e flow rate , and then follow up by solving for the horsepower.

30

Basics of Hydraulics

Chap. 2

Drum rad ius

w Figure 2-10 Hoisting drum .

In English units

_ (1714)(20 rev /min)(2000 Ibf-ft) _ 65 II . Q(5252)(2000 lbf/in .") - . ga min and the horsepower would be

FHP = (2000 Ibf/in . 2 )(6.5 gal/min) = 76 h (1714) . P The metric equivalent would be solved either by knowing the va lues for the pre ssure and torque in SI units , or by converting the pre ssure to kPa and torque to N . m. Solving the problem that way

_ (44 760)(20 re v/min)(2712 N . m) _ 2471 I' I ' Q(7122)(13 793 kPa ) . iters rnm and the horsepower would be

FHP

= (13793

kPa)(24 .71 liters/min) (44 760)

= 76 .

h P

which is the sa me an swer. Example 10 Another problem is to consider how much torque might be expected from a hydraulic motor turning at 750 rpm , operating at 1000 Ibf/in. 2 , and receiving fluid at 5 gal l min (see Fig . 2-9). Here we are asking about expected torque. The problem could just as easily de scribe a pump as well as a motor, except for losse s which will be de scribed later. Now look at the solutio n. In Engli sh unit s , using Eq . 2-15

T

= 5252

x p x Q 1714 x N

= 5252

x 1000 x 5 1714 x 750

= 20 4 Ibf-f .

t

Lo sse s will be co vered in the discu ssion of pumps, mot or s, a nd c ylinders. but a few words here will expl ain the general concept. Simpl y put , it take s more than the Sec. 2-9

Torque Horsepower and Fluid Horsepower Relationships

31

requ ired fluidpower output to turn a hydra ulic pump input shaft. Ta ke anothe r look at the motor pump unit in Fig . 2-6 . There are losse s in the dri ving motor, in the coupling , a nd in the pump itself where slippage and friction eat away at th e input. The sa me is true of hydraulic motors whic h receive more fluid po wer th an the y deliver as mechanical power at the output shaft. It comes do wn to thi s. If a hydraulic pump is 85% efficient it will take about 118% at the mechanical input to give 100% of the fluid horsepower specified at the output. The same is true of a hydraulic motor which receives 100% of the requirement in fluid horsepower at the input , but will deliver only 85% of that to the output shaft as mechanical horsepower. It's not a great problem as long as it' s understood that in actual applications the losse s must be figured in.

2-10 SUMMARY

What makes a fluid power system is the pressurized fluid. When the fluid move s a cylinder, work is accomplished. If the movement is timed , the horsepower can be calculated using James Watt's horsepower formula, the fluid horsepower formula, or the torque horsepower formula. Torque and twisting effort mean the same thing . Torque and work have the same units but conceptually they're different. Work requires movement, torque does not. If the shaft moves , torque horsepower can be calculated. Three advantages of fluid power systems are: (I) tremendous force multiplication occurs, (2) no harm is done to the system should it stall, and (3) torque output continues even when the hydraulic motor is stalled.

STUDY QUESTIONS AND PROBLEMS 1. The cyli nder in Fig . 2-11 is used to raise the weight 9 in. If the bore of the cylinder is 21/2 in. and the pressure is 1800 lbf/in.", how much work is accomplished ? 2. An automobile lift raises a car weighing 3000 lbf', 6 ft off the sho p floor (Fig. 2-12 ). Assuming the lift itself weighs 500 lbf a nd th at friction is negl igible , wha t is the work necessary to raise the car in ft-Ibf a nd m-N ? 3. If the bore diameter of the lift c ylinder in Problem 2 is 6 in. , wh at will be the pressure in the system when it comes to re st in the raised position ? 4. In Problem 2, what would be the power expended if the lift rai se s the car in 15 sees? Compute the answer in ft-Ibf/min and hp . 5. If the auto lift cylinder in Problem 2 has a diameter of 8 in .. how fast would it have to desce nd for the flow rate to become 10 gal /min ? (Clu e: set t = I.) 6. What would the return flow rate in Problem 5 have to be set at for the auto to descend in 10 sees? 7. A double-acting cylinder with a 2.5-in . bore , a I-in. diameter rod, and 14-in. stroke cycles 25 times per minute. What is the flow rate ? 8. What is the flow rate to a c ylinder with a 3-in. bore and 22-in. stro ke that full y extend s in 8 see s? 9. How man y seconds should it take a hydraulic cylinder with a 4-in . bore a nd 24-in. stro ke to fully extend if it receives fluid at 9 gal/min ?

32

Basics of Hydraulics

Chap. 2

,

.

.' '.

, '

- - Fluid Input

Figure 2-11

Proble m I.

10. What is the bore of a cylinder that extends 18 in. in 6 seconds while receiving fluid at 4 gal /min ? 11. How man y in./ sec should a cylinder with 1.5-in. bore extend if it receives fluid at I gal / min? 12. What is the minimum displacement of a gear motor turn ing at 800 rev /min if the flow rate to the motor is 6 gal/min ? 13. What flow rate would be required to turn a gear motor with a displacement of 2 in.Vre v at 1000 rev /min ? 14. What is the displacement of a motor that turn s 300 rev /min when the flow rat e is 10 gal/ min?

15. If a hydraulic motor with a displacement of 1.65 in.Vrev rece ives fluid at 20 gal/ min, what would be its maximum output speed? 16. A low-speed , high-torque piston motor with 5 cy linde rs having a bore of 1.5 in. and stroke of 1.75 in. turn s at 200 rev / min. What is the flow rate to the mot or? Chap. 2

Study Questions and Problems

33

Figure 2-12 Problem 2.

17. What size cylinder connected to a 5 gal/min (22.7 I/min) pump would be required to limit the extension velocity to 2 It/sec? 18. Compute the flow rate to a cylinder with a 4-in. bore extending at 3 ft/min. 19. Compute the fluid horsepower of a system with a 5 gal/min pump transferring fluid to a cylinder at a pressure of 1800 lbf/in". 20. A hydraulic cylinder is used to lift a load of 750 lbf a distance of 4 ft in 3 sees. Compute the power and fluid horsepower. 21. A wheel motor similar to that in Fig. 2-13 must exert a traction force of 1500 lb. If the outside diameter of the wheel is 30 in., what would be the torque from the motor? 22. Ignoring the friction , if the wheel motor in Problem 21 receives fluid from a 7 gal/min pump, and the pressure is limited to 2500 lbf/in .", how fast would the wheel motor turn and what would be the ground speed of the vehicle? 23. A 5-in. 3 displacement motor is to be tested on a bench at 3000 rpm at 2500 lbf/in", Assuming 100% efficiency , estimate the FHP output of the motor? 24. A cylinder with an area of 4 in.? raises a load 2000 Ibf, 2 ft using fluid supplied by a single-acting hand pump with a cylinder area of 0.25 in.? and a 3-in. stroke . Assuming no losses , compute the following : (a) pressure in the system (b) number of strokes of the hand pump (c) operator force exerted on the pump piston

34

Basics of Hydraulics

Chap. 2

Hydraulic wheel motor

Figure 2-13 Problem 21.

25. How long would it take a pump delivering 25 gal/min to extend a 5-in. cylinder with a 30-in. stroke? 26. In Problem 25, if the rod in the cylinder were 1.5 in., how long would the cylinder take to retract? 27. An operator pump s a foot-operated single-acting floor jack through a IO-in. stroke at a rate of 40 strokes per minute . The 2000 lbf car being raised is observed to move at the rate of 12 in. per min. What is the multiplication of force ? 28. A hydr aulic sys tem powered by a 10 hp motor operate s at 2500 lbf/in". At maximum capacit y, and ass uming no losses, what is the flow through the system in gal/min and liters/min ? 29. A pump with a 10 hp rating operates at 1500 rev /min . What is the torque in Ibf-ft and N-m? 30. Assuming 100% efficiency, what displacement motor would be required to deliver 200 Ibf-ft of torque at 1000 rev/min if the pre ssure is limited to 2500 lbf/in"?

Chap. 2

Stud y Quest ions and Problems

35

_ _ _ _ _ _ _ _ _CHAPTER

3

ENERGY IN HYDRAULIC SYSTEMS 3-1 INTRODUCTION

H ydr aulic fluid flowing in a high pre ssure sys tem contains p ot ential ene rgy due to its initial elevation and pressur e energy de velop ed by the pump , and kin etic or flow energy which res ults from the movement of the fluid through the plumbing and compone nts . Energy is added to the sys tem through the pump and subtracted from the sys tem through cy linde rs and hydraulic motors moving the load resistance. And since energy ca n neith er be cre ated nor destroyed , the tot al energy of the sys tem rem ain s con stan t. Thi s is the basis for the Bernoulli equation. The pot enti al ene rgy due to ele vation (EPE) is mea sured in unit s of head or pressure . It result s from the distance the reservoir is mounted above or below the pump inlet , and the specific gravity (Sg) of the fluid. For water and high water content fluids, the Sg is approximately 1.0. Petroleum ba se hydraulic oils have an S g in the rang e of 0.75 to 0.90; whereas the Sg of synthetic fluids such as phosphate esters has a range of 1.02 to 1.20. In the English system of mea surement , this est ablishes the specific weight of the fluid between 35 and 75 lb/ft", or 7 and 10 lblgal. The pressure energy (PE) of the fluid added by the pump worki ng against the load resistance has the same unit s as the energy due to elevation, but it resu lts from energy con sum ed by the prime mover driving the pump . Most of the energy in a hydr aulic sys te m is att ributed to the PE supplied by the pump, and thi s is where the energy co sts are incurred. The kinetic energy (KE ), which is usuall y a small portion of the total , accounts for flow losses. It is measured as a head loss or pre ssure drop across restrictions , valves , fittings, and hydr aulic tubing. Thu s, the KE term is a measure of the efficienc y of the sys tem. 36

3-2 PRESSURE AND HEAD

Hydraulic head is the heig ht in feet or meters that a liquid can be elevated for a given pressure which is measured in lbf/in .? or Pascals (N /m 2 ) . Hydraulic head is sometimes referred to as pressure head. For every pressure given for a specific liquid, there is a height that can be assigned that indicates the elevated length that a column of the sa me fluid can attain. The term head receives widespread use in hydroelectric applications where the turbine is po sitioned many feet below the surface of the re servoir that provides the motive force . It also provides a means of expressing the potential energy of the liquid fluid direc tly if the specific weig ht of the fluid is known . Frequentl y it is de sirable to express head as pre ssure , and conversel y, pressur e as head . From the definition of pre ssure , we ha ve Pressure (p )

Force (F) Area (A)

=

The pressure exerted by a I ft3 cube of water with a specific weight of 62.4 lbf/ft? (Fig. 3-1) is f f - 62.4 lbf/ft! _ lbf/i 2 t' 144 in. 2/ft2 - 0.433 m . lor each to head ( h )

p -

Thi s indicates that the force on I in.? of the lower surface area of I ft3 of water in the exam ple is 0.433 Ibf. Since the water occupies 1 ft ', 1 in.? of the lower surface area ha s a dimension of the base of a co lumn of water 1 ft high . The force exerted by each of the se columns with a l-in.! ba se, mu ltipl ied by a I-ft length, against the

/

12"

Water 62.4 lb.

1" 1"

J;2

For ce on each = 0.433 lb.

If-o.--.

sq. in.

/ I 1

"......"

- -12 "

.~

------i~~~

12"

Figure 3-1 Pressur e, head and specific weight.

Sec. 3-2

Pressu re and Head

37

base of the container is 0.433 lbf, and is the equivalent of 1 ft of head . So that p

= h x 0.433

(3-1)

or solving the expression for head H ea d (h)

= Pressure (p) 0.433

p h - 0.433

and h

= p x 2.31

(3-2)

In metric units since 1 m3 cube of water weighs 9802 N p = 9802 h Pa(N/m 2 )

(3-3)

and h

=

9:02 = 1.02 x 1O-4p meters

(3-4)

Notice in developing the relationship of pressure, defined as a force per unit area, that the specific weight of water was used as the force per unit cubic foot and cubic meter. In fluid power systems, the medium is oil which usually weighs less than water and because of this difference in specific weight, expressions to convert pressure to head must be altered. Water has a specific gravity of 1. Oil has a specific gravity in the range of 0.75 - 0.90, and 1 ft3 of this fluid would weigh about (0.80 x 62.4) 50 lbf rather than 62.4 lbf as would water. Expressions for converting pressure to head, and head to pressure, thus become F

Sg x 62.4 144

p

=A=

p

= Sg x

h

=

h

=

and h

x 0.433

(3-5)

Rearranging P

Sg x 0.433

and P x 2.31 Sg

(3-6)

In metric units p = 9802 x Sg x h 38

Energy in Hydraulic Systems

(3-7) Chap. 3

and h

=

1.02

X

1O-4p

(3-8)

Sg

Several problems requiring the solution for pressure in lbf/in. ? (Pa) , or head in ft (m), can be solved using these formulas .

Example 1 Compute the reading on a pressure gauge inserted at the base of a column of fluid 50 ft high with a specific gravity of 0.85 (Fig. 3-2).

Mx g =Wt

1

Height = h

_J Solution

Figure 3-2 Example I.

From Eq. 3-5 p = Sg x h x 0.433 p = 0.85 x 50 x 0.433 = 18.4 lbf/in. ?

Example 2 Convert to head in feet and meters a pressure reading of 1500 lbf/in .? taken from a hydraulic pressure gauge monitoring a system using a hydraulic oil with a density of 0.78. Solution

Substituting in Eq. 3-6 , h = p x 2.31 Sg

and

In metric units, a pressure reading of 1500 lbf/in. ? first must be converted to Pa using the conversion factor' lbf/in. ? x 6895 = Pa

and then solve Eq , 3-8 h I

= (1500 Ibf/in.2)(~~~:)(1.02

x 10-

4 )

= 1352 m

The exact conversion factor is lbf/in.? x (6.8947572 x lOl l .

Sec. 3-2

Pressure and Head

39

Example 3

Co nvert a hydraulic pressure reading of 13.79 megapascals (MPa) to meters if the oil has a Sg of 0.83. Solution

Sub stituti ng in Eq . 3-8 II = (1.02 x 1O-4)(~~8;90 000 N /m

2 )

= 1695 m

3-3 CONTINUITY EQUATION For purpo ses of making most calcu lat ions , the flow in fluid power systems is co ns idered to be steady. The flow is steady when the velocity at any period in time is constant. The velocity at successive places in the conductor , however , ma y change; for example, where the cross section area of a tube or hose is larger or sma lle r. To approximately fulfill the conditions of steady flow, changes with res pect to time in pressure, temperature , fluid den sity, and flow rate mu st be almost zero. T he continuity eq uation governi ng flow in a fluid co nductor is derived from the pri nci ple that duri ng conditions of steady flow, the mass of fluid flowing past a ny point in the co nductor is the same . For practical purposes and in the case of hyd raulic fluid s that are nearly incompressible (3-9)

Where the flow rate ( Q) is in gal /min , the area ( A) is in. ", and the velocity (v) is in feet per second (ft/ sec), Q(gal /min) = A(in .2) x v(ft/sec) x 12(in./ft) x 60(sec /min) x l/23I(gal /in .3 ) Q(gal /min)

= A(in .2)

x v(ft / sec) x 3.12

and Q = 3.12A v

(3- 10)

The continuity equation is used to solve for the veloc ity , area , or flow rate at any point in a system if two of the three quantities are given . Example 4

Referring to Fig. 3-3 , fluid is flowing in a 2-in. diameter pipe at the rate of 40 gall min. What is the velocity of the fluid in that portion of the conductor? What wou ld be the velocity of the fluid in the 4-in. diameter manifold? Solution (a) The 2-in. diameter pipe has a cro ss section area of 3.14 in 2 • The fluid velocity is computed by rearranging Eq . 3-10. Q =3 .12Av

v-

Q

- A x 3.12

40 v = 3.14 x 3.12 = 4.08 ftl sec

40

Energy in Hydraulic Sys te m s

Chap. 3

Q = 40 gal/min -

I.D . = 4 in.

Pip e

Manifold

Figure 3-3

Example 4.

(b) Th e 4-in . diamet er pipe has a cross sec tio n of 12.56 in2 • Solving the co ntinuit y Eq . 3-9 for V2: A ,vl

= A 2V2 A ,vl

V2=~

and V2 =

3. 14 x 4.08 12.56

1.02 ft/sec

Noti ce that dou bling the diameter reduces the flow rate to one-fourth of its original value . Example 5 Fluid is flowing in a 2-in. diamet er pipe at the rate of 18 gal/mi n (Fig. 3-4) . What is the minimum diameter to whic h the pipe might be reduced wit hout exceedi ng the maximum reco mme nded veloc ity of 10 ft/sec? Solution First, the velocity of the fluid in the 2-in. diameter pipe is determined using Eq . (3- 10): Q =3 . 12Avl VI

=

VI

= 3. 14 x 3.12 = 1.84 ft/sec

AI

Q x 3. 12 18

I.D.

'r = 2 in.

\ \

Q = 18 gal/min -

t

I I

Figure 3-4

Sec. 3-3

Continu ity Equation

-------~lI.D .=? ~------....I. _-

Example 5.

41

Knowing that the area of a circle is computed from

the value for the reduced diameter of the pipe is computed from Q = A 2 X V2 X 3.12

Q A - 1TDl 2 - - 4- - v2 X 3. 12 2 2

=

D2

=

D

1T

X

4 x Q V2 x 3. 12

and

~

4 x 18 3.14 x 10 x 3.12 = 0.857 in.

Similar values are obtained using flow nomograms, such as that shown in Table 3-1, that have been derived using the continuit y equation. Knowing two values allows for the solution of the third value by placing a straightedge across the two known values and reading the third quantity directly. Example 6

Compute the velocity in a I-in. I.D . suction line (actual size) of a hydraulic system in which 10 gal/min is being pumped . Solution Using Table 3-1 and placing a straightedge across the values for the 10 gal/min flow rate in the left column and the I-in. actual I.D . pipe size in the center column, the velocity at the suction line is read as 4.1 ftlsec in the right column. This is approximately equal to the computed value.

3-4 POTENTIAL AND KINETIC ENERGY The energy available from the prime mover is converted to a more useful form through the action of the fluid hydraulic pump. The prime mo ver is itself a conversion device th at usuall y consists of a dri ve motor po wered by electricity or chemical energy sto red in hydrocarbon fuel to power a combustion engine. Although energy is also av ailable from other sour ces, electric motors a nd internal combustion engines provide nearly all th e moti ve force to rai se the energy level of fluid power sy stems. The total energy av ailable from fluid in a syste m consists of the potential energy due to its initial elevation (EPE), the pressure potential energy (PE), and the kinetic energy (K E) due to the movement of the fluid . That is Total Energy

=

Elevation Energy (EPE)

+ Pressure Energy (PE)

+ Velocity Energy (K E) and Total Energy 42

= EPE +

PE

+ KE

Energy in Hydraulic Systems

(3- 11) Chap. 3

TABLE 3-1

20 ,000

CONTINUITY EQUATION NOMOGRAPH (ENGLISH UNITS ).

100 90 80 70

Based on for mu la

60 Area (sq in) . .

50 10,000 9000 8000 7000 6000

G.P.M. x 0.3208 Velocity (ft./sec.)

=

40 5·5 4 ~ .4 ~ 4·4

30

3 ~-3 ~

5000

3-3

20

2 ~ -2 ~

4000 15 3000

2000

10 9

2

- 8

1 0.9 0.8 0. 7 3 8 0.6 4 - -3---'-0.5 4 0.4 1 5 "2 - 8 0.3 1-

7

6 5 1000 900 800 700

4

1

8- 2

3 1

16

_

'" Ol u s:

~ l Ol

"0"0

0.75

Ol Ol

:;

~

c

Ol

0.5 100

'E ~

QJ

N

'U; OJ

"E

"0

:0

.2

o I

~ I

"iii

:::l

0.4

e.

~

~

u,

u,

.2

Sec. 3-4

.2

c:.o ;)

._

co

+oJ

e. ~ 'o.

c'"

c

c:

"0 '" 'Vi OJ

.9- ~ ~

Ol

o

Ol

~ .r= 0 ._

200

'" c

s

'" c

'E o z

Potential and Kinetic Energy

5

6

7

5

2

4

0.2 7

8

o c:

ill

Maximum recommended velocity for intake lines

4" - - --...-- 0.1 0.09

300

"5

16

3

316

400

e.

1

9

3

600 500

--

7

0.08 0 07 - 0.06 . 0.5 - 0.4 - 0.3 - 0.2

8 9 10 Maximum recommended velocity for pressure lines .S

0.1 0.009 g- 0.008 I 0.007 Ol 0.006 0 -- 0.00 5 s: 0.004 ~

'" .0

15 20

30

c

o 40

al

'" ~

Ol

;:

"0

50

e.

Ol"

e. '0.

'0 Ol '"

~

I

Z-

'u o ~

>

43

Mx g = Wt

Height = h

Figure 3-5 Potential energy as a forc e available to act through a distance (h) .

The potential energy of a fluid can be equated to the force it exerts toward the center of gravity (Mg ), and the distance (h) through which it is available to move on command (Fig. 3-5). This is expressed by the formula Potential Energy = M x g x h

(3 -12)

or, if the weight of the fluid rather than its mas s is given , Potential Energy

=wx

in ft-lbf

h

(3-13)

Where it is more de sirable to use pressure (lbf/in. ") than head (h) , Eq . 3-6 can be used h

=Px

2.31

Sg

and substituting .

Potential Energy

=

IV X

P

x 2.31

Sg

in ft-lbf

(3-14)

When fluid flows in a hydraulic system, potential energy is converted to the energy of motion , kinetic energy (F ig. 3-6). Potential energy and kinetic energy are mutually convertible . As the potential energy of the fluid is dis sipated by falling from a higher point to a lower point , its kinetic energy is increased by an Velocity = 0

.

Average velocitv =

Velocity

44

=

v

2"v

"'--.....l~==l= Discharge

Figure 3-6 of mot ion.

Kinetic energy as energy

Energy in Hydraulic Systems

Chap. 3

equal amount. This is a restatement of the law of conservation of energy which says, in effect , that energy ca n neither be crea ted nor destroyed . Ifa body of fluid at rest at some height is dropped through a distance (h), its velocity is (v) feet per second (ft/sec). The acc eleration due to gravity (g) is approx imately con stant. The ave rage velocity of the falling bod y of fluid is (v /2) ft/ sec since it has a starting velocity of 0 ft/ sec, and a terminal veloc ity of (v) ft/ sec. The time taken for the bod y to cover the distance equ als the distance (h) divided by the average velocity (v / 2). That is, h

2x h

v

v

/ =- = - -

2 where the time (I) is in sec. The acceleration constant (g) due to gravit y equ als the change in velocity divided by the time. That is, v -

g = --

0

/

v

and v2

-

2

= g

x h

From the potential energy formul a, Eq . 3-12, Potential Energy

=M x g x h

The value (v 2/2) can be substituted for (g x h), and the potential energy formul a converted to the kinetic energy formul a. That is, . . Kinetic energy

=

M

X

2

v2

Converting mass (M) to weight (w), the formul a for kinetic energy becom es KE

=wx

v

2 x g

2

in ft-Ibf

(3-15)

and the expression for the total energy in the system now becomes Elevation + Pre ssure + Kinetic

Total Energy

=

Total Energy

= (w

w x v2

x h) + (w x P x 2.3 1) + -=2,-----X g

(3-16)

where each of the se term s is in ft-Ibf. Let' s look at three examples to see how each of the energy terms fits into the general formul a . Sec. 3-4

Potential and Kinetic Energy

45

Example 7 A hydraulic sys te m has a 200 gallon re servoir mounted abo ve the pump to produce a positive head at the inlet. Thi s will prevent the pump from cavitating, particularl y at start -up. If the sta tic pre ssure at the inlet is to be 10 lbf/in ", and the fluid has a Sg of 0.87, how high mu st the average fluid level be above the pump inlet, a nd what is the energy of elevation of the fluid in the reservoir? Solution The problem , which is illustrated in Fig. 3-7 , as ks first for the elevation pressure and then the total elevation energy (the first term in Eq , 3-11). The ele vation pressure can be computed using Eq . 3-6 as follo ws:

h = p x 2.31 Sg 10 lbf/in. ? x

2.31Ibf;~n.2

h = - - - - - ---'-'--'-'-0.87

and

h = 26.6 ft The units of the seco nd term may at first look confusing until one notices that they are derived from 1/0.433 in Eq. 3-6. That is , 1 ft 0.433 Ibf/in .2 - Ibf/in. 2 ft

2.31 1

Finally , solving for the elevat ion energy using Eq. 3- 13 EPE

=

IV

x h = 200 gal x 0.87 x 8.34 Ibf/gal x 26.6 ft

~--

1.4--

Energy elevation

-

Flu id level

Reservo ir

1

1 Figure 3·'

46

Example 7.

Energy in Hydraulic Systems

Chap . 3

and EPE = 38,601 ft-Ibf

Example 8 If friction and kinetic or flow energy in Example 7 are ignored , how much pressure would the pump have to account for to lift a 1000 Ibf load using a 2-in. bore cylinder. Solution Here the problem illustrated in Fig. 3-8 asks that the total pressure energy required to lift the load be equated to the combined pressure developed by the energy of elevation (EPE ) and the pump. From Eq . 3-11 Total Energy = EPE + PE + KE and since friction and kinetic energy are ignored , the KE term drops out and the energy developed by the pump equals PE = Total Energy - EPE

Notice that each of the energy terms in Eq. 3-16 contains the weight of the fluid which for this example does not change and can be cancelled out allowing us to solve only for the pressure to lift the load .

_ F _ 1000 Ibf _ . 2 Total Pressure - A - 3.14 in.2 - 318.5 Ibf/m . Finally, subtracting the pressure of elevation (Example 8) from the total pre ssure , p = 318.5 lbf/in .? - 10.0 lbf/in. ? = 308.5 lbf/in.?

This assumes, of course, that the oil is not returned to the reservoir.

1000lbf

t Pressure

2-1n. bore cylinder

Figure 3-8 Example 8.

Sec. 3-4

Potential and Kinetic Energy

47

Example 9

In Fig . 3-9, if frict ion is ignored, how much kinet ic e nergy would be di ssip ated if 25 gal/min we re flowing thro ugh a pump ou tle t with an area of I in"? Rota t ion

Flow rate = 25 gpm through l oin? out let area

Figure 3-9 Exa mple 9.

Pump

Solution He re th e probl em ask s for th e valu e of the kinetic e nergy term in Eq . 3- 11, and it mu st be rem ember ed th at energy is a sum tot al , given in units of force and distance-for example, ft-lbf, or m . N. Thus, the term II' in Eq. 3-15 mu st be in for ce units-that is, th e tot al weight of th e fluid for so me specified time-for example , a minute o r an hour. Using I min as th e tot al time and 0.87 as th e S g of the fluid , II'

= 25 gal/min x Sg x 231 in. Vgal x 1/ 1738 ft3/in. 3 x 62.4 lbf/ft !

II'

= 181.4 Ibf

a nd So lving the continuity equa tio n for the velocity 25 gal/min x 231 in.3/gal x 1/12 ft /in. x 1/60 min/ sec v = ;; = 0.785 in.?

Q

and

v = 10.2 ftlsec Fin ally , so lving Eq . 3- 15 for th e kinetic e nergy 2 II' + v KE = -2 xg

KE

=

(184.4 IbO x (10.2 ft / sec)2 = 298 ft-Ibf 2 x 32.2 ft /sec?

Not ice th at th e va lue of the kinetic energy term is small, particularly when the velocity of th e fluid is kept within limit s. When the ve locity increase s much above 10 It/sec , friction loss generates unnecessary heat in the sys te m.

3-5 BERNOULLI'S EQUATION

Con se rvation of energy in a fluid power system is de scribed by Bernoulli' s theorem which defines the tot al energy of the sys tem as constant (Fig. 3-10). For example , ass uming no work is done or energy dissipated to the surroundings, in a pipe where fluid flows in a streamline through a rest riction, the energy level of the 48

Energy in Hydraulic Systems

Chap. 3

High pressur e

High pressure

Low pressure

=:::7~~1

t

High veloc ity

Low velocity

Figure 3-10 Bern oulli' s theorem .

fluid through the re striction will remain constant even though both th e pressure and the velocity of the fluid change . Bernoulli ' s theorem explains thi s by equa ting the energy at any two points in the sys tem. That is Total Energy(point I )

Total Energy(point Z)

=

and (w )

x h) +

W)

X PI

Sg

x 2.31

+

WI X VIZ

2 x g

= ( Wz x

h)

+

Wz X P Z

Sg

x 2.3 1

(3- 17)

For mo st calculations , the we ight of the fluid doe sn 't cha nge a ppreciably a nd will cancel out. And if the potential energy du e to elevation is equated to Z , th e equation be comes Z 1+

PI

x 2.3 1 Sg

VIZ

Z

+ -2 x-g= z +

PZ

x 2.3 1 Sg

vz z

+ -2 x- g

(3- 18)

Solving problems using Bernoulli ' s equa tio n requires a sys te matic approach. In general , the follo wing ste ps are follo wed :

1. Diagram the system indic ating at which points solutio ns are to be made a nd the direction of flow. 2. Determine the datum plane Z z and the ele vation of ZI . If these are a pproximatel y equal-for ex ample , when fluid flows horizontall y in a hydraulic syste m- the y can be cancelled out. 3. Set up Bernoulli' s eq uation with each energy component in the sa me units, including en ergy added by pumps and e nergy subtracted by motors. 4. Flow losse s should be subtracted . S. If the velocity of th e fluid is not known , solve the continuity equat ion. 6. Finally , equate the tot al ene rgy added a nd subtrac ted a t point I in th e syste m to th e to tal en ergy a t po int 2 in the syste m. T ypi cally, the equation is solved for pressure or head , including losse s, from po int I to point 2 Sec. 3-5

Bernou lli's Equation

49

-

~--------------------

Example 10 In Fig. 3-11, fluid with a specific gravity of 0.91 is flowing horizontaIly at the rate of 500 gal/min from a pipe with an inside diameter of 3 in. to one with a diameter of I-in. The pressure at point I is 1000 lbf/irr', Assuming no work is done or energy dissipated from the system, compute the pressure at point 2. Solution The direction of flow is from PI to P 2 , and since the flow is horizontal , thus introducing no change in elevation , the quantities ZI and Z2 can be canceIled from Eq. 3-18 . The energy equation for the system then is PI x 2.31

V,2

"-'-----'-+2 -x -g = Sg

P2

2 x 2.31 V2 +2-x -g Sg

The velocities VI and V2 are not known and must be determined by solving the continuity equation. The velocity VI is determined from Q = AI V

I -

X VI

x 3.12

Q

(Ad(3 .12)

and VI

=

500 (7.065)(3.12)

Solving the continuity equation for

= 22.68

ftlsec

V2

and V2

=

500 (0.785)(3 .12)

= 204.15

FinaIly, solving Bernoulli 's equation for (1000)(2.31 ) 0.91

P2

+ _(2_2._68_)2 = (p 2)(2.31) + 64.4

(2538.46 + 7.99) =

0.91

(P26~~t)

ftlsec

where g has the value 32.2 ft /s? ~ (2~04-,-.:-:..15~ )2

64.4

+ 647.16

and

= 0.91(2546.45 - 647.16) = 7481bf/ · 2 2.31

P2

In .

1000 Ibf/in 2

- t3 in . 1.0.

~ '-----_----.-/

50

1 in. 1.0.

-----~ -t-

Figure 3-11 Example 10.

Energy in Hydraulic Systems

Chap. 3

Here you will notice that a decrease in the size of the pipe is accompanied by a corresponding increase in velocity since the flow rate in the system is constant. The pressure also drops because the energy transfer through the system remains constant. If the size of the conductor were increased rather than decreased, the velocity would have decreased with a corresponding increase in pressure . Finally , the kinetic energy term or velocity head in Bernoulli's equation is seen to increase or decrease as the square of the velocity . Potential energy that is added to the system in terms of head by the pump (H,,), energy extracted (H,) , and lost (Hd also must be accounted for when computing actual rather than ideal flow conditions. These take the form of potential energy and are written in terms corresponding to head or pressure . The complete equation for Bernoulli's theorem then becomes ZI

+

PI

x 2.31 Sg

Ul

2

+2x

g

+

Z

H" =

2

+

P2

x 2.31 Sg

2 U2

+2x

g

+

H,

+

HI

(3-19)

Exa mple 11 Fluid flows from a pump at 1000 lbf/in .? at the rate of 200 gal/min horizontally to a motor operating at 750 lbf/in .? (Fig. 3-12). Back pressure on the motor discharge port is 350 lbf/in", The fluid has a specific gravity of 0.85. Compute the energy extracted from the fluid in terms corresponding to head. Solution Flow is horizontal, introducing no change in elevation so the quantities ZI and Z2 cancel out of Eq . 3-19. Since the fluid flow rate and pipe size are constant, the velocity head terms are equal and also cancel out of the equation . If the pressure at the inlet to the pump is considered to be 0 lbf/in .? gauge, the potential energy term on the left side of the equation also becomes O. The energy equation for the system then becomes H " --

P2

x 2.31 Sg

H

+, +

H I

Computing the head associated with the potential energy added to the system

H = P x 2.31 " 0.85

H" =

10000.~52.31

= 2717.65 ft

7501bf/in 2

Q = 200 gal/min

3501bf/in 2

Figure 3-12 Example II .

Sec. 3-5

Bernoulli's Equation

51

Finally , solving Bernoulli' s equation for head ass ociated with the energy that can be extracted from the sys tem 2717.65 = 951.18 + H , + 679.41 H,

= 2717.65

- 951.18 - 679.41

=

1087.06 ft

3-6 TORRICELLI'S THEOREM

Torricelli ' s theorem is a special case of Bernoulli ' s equation derived for an ideal system. In Fig. 3-13 a large tank holds water that follows out a small pipe at the base. Bernoulli 's equation for the system is 2, +

PI x 2.31

Sg

V, 2

+ 2 x g =

2

P2 2

+

2 x 2.31 V2 Sg +2x g

Since the system is considered ideal, losses are negligible and HI equals O. The elevation (h) equal s (2, - 2 2 ), and if 2 2 is zero, h equals 2 1 • .....-----:- p = 0 psi ~---='=---1

1

Elevation = h = Z,

Z"o

1

i

~PSi

Ibf / in

2

Figure 3-13 To ricelli' s theorem.

Ifthe top of the tank is open to the atmo sphere , the gauge pre ssure is 0 at the surface and PI equal s O. The pressure at the surface of the outlet near the tank is also 0 since the free jet is discharged into the atmo sphere . Where the area of the surface of water in the tank is sufficiently large , the downward velocity of the surface is negligible and the velocit y head ( vd also becomes O. Bernoulli ' s equation for this case then becomes 2 V2

2, = h = - 2 x g

and V2

=

v'2gh

which is a statement of Torricelli's theorem . Example 12 Compute the velocity of a fluid flowing from the base of a water tank located 40 ft belo w the water level (Fig. 3-14). 52

Energy in Hydraulic Systems

Chap. 3

----- ---_.-1 h = 40'

Lt-=-_J Solution

Figure 3-14 Example 12.

Using Torricelli' s theorem

v

= Y 2g h

v = Y2

x 32.2 x 40 = 50.7 ft/ sec

3-7 SUMMARY

The potential and kinetic energy in hydraulic systems can be accounted for in terms of the specific gravity of the fluid, head or pres sure, and the ve locity. Potential energy is influenced by the pre ssure resulting from ele vation position and from input energy added through the hydraulic pump. Kinetic energy is the energy of motion and is influenced primarily by the velocity of the fluid through the sy stem . In hydraulic systems, the primary factor affecting the energy level of the system is the pressure resulting from input to the system from outside sources through the pump . Bernoulli' s equation sets the energy level of the system at any two cross sections through the conductor equal to each other. It is derived from laws governing the conservation of energy and is used, typically, to determine the amount and source of energy extracted from the sy stem as a result of friction losses , energy dissipated moving the load resi stance , and energy discharged or wasted as heat. Torricelli 's theorem is a special application of Bernoulli 's equation. Stated as v = Y 2gh , Torricelli's theorem is useful to determi ne flow through orifices and restrictions. The basic formula assumes frictionless flow. In practice , flow coefficient between 0.5 and 1.0 are introduced to account for friction losse s associated with a particular orifice configuration.

STUDY QUESTIONS AND PROBLEMS 1. A 250 gal reservoir must be mounted in a 4 ft x 6 ft space. What is the height of the reservoir in inches ? 2. How tall wo uld a building have to be to deliver oil with a S g of 0.82 from the roof to ground level at a pressure of 500 Ibf/in 2 ? How might this be achieved in a building half thi s tall ? 3. What pressure would a pump have to overcome to deliver oil with a S g of 0.90 to a tank with fluid surface at a height of 120 ft above the pump ? Chap. 3

Study Quest ions and Problems

53

4. Compute the pressu re readings on a dual gauge in lbf/in .? and kPa inserted in the base of a storage tank 40 ft high , full of oil that has a S g of 0.87. 5. A pre ssure read ing of 2250 lbf/in. ? is observed on a hydraulic gauge . If the fluid has a Sg of 0.85, what would be the pre ssure in MPa ? 6. A 200 gal hydraul ic reser voir is placed on the sec ond floor with the fluid level 12 ft above the pump inlet. If the petroleum base fluid has a Sg of 0.89, what would be the pre ssure at the pump inlet in lbf/in .? and ft of head ? How much would the pre ssure increase for a synthetic fluid with a S g of 1.15? 7. In Problem 6, assuming no friction or pumping losses , what would be the absolute and gauge pre ssure readings at the pump inlet if the reservoir were mounted so that the fluid level wer e 2 ft below the inlet ? See Fig. 3-15. (Clue : atmos pheric pres sur e = 0 lbf/in. ? gauge = 14.7 lbf/in .? absolute.) p= - ?

I 2' Elevation

Pt,

-------- ----------------------------------------- --------- -- -----Figure 3-15 Problem 7.

8. At a maximum velocity of IS ft /sec, how man y gal/min of fluid will flow through a pipe with an J.D. of 4 in? 9. A I-in . J.D . hydraulic tube delivers 35 gal/min . What is the fluid velocity in ftlsec? 10. What hose J.D. would be required to deliver 50 gal/min , assuming no friction losses at a maximum velocity of 8.5 ftl sec? 11. How large would th e J.D . ofa tube have to be to deliver 15 gal/min if the fluid velocity were held to 10 ftl see ? Assume no friction losses. 12. A flow meter in the return line to a hydraulic reservoir read s 18 gal/min. How large would the suction line have to be to limit the velocity to 3 ftlsec ? 13. Fluid is flowing from a larger pipe with an J.D. of 8 in. to a smaller pipe at the rate of 500 gal/min (Fig. 3-16). What would be the minimum J.D. of the smaller pipe to limit the velocity to 10 ftl see ?

54

Energy in Hydraulic Systems

Chap. 3

Q = 500 gal/min

-

--

4·in .

Pt, Figure 3-16 Problem 13.

14. It is noticed that doubling the I.D. of a pipe reduces the velocity by a factor of 4. If instead , the objective were to reduce the velocity to half its original value , by what factor would the smaller diameter pipe be increased ? 15. Hydraulic oil with a Sg of 0.91 is pumped horizontally in a 2-in. I.D. distribution line at the rate of 200 gal/min. A pressure gauge mounted in the pump outlet reads 2000 lbf/in.", but some distance downstream another gauge mounted in the line reads 1750 lbf/ irr' , Assuming the gauge readings are accurate , how does the Bernoulli formula given by Eq . 3-19 account for th is difference? 16. Fluid is flowing horizontally at the rate of 500 gal /min from a pipe with a 4-in . I.D . to a smaller one with a 2-in. I.D. If the fluid has a Sg of 0.88 , and the pressure is 2000 lbf/in. ? at Point I in the 4-in pipe, compute the pre ssure at Point 2 in the 2-in. pipe. 17. A hydraulic pump delivers 55 gal /min of oil with a S g of 0.95 at 2000 lbf/in .? through a line with a cross sectio n of2 in 2 • If the line is reduced to a cross section area of 1/2 in . 2, what change would there be in velocity and pre ssure ? 18. Compute the velocity of a fluid which would flow from the base of a water storage tank where the level of the surface is 35 ft above the ground. Con sult Fig . 3-17 . Pt1

t====--.=-_--:.--_-_~ _-_-_-_-_-: !--

~~=~~=~~*~=~

I~~~~~~~~~~~~

-=-=~-=-=-=~-:::ptJ=-=-_- -=.....,.

=-=-=-=-=-=-=-=-=---~=-= - . ~-::..~. . . ,. --------------------t-:

Figure 3-17 Problem 18.

19. In Problem 18, if the hole in the base of the tank has a 3-in. I.D., and there are no friction losses, what flow rate could be expected? 20. Wha t de livery in gal /min would be re quired to shoot a water jet approximately 25 ft in the air from a vertical tube with a 1.5-in . I.D .?

Chap. 3

Study Questions and Problems

55

-

CHAPTfR

4

HOW FLUIDS FLOW 4-1 INT RODUCTION

When a fluid flows, its volume is displaced as it changes location from one place to another in the sys tem. The time it tak es the fluid to change location determines the flow rate, and thi s, in turn, determines how much power the fluid transmits through the sys tem. For example , doubling the displacement and flow rate would double the fluid power transmitted through the sys te m-assuming, of course , that the pre ssure rem ains con stant. In addition to Bernoulli ' s theorem which gives us the basic energy equation , there are a number of laws that govern the flow of fluids. These serve as the basis for several equ ation s used to make sys tem calcul ation s. From the principle of conse rvation of mass which says, in effect , that the mass of the fluid remain s con stant , we have the continuity equation. The law of conse rvation of energy which says, in effect, that energy can neither be created nor de stroyed , allows us to convert energy from one form to another and develop several equations that govern flow. Finally, the concept of mom entum which de scribes the change in the velocity of the mass of fluid, is the basis for equations governing friction . Here , theory and practice are brought together. Because the flow of fluids is complex , the law s and equations that govern flow through fluid power systems must be confirmed through experimentation. Thu s, man y of the formulas that have their basis in theory are modified slightly with flow and friction factors to de scribe conditions as they actuall y exist in practice. 4-2 APPLICATIONS OF THE DISPLACEMENT FORMULA

While displacement can be calculated from the bore diameter and stroke length , it is more common to use tabled values to solve this formul a. Table 4-1 list s the bore are a diameter, with the area of the blank end of the cylinder down the left margin 56

TABLE 4-1

DISPLACEMENT OF CYLINDERS IN CUBIC INCHES

Bore area

Stroke (in.)

-

(in.)

(in.')

3

6

12

18

24

30

36

42

48

54

60

I

0.785 0.994 1.77 3.14 4.91 7.07 9.62 12.56 15.90 19.63 28.26

2.36 2.98 5.31 9.42 14.73 21.21 28.86 37.68 47.7 58.89 84.78

4.71 5.96 10.62 18.84 29.46 42.42 57.72 75.36 95.4 117.8 169.6

9.42 11.93 21.24 37.68 58.92 84.84 115.4 150.7 190.8 235.6 339.1

14.13 17.89 31.86 56.52 88.38 127.3 173.2 226.1 286.2 353.3 508.7

18.84 23.86 42.48 75.36 117.8 169.7 230.9 301.4 381.6 471.1 678.2

23.55 29.82 53.10 94.2 147.3 212.1 288.6 376.8 477 .0 588.9 847.8

28.26 35.78 63.72 113.0 176.8 254.5 346.3 452.16 572.4 706.7 1017

32.97 41.75 74.34 131.9 206.2 296.9 404.0 527.5 667 .8 824.5 1187

37.68 47.71 84.96 150.7 235.7 339.4 461.8 602.9 763.2 942.2 1356

42.39 53.68 95.58 169.6 265.1 381.8 519.5 678.2 858.6 1060 1526

47.1 59.64 106.2 188.4 294.6 424.2 577.2 753.6 954.0 1178 1696

U I! 2 2! 3 3! 4 4! 5 6

U1

"""

and the stroke in inches ac ross the top . The numbers in the Table represent the cubic inch displacement of the cylinder. The Table can be used to determine the working volume of the c ylinder on the extension stro ke . If the displacement on the return stroke is required , the displacement of the rod must be subtracted from the di splacement on the extension stroke. Example 4-1 A ram with a 4-in. bore diameter cylinder and 2-in. rod (Fig. 4-1) has a stroke of 24 in. What would be the displacement of the cylinder through the extension and retraction strokes?

4·in. Bore diameter

Figure 4·1

Example

I.

Solution On extension , the displacement can be read by finding the place in Table 4-1 where the bore diameter and stroke length intersect. Finding the 4-in. bore diameter down the left margin and the 24-in. stroke across the top , the displacement is given as 301.4 in'. On the return stroke the displacem ent of the ram would have to be subtracted from this value. In this case Table 4-1 can be used again. Locating the intersection of a 2-in. bore and a 24-in. stroke, the displacement is found to be 75.36 in'. Subtracting this value from the bore displacement of 301.4 in.? results in a value of 226.04 in.' on the return stroke . Thu s the combin ed displacement of the extension and retraction strokes would be

and

V,

= 301.4 in.' + 226.04 in.! = 527.4 in.!

where V, = total volume (in.') V, = exten sion volume (in .')

V, = retraction volume (in.3 )

4-3 APPLICATIONS OF THE CONTINUITY EQUATION It is common practice in sizing components to so lve the continuity equation with nomographs. The continuity equation is so lved with the conversion constants inserted so that so lutio ns are given directl y in the de sired units. For example , Fig. 4-2 so lve s the continuity equation in English units: (i

A rea m .

58

2) _ Gal/min x 0.3208 - V e Iocity . (f t / sec ) How Fluids Flow

Chap . 4

20,000

100 90 80 70

8ased on form ula

60

Area (sq. in.) = G.P.~. x 0.3208 Velocity (ft./sec.)

50 10,000 9000 8000 7000 6000

40 5·5

4~-4 ~

30

4·4

3 ~ -3 ~

5000

3-3

20

2~-2~

4000 15

2

3000

1

2000

12" 1 1-

10 9 8 7

4

2

1~

1

12" 11 4

1

7

6

3

4

8 3

5 1000 900 800 700

4 1

2"-8

4

3

1

3

8

4 -2

5 16

1

ii

300

_ s: u'"

1

4

'"

0 .::

:;; I 0; '"

200

.

'E

c

c.

~

s: u

.s u :.0 :J u I

;:

..S!

u..

0.5 100

0.4

.'"

'E

'"

.~

'"c. '0.

c.

"E

'" c

"0

..S!

"iii

Cl

I

3 2 2 1 0.9 0.8 0 .7 0.6 0.5 0.4 0.3

3 Maximum recommended velocity for intake lines

5 6 7 8 9 10

0.1 0.09 0.08 07 0.06 0 . 0.5 0.4 0.3 0.2

Maximum recommended velocity for pressure lines

15 20

"0"0

0.1 0.009 li 0.008 I 0.007 '" - 0.006 '" 0 s: 0.005 "0 - - 0.004 c

40

~

50

"0 '" ·iii Q)

C:.J:l . - :J

- '"

"iii:J

"

uc.

Flow nomograph (English units).

Applications of the Continuity Equation

59

All that is required is to determine which two of the three variables are given by the problem and then use a straightedge across these two values to read the third qua ntity directly. Example 2

Compute the velocity in a I-in. I.D . suction line ofa hydraulic system through which 10 gal/min is being pumped. Solution Using Fig. 4-2, and placing a straightedge across the values for the 10 gal/min flow rate in the left column and the I-in. actual I.D. pipe size in the center column , the velocity in the suction line is read directl y as 4.1 It/sec from the right column. Values thus determined will be approximately equal to the computed values . Example 3

What size standard pipe would be required to deliver 40 gal/min if the velocity in the line was not to exceed 7 ft/sec ? Solution Placi ng a straightedge acro ss the values for the 40 gal/min in the left colu mn of Fig. 4-2, and the velocity of? It/sec in the right , the actual I.D . of the pipe is give n at 11/ 2 in., which is slightly less tha n a 11/2 in. standard pipe.

4-4 VISCOSITY

The viscosity of a fluid determines how easily it will flow and is required to solve most flow problems. Viscosity is a meas ure of the internal resistance of a fluid to shear and is related to the internal friction of the fluid itself. Thick fluids flow more slowly than thin fluids, because they have more internal friction. Viscosity numbers are assigned to describe the relative differences in the ability of a fluid to flow in comparison with other fluids. Higher numbers are assigned to thicker fluids, lower numbers to thinner fluids. The oil film between the moving and stationary plates may be thought of as a series of fluid layers separating the two parts. Oil adheres to both surfaces. The velocity of the oil at the stationary plate is zero and the velocity of the oil at the movi ng plate equals the velocity of that surface . Between the two surfaces, the velocity of the oil varies on a straig ht line between zero at the stationary surface , and the speed of the moving plate at its surface. What happens is the moving plate cau ses the adjacent layers of oil between the two plate s to shear and the force necessary to move the plate is what we call viscosity (Fig. 4-3). There are really two ways to describe the visco sity of a fluid: one is the absolute or dynamic viscosity desig nated by J1. (Greek letter mu), and the other is the kinematic viscos ity designated by v (Greek letter nu). Absolute viscosity equals the force required to move a flat surface with an area of one unit at a velocity of one unit , when the two are separated by an oil film one unit thick. Absolute viscosit y is illustrated in Fig. 4-4. Kinematic viscosity, which is used freq uently in hydraulic calculations , is simply equal to the absolute viscosity divided by the mass den sity of the fluid . That is, J1. v=-

(4-1)

p

60

How Fluid s Flow

Chap. 4

Veloc ity (v)

~

Oil film thi ck ness (v) Figure 4-3

Viscous friction. Unit velocity

____ Force

Moving surfac e

Oil fi lm {

St at io na ry surfa ce

Slope

Unit distance

Iu

~~~~~~~~~~~~~~~~ Figure 4-4

Absol ute viscosity.

Bec au se viscosity is calculated in all three sys tems of unit s: English (ft, slug, sec), traditional metric (em, gm, sec) and SI metri c (rn, kg, sec), it is sometimes difficult to rem ember which viscos ity is being descr ibed . Figure 4-5 can be used as a refer ence to keep them separate. Becau se the reyn and poise are large unit s of absolute viscos ity, the micro- reyn (one millionth of a rey n or 10- 6 rey n) and centipoise (one hundredth of a poise or 10- 2 poise) are used to simplify calculation s. To follow through from absolute to kinematic visco sity, the equivalent kinem atic unit of viscosity to the centipoise (cP) is the centi stoke (cSt). Absolute viscosity

Kinematic viscosity

English units

tb-sec/tts lb-sec/in" (reyn) Saybolt Seconds Universal (SSU)

ft2/sec (no name) in2/sec (Newts)

Traditional metric units

dyne-sec/ems (poise, P)

crnvsec (Stokes, St) Stokes x 10- 2 = cSt

SI metric units

N's/m 2 (Pascal-second)

poise x 10- 2

Figure 4-5

Sec. 4-4

Viscos ity

= cP

mvsec (no name)

Viscosity units .

61

When absolute viscosity is converted to kinematic viscosity, using Eq. 4-1, care must be exercised to assure that the units are correct. When p. is given in reyns, Ibf-sec) Ib~-sec) (~) p. ( in.? p. m.2 ft

(

v(in. 2I sec) = - - - -

slUgs) p ( in. !

p (

lbf-sec 2 x _1_) ft in.'

and - 12p.(reyns) p (I s ugs)

v (N ewt s) -

When p. is given in poise, p. (

d yne-sec) ern?

p. (

= -----

v(cm 2I sec)

p

and v(Stokes) =

(~:3)

p (

dyne-sec) ern?

d y ne-sec 2) em

(_I) ern!

p.(poise) ( I 3) P gm ern

When p. is given N . s/ rn", P.

(N~2 s)

v (m 2I sec) = - - - -

p

(~)

p

(N~ S2) (~3)

and _ p.(Pascal-second)

v ( no name ) -

p (kg /m 3 )

The conversion from kinematic viscosity in cSt to Newts is given by (Newts)

= (cSt) x

0.001552

(4-2)

and from cSt to the SI metric system (no name ) by (m 2Isec) = (cSt) x 10- 6

(4-3)

The kinematic viscos ity in cSt and absolute viscosity in cP can be determined by ASTM te st procedure 0 445 which mea sures the time required for a fixed amount of oil to flow through a calibrated capillary instrument using gravity flow at one of the two con stant temperatures. The equipment and procedure are shown in Figure 4-6. The time is measured in seconds and then multiplied by the calibration con stant for the viscometer to obtain the kinematic viscosity of the oil sample in cSt. The ab solute viscosity is derived by multiplying the kinematic viscos ity by the den sit y of the oil. One of the more common units of absolute viscosity is the Saybolt Second Universal , abbrev iated SSU , and it is mea sured using ASTM te st procedure 0-88 with equipment like that shown in Figure 4-7. The resistance of the fluid to flow is 62

How Fluids Flow

Chap. 4

(a) Met hod of charging sampl e

(b) Place in constant temp eratu re bath

Figure 4-6

Kinematic viscome ter.

Figure 4·7

Sec. 4-4

Viscosity

(c) Ad just head level 5 mm above sta rti ng mar k

Saybolt viscome ter.

63

measured as the time that it takes a 60 ml sample of oil to drain through a standard orifice at a constant temperature of 100°F (37.7°C) or 210°F (98.9°C). The elapsed time is the SSU viscosity for the fluid at the given temperature . For thicker fluids, the same test is repeated using a larger orifice to derive the Saybolt Seconds Furol (SSF) viscosity . The conversion from kinematic viscosity in cSt to the equivalent viscosity in SSU is read from charts in ASTM Procedure D 2161. Basic conversion values are given for 100°F. For temperatures other than 100°F, kinematic viscosities are converted to SSU viscosities by using temperature correction factors. A number of other conversions are made among the various measures of viscosity using the conversion tables given in Appendix C. Approximate conversions between the four measures of viscosity most commonl y used are given by Fig. 4-8. By placing a straightedge horizontally across

.,..

.,..

.><

.><

.,..c

·2c

u

u

S

S . (j)

.,

.,

1 x 10- 2

10000

5 x 10- 3 2 x 10- 3

5000 2000

1 x 10- 3 5 x 10- 4 2 x 10- 4 1 x 10- 4

1000 500

5 x 10- 5 3 x 10- 5

50

2 x 10- 5

20

1 x 10- 5

10

200 100

.

..,

~ .,

~

z

z

15 10 5 3 2 1 0.5 0.3 0.2

10000 5000

3000 2000 1000 500 300 200

0.1 30

5 x 10- 6

5

4 x 10- 6

4

3 x 10- 6

3 2

2 x 10- 6

:::l

~

0.05

100 80 70 60

0.03 0.02

0.01 0.008 0.007 0.006 0.005 0.004

50 40

0.003

1.75 1.5 x 10- 6

1.5

1.25 x 10- 6

1.25

0.002

1 x 10- 6

Figure 4-8 Visco sity conversion chart.

64

How Fluids Flow

Chap. 4

the chart , relative comparison s can be made between them. For example, if it is known that a hydraulic pump requires a fluid with an absolute viscos ity in the range of 100 to 3000 SSU , an equivalent oil with a kinematic visco sity of just over 20 to 100 centistokes would be required. Temperature affect s visco sit y inversely. Hydraulic fluids thin out at high temperatures and become thick at low temperatures. How much the viscosity change s with temperature can be det ermined for the ra nge of expected temperatures encountered by the particular applica tion. The most critical component affected is usuall y the pump. If the viscos ity number is too high , the oil will be too thick and the pump will cavitate or break. If the visco sity number is too low , excessi ve leakage and wear will occur. Most pump manufacturers specify a range of visco sit y no higher than 4000 SS U, including cold sta rts, or lower than 65 SSU at machine operating temperatures. • 4-5 NONCOMPRESSIBLE FLOW IN PIPES

For practical purposes , the flow of noncompressible fluids in pipe s is con sidered to be laminar or turbulent . Laminar flow is governed by the viscous nature of the fluid and the fluid flows in a streamline. When the flow is turbulent , the flow is mixed up and disorganized . In circular pipe s where the flow is governed by the viscous nature of the fluid, layer s of unit thickness called lamin ae ass ume the shape of thin shell concentric tubes sliding one over another success ively (Fig. 4-9). Viscou s friction between the layers dissipates the potential energy and returns it to the fluid as heat. Here we are assuming that steady conditions pre vail. Th at is, that the velocity and volume of fluid flowing past a fixed point in the conduit are con stant. The distribution of the se thin shell tube s is parabolic , and they tele scope as the y flow in a circular conduit. Near the center of the strea m, the velocity of the thin shell tubes is gre ater, even though each has a velocity that is con stant along its length. At the center of the stream, then , the velocity of the fluid is at the maximum; whereas at the inside wall of the conduit the velocity is zero, and this is where the greate st shea r and internal friction occur.

Parabol ic flow d istr ibution

/'

Flow -

Figure 4-9

Viscous flow.

Turbulent flow, on the other hand , is characterized in a disorganized manner where the smooth flow of the streamline is disrupted (Fig. 4-10). Agitation within the flowing fluid strea m is such that it is difficult to determine the flow path of Sec. 4-5

Noncompressible Flow in Pipes

65

Q

Figure 4·10

Tu rbulen t flow.

indi vidual particles, and experimenta l rather tha n theo re tica l equ at ion s are used to solve flow problem s. Bet ween laminar and tur bulent flow is a spa n of critical veloc ities where th e flow is neither smooth and lamin ar nor disrupted and turbulent. Belo w th e critica l velocity the flow is laminar , and above th e critica l veloc ity the flow is turbulent. 4-6 REYNOLDS NUMBER

Osborne Reynold s' -? (1842-191 2) discovered th at viscous flow in pip es was related to the dimensionless ratio N R (called Reynold s number) of the product of th e fluid velocity (u), the diameter of the pipe (D), and the den sit y of the fluid (p ) , to the absolute viscosity (J-L) of th e fluid. Reynold s number is express ed as N

R

=

vDp

(4-4 )

J-L

If the kinematic visco sit y rather than th e absolute viscos ity is used , NR

= uD

(4-5)

v

To be dimen sionless, the unit s of the factors in the Reynolds equation mu st cancel. If th e ve loc ity is given in ft/sec, the diamet er of the pipe (D) in inch es , th e mass of the fluid ( p) in slugs/in' and the absolute viscos ity (J-L) in re yn s, N _ v(ft/s ec)D( in.)(s lugs/ in.3 ) R J-L (reyns) and

N

= R

v(ft/sec )D (in.)Obf-sec 2/ft-in. 3 ) J-LObf-sec /in. 2)

Where th e kinematic viscos ity in Newt s ra the r than the ab solute viscosity is given

12J-L(re yns)

v(Newts)

= P(Is ugs I'm, 3) = v (ft / sec)

N R

D (in.)(12 in./ft ) v(in.2/sec)

I For a det ailed accounting of Osborne Reynold s' experiment , see G.A. To katy, A History and Philosoph y of Fluidmecha nics. Henl ey-on-Thames , Oxfordshire: G.T. Foulis and Co. Ltd. , 1971 2 Or see Reynolds' or iginal paper , " An Ex peri menta l Investigat ion of the Circumstances Whic h Determine Whether the Motio n of Water Shall Be Direct of Sinuous, and the Law of Resistance in Parallel Cha nnels," Philosophical Transactions of the Royal Society , 174, Part Ill , 935 (1883), or Scie ntific Papers , London : Ca mbridge Unive rsity Press, 1900-1 903, Vol. II , pp. 5 1- 105.

66

How Fluids Flow

Chap. 4

and

N R

=

12(v)(D)

(4- 6)

v

Re ynolds apparatus (Fig. 4-11) passed water through horizontal tubes of different diameters from 1/4 to 2 inches at increasing velocities to find the velocity at which a dye bled into the stream would show a transition from laminar to turbulent flow. Thi s work, which sta rted about 1880, and work by other investigators which replicated similar experiments through 1910, indicated that the upper and lower c ritica l limit s for the ch ange from lamin ar to turbulent flow lie between N R of 2000 and 4000 . Below 2000, the flow is laminar, wha tev er its previous sta te, and this is consid ered to be the lower critical limit. Above 4000 , the flow becomes or remain s turbulent. Th e range of critical velocities, then , is considered to be 2000 -s N R :::; 4000.

Figure 4·11

Reynold s apparatus .

Example 3

Oil with a kinematic visco sity of 0.05 Newts is flowing in a l-in . pipe at the rate of 100 gal/min (Fig. 4-12). Is the flow lamin ar or turbulent? Solution unit s)

The Reynolds number will be computed from Eq. 4-6 (with compatible N

= 12(v )(D ) R

v

Fir st , the velocity (v) is computed fro m

-

Q

v - (A)(3. 12)

Sec. 4-6

Reynolds Number

67

Q = 100 gal/min--

Fluid viscosity = 0.05 Newts

Example 3.

Figure 4·12

and 100

v = (0.785)(3.12) = 40.83 ftlsec Solving Eq, 4-6 for N R N = (12)(40.83)(1) = 97992 R (0.05) . and the flow is turbulent. Example 4 An oil with a specific gravity of 0.85 and an ab solute viscosity of 27 cP is flowing in a l-in, I.D. pipe. Compute the range of critical velocities. Solution There are two ways to approach the problem. One is to solve it directly in absolute viscosity unit s given , using Eq, 4-4 ; and the other is to convert the absolute viscosity to kinematic viscosity in English units and solve Eq , 4-6 for the velocity in ft /sec. Both solutions will be given here . Reynolds Eq . 4-4 is solved for the velocity .

and

NRM v =-Dp At the upper end of the critical velocity range N R = 4000, and at the lower end of the critical velocity range N R = 2000. Also notice that in cm, gm, sec, system of units, the de nsity and Sg are numerically equal. And since the absolute viscosity is given in cP, this must be converted to poise using cP x 10- 2 = P. Solving Eq , 4-4 for the velocity when N R = 4000, v = .,.,.....,..:...:(4..:..000~)(~0..=..27:.-P:....;)~ (I in.)(0.85 gm/crn")

(4000) (0 .27 dyne-Sec )

em?

v=--------------2 in. x 2.54 ~m) (0.85 dyne-sec x ~) I ( m.

em

cm

and

v = 500.23 ern/sec 68

How Fluids Flow

Chap. 4

~'i): O.85~ ' -!:::================:7 L

Velocity

===e=-

Fluid specific gravity absolute viscosity 27 cP

-

Figure 4-13

Exa mple 4.

or cm ) ( 500 sec

v=-- - - - - -

( 2.54 ~m x 12 in.)

16.4 ft/ sec

ft

In .

Following through then to solve for the lower critic al velocity, (2000) (0.27 dy ne-sec ) cm 2

v = - - - - - - - - - -- - -2 - - -

(I in. x 2.54 ~m) (0 .85 dyne-ernsec x _1_) ern! In.

and

v = 250.12 ern/ sec or

v=

( 250. 12 cm ) sec x 12 in.) ( 2.54 ~m In . ft

= 8.2 ft/ sec

When English unit s are used to solve the problem , Reynold s Eq. 4-6 (with compatible unit s) is solved for the velocit y, N

_ 12vD v

R -

and

Absol ute visco sity in cP is converted first to kinematic viscosity in cSt and then to Newts so the unit s in the Reynolds formula are compatible: v

=

27 p = 0.85 = 31.76 cSt

!L

Notice here that the value of the Sg was used for the den sity. Converting cSt to Newt s using Eq. 4-2, v(Newts) = 0.001552 x v( c St)

Se c. 4-6

Reyno lds Number

69

and IJ

= 0.001552 x 31.76 = 0.49 Newts

Substituting in the rearranged Reynolds formula and solving for the upper critical velocity limit, = (4000)(0.049) = 164 ft l v (12)(1) . sec Finally, substituting and solving for the lower critical velocity - (2000)(0.049) - 8 2 f I v(12)(1) - . t sec Thus, in both cases the critical velocity for the fluid is from 8.2 ft/sec to 16.4 ft/sec.

4-7 SUMMARY

A fluid power system transmits power by displacing the fluid from one place to another in the system. Displacement measured over time becomes flow rate. For a given flow rate, the continuity equation, Q = A v, relates the cross section area of the conductor to the velocity of the fluid. For the continuity equation to be valid, the flow must be steady. That is, the velocity and volume of fluid flowing through any cross section area of the system must remain constant over time . This condition would allow the velocity of the fluid to change as the cross section of the conductor increases or decreases, but would not allow surging that increases or decreases the mass flow rate. The continuity equation is widely used to solve for the conductor size when the required flow rate for the system is given . And in practice , the velocity in pressure lines is limited to 10 ft/sec to keep flow losses low. Noncompressible flow in pipes is affected by the velocity in another way. At low velocities, the flow is laminar and the fluid flows in a smooth streamline . Above a certain velocity, the flow becomes disorganized, unpredictable, and turbulent. Osborne Reynolds found that laminar and turbulent flow were related to a dimensionless ratio which became known as Reynolds number, N R. At values below N R = 2000, the flow is laminar, whereas above N R = 4000 the flow is turbulent. Between N R = 2000 and N R = 4000 the flow is neither laminar nor turbulent and describes what is known as the critical range . Most fluid power applications involve flows above the critical range .

STUDY QUESTIONS AND PROBLEMS 1. What are the extension and retraction displacements of a double-acting hydraulic cylinder that has a 2-in. bore, a l-in . rod, and an 18-in. stroke? 2. What is the displacement of a single-acting ram with a 4-in. bore and a 30-in. stroke? 3. What is the displacement of a 5-cylinder piston motor with a bore diameter of 2.0 in. and a stroke of 3.0 in.? 70

How Fluids Flow

Chap. 4

4. A pump deliver s 10gal/ min through a standard 3/4-in. pipe. What is the flow velocity? 5. What is the smalles t intake line that could be used for a pump rated at 25 gal/min if the inlet velocity is limited to 4 ftl sec ? 6. What flow rate will a standard 2-in. pipe deliver if the velocity is limited to 10 ftl sec ? 7. A double-end rod cylinder with a 2-in. bore and I-in. dia. rod cycles through a 12-in. stroke at 80 cycle s per minute . What flow rate is required from the pump? 8. An automatic log splitter with a 5-in. bore cylinder and a 1.5-in. rod cycles through a 30-in. stroke 6 times each minute. If the flow velocit y is limited to 5 It/sec, what inside diameter hose would be required at the blank end of the cylinder? 9. What nominal standard pipe size would be required for an intake line to deliver 30 gal/min if the velocity is limited to 3 ft/sec ? 10. If the maximum recommended velocity were 30 ft/sec , what would be the approximate flow rate that would ensue from a 7/16-in . tube ? 11. What would be the actual I.D. ofa pipe that would deliver 7.5 gal/min at a velocity of3 ftlsec? What is the cross section area of the pipe and what would be the nearest nominal pipe size ? 12. The area of the cylinder in Fig. 4-1 is based on the formula . 2 _ Flow rate (gal/min) x Constant (0.3208) Area (10. ) Velocity (ftl sec)

13. 14. 15. 16. 17. 18. 19. 20.

If the constant were to be expanded to 7 digits past the decim al point rather than the 4 given in the figure, what would the last 3 digits be? A hydraulic oil with a Sg. of 0.85 has an absolute visco sity of p. = 425 poise. Convert this value to kinematic viscosity in centi stokes. Convert the kinematic visco sity of 1.5 Newt s to cSt and SI metric units. An oil has a kinematic visco sity of 160 SSU. Con sult Appendix C and list the equ ivalent viscosity con ver sion s in cSt , ft2/sec and SI metric (m2/s) . Using Fig. 4-8 , convert a kinematic visco sity of 100SSU to Newts, cSt , and SI metric unit s (m2/s) . Oil with a Sg. of 0.86 and an absolute visco sity of 12.5 x 10- 5 lbf-sec /ft? is flowing through a 2-in. I.D . pipe at 65 gal/min . Is the flow laminar or turbulent? Oil with a kinematic viscosity of 0.07 Newts is flowing through a 2-in. I.D . pipe at 500 gal/min. Is the flow laminar or turbulent? Compute the critical velocity range for a fluid with a viscosity of 0.045 Newts and a specific gravity of 0.91, flowing through a 1/2-in. I.D . tube. Compute the critical velocity range of an oil with a Sg. of 0.90 and an absolute visco sity of 27 cP flowing in a I-in. I.D . pipe .

Chap. 4

Study Questions and Problems

71

_ _ _ _ _ _ _ _CHAPTER

5

FRICTION LOSSES IN HYDRAULIC SYSTEMS 5-1 INTRODUCTION

When fluid is pumped through a fluid power system, a certain amount of the energy in the fluid is lost to friction. Major losses occur as the fluid flows through pipes, hoses, and tubing , while minor losses occur at valves, fittings, bends, enlargements, contractions, and orifices. Major losses are calculated for a given length of pipe. Minor losses, on the other hand , first must be converted to losses through an equivalent length of straight pipe using various experimental friction factors. To arrive at the total loss for a circuit, the major and minor flow losses are combined and substituted in one of the formulas that determine the pressure and horsepower losses associated with pumping the fluid. The friction generated from the flow of fluid through pipes brings to mind the notion that the fluid rubs the boundary surface as if a bullet were being pushed through the barrel of a gun, generating heat losses at the boundary. This is not what happens. What occurs is a rubbing action between the fluid particles themselves as they move through the pipe. When the flow is laminar, fluid layers move one past another, like a telescope, generating friction that can be related to the viscosity of the fluid. At the outside boundary, the fluid velocity is zero; at the center of the fluid stream, the velocity is maximum; and through the cross section of the stream, the friction gradient is linear. I The transition from laminar to turbulent flow introduces another important cause for friction and that is the cross flow and intermingling of particles as the total mass moves through the length of the pipe. The effect of this intermingling of flowing particles is to void the relationship between the relative roughness of the boundary surface and the Reynolds number, which changes the method by which the friction factor is calculated. Again, while the friction is not between the flowI

72

This is strictly true only for Newtonian fluids .

ing fluid and the wall itself, the rate of shear and heat generated are, in fact , greatest near the wall, and here is where most of the energy transfer occurs. 5-2 DARCY-WEISBACH AND HAGEN-POISEUILLE FORMULAS

The basic equation that governs viscous noncompressible flow in pipes is (5 - 1)

where hi is the head loss required to pump the fluid,f is a dimen sionless friction factor, L is the length of the pipe , D is the internal diameter of the pipe , v is the velocity of the fluid, and g is the acce leration due to gravity . This is known as the Darcy-Weisbach formula for viscous flow. The basic formula says , in effect , that the head loss associated with pumping a fluid thro ugh a system is a function of the velocity head (v 2 /2 g ), the length to diameter ratio of the pipe (LID), and a friction factor (f) that is de rived experi- men tally. Example 1 Oil with a Sg of 0.91 is flowing thro ugh a 2-in. J.D. pipe at the rate of 200 gal/min (Fig . 5-I). If the friction factor is 0.05, what would be the pres sure drop through 800 ft. of pipe ? Solution

Fir st , the velocit y of the fluid is determined from the conti nuity equation.

-

Q

v - (A)(3.12)

and (200)

v = (3.14)(3.12) = 20.4 ft/ sec Substituting in the Darcy-Weisbach formula to solve for head loss hf =

f

(~)(~;)

h = (0.05) f

(800 ft)(20.4 ft/sec )2 2 in. x 1 ft) ( 12 in. (2 x 32.2 ft/sec)

-\

Fluid 59 = 0.91 Fricti on factor f = 0.05

Figure 5·1

Sec. 5-2

Example I.

Darcy-Weisbach and Hagen-Poiseuille Formulas

73

.and

hJ = 1551 ft

Converting the head loss hJ to pre ssure loss hpsif

hpsif = (0.433)(0.91)(1551 ft ) = 611 lbf/in .! The value of the friction factor in the Darc y-Weisbach formul a is largel y determined by whether the flow is laminar or turbulent. Fo r laminar flow, the friction factor has been determined experimentally to be f = 64 NR

(5-2)

When this value is substituted in the Darc y-Weisbach formul a for head loss due to friction , the resulting equation is one form of the Hagen-Poi seuille formul a for losses due to friction when laminar flow prevail s. That is,

and

32Lv 2 'f = NRDg

h

(5-3)

If the value for N R is substituted in Eq. 5-3 N R = vD v

h = (~)(32 LV2) f vD Dg and

h 'f

32vL v = D2g

(5-4)

Thi s is the more common form of the Hagen-Poi seuille formul a. Example 2

Oil with a S g of 0.85 and a visco sity of 0.08 Newt s is flowing in a I-in. I.D. pipe 400 ft long , at the rate of 15 gal/min (Fig. 5-2). Determine first that the flow is laminar, and then compute the pre ssure drop in lbf/in ",

-I

_------400' ------~

Fluid 59 = 0.85 Kinemat ic viscosity = 0.08 Newts

Figure 5·2

74

Exa mple 2.

Friction Losses in Hydraulic Systems

Chap. 5

Solution

The flow is considered to be laminar if the Reynolds number is less than

2000. First, the velocity is determined to be v-

Q

- (A)(3.12)

and (15) v = (0.785)(3.12) = 6.12 ft/sec Substituting in the Reynolds formula, N = 12vD R v

and N = (12)(6.12)(1) = 918 R (0.08) so the flow is laminar. Substituting in the Hagen-Poiseuille formula for pressure drop when laminar flow prevails, h

f =

32vLv D2g

h = (32)(0.08 in.2/sec)(4oo ft)(6.12 ft/sec)

(I-in.)2(32.2 ft/sec ")

f

and

hf

=

194.6ft

Finally, solving for the pressure loss where hpsif = (0.433)(Sg)(hf ) hpsif = (0.433)(0.85)(194.6 ft) = 71.6 lbf/in.?

Notice that in the solution of the Hagen-Poiseuille formula that the units must be compatible such that hf is in units of ft or m. 5-3 f-FACTOR FOR TURBULENT FLOW

During laminar flow, the friction factor is relatively independent of the surface condition of the inside diameter of the pipe. When the flow is completely turbulent, the f-factor is read from Fig. 5-3 by locating the place of intersection of the Reynolds number and the relative roughness of the conductor surface, and then reading the friction value from the left or right margins. The relative roughness of the pipe wall is computed as the dimensionless ratio of the absolute roughness e (Greek letter epsilon) to the inside diameter of the pipe D. That is, Relative roughness

=;

(5-5)

where the absolute roughness equals the average projection of the surface imperfections on the inside wall of the pipe. The absolute roughness of pipes made from Sec. 5-3

f-Factor for Turbulent Flow

75

10J

....en

_____

0.100 I 0.090 0.080

IIIII11I

'2

'Y

t

IIII1

4

Stroke x 2.0

~

Stroke x 4.0

5. Compute the stop tube length if one is applica ble (Fig. 8-5). Stop tub es are used on the cylinder rod s bet ween the piston and the end cap to pre vent overextension of the rod when the cylinder is fully ex tende d . Misalignm ent , due to long stro kes and unsupported weight, so metimes cau ses rod jackknifing or buckling if the piston rod is allowed to extend the piston fully to the

Figure 8-5

Sec. 8-2

Cylinders

Stop t ube length .

167

TABLE 8-3

PISTON ROD DIAMETER FOR MAXIMUM LOADING *

...en

Piston rod dia mete r in.lmm

co

Load (lbf) 100 150 250 400 700 1,000 1,400 1,800 3,200 4,000 5,000 6,000 8,000 10,000 12,000 16,000 20,000 30,000 40 ,000 50,000 60, 000 80,000 100,000 120,000 140,000 160 ,000 200,000 250,000 300,000

V l6 65/165 50/127 38/97 29/74 21/53 17/43 14/36 10/25

1/25

100/254 78/198 58/ 147 48/122 40/102 30/76 25/64 22/56 20/5 1 18/46 15/38 12/30 11 /28

Ii!l35

155/ 394 112/284 94/239 78/ 198 58/ 147 50/ 127 44/ 112 39/99 35/89 30/76 26/66 23/58 18/46 14/ 36

-

* Tab led

values are stroke in in.l c m.

W 44

190/483 160/406 132/ 335 98/249 85/2 16 75/ 191 67/99 60/ 152 52/132 46/1 17 42/107 35/89 29/7 4 19/48 10/25

2/5 1

210/533 175/44 5 130/ 330 11 0/ 279 98/249 88/224 78/ 198 671170 60/ 152 54/137 46/117 40/ 102 30/76 21/53 12/ 31

21/64

275/699 210/533 170/432 158/40 1 140/356 127/ 323 110/ 279 96/244 92/234 76/ 193 68/ 173 52/132 40/102 30/76 13/33

3/76

300/762 257/653 228/579 200/508 182/462 156/396 138/351 125/3 18 108/274 95/24 1 77/ 196 64/ 163 54/ 137 44/ 112 28/71

31/89

300/762 2801711 252/640 220/559 192/488 178/452 152/ 386 136/345 110/279 94/239 84/2 13 74/188 57/ 145 43/ 109 29/7 4

4/102

300/762 290/737 260/660 236/599 200/508 180/457 145/368 125/318 111/282 101/257 85/2 16 70/178 58/ 147 44/ 112 10/25

4!1 114

300/762 295/749 254/645 228/579 183/465 159/404 140/ 356 130/ 330 110/279 97/246 84/2 13 73/ 185 62/ 157 38/97

5/120

300/762 275/699 220/559 190/483 170/432 155/394 135/343 120/305 11 0/279 97/246 85/2 16 64/ 163

51/ 140

Loa d (k N)

300/762 275/699 240/6 10 2 12/538 192/488 168/427 148/ 376 136/ 345 124/315 112/284 93/236 69/ 175 10/25

0.45 0.67 1.11 1.78 3. 11 4.45 6.23 8.00 14.23 17.79 22.24 26.68 35.58 44.48 53.38 71. 17 88.96 133.44 177.92 222.40 266.88 355.84 444.8 533.76 622.72 7 11.68 889.60 1112.0 1334.0

end of the cylinder. Stop tubes spa ce the piston away from the cylind er end to prevent thi s condition and increase bearing life . Oversized cylind er rod s are not considered an acceptable substitute for cylinder rod stops since they are more expensive , add weight to the system, and if misalignment doe s occur, promote be arin g and seal wear bec au se of increased stiffness of the oversized rod. Stop tub e length is computed using the formula S top T u be Lengt h

=

Corrected Length - 40 in. 10

(8- 3)

Additional bearing surface at the rod support end of th e cy linder and/ or double piston construction may be necessary for unu sually long stro kes or heav y load s. Example 3 Compute the diameter of a cy linder rod to be used in a cy linde r with a bore diameter of 3 in ., op erating at a maximum pre ssure of 2000 lbf/in. ", cen ter trun nion mounted with a 20-in. stro ke (Fig . 8-6) . Compute the length of the stop tube if one is applicable .

Solution 1. Th e co lumn strength factor from Table 8-2 equ als 1.5. 2. The corrected length of the rod is then computed from Corrected Length

= 20 x 1.5 = 30 in.

3. Th e thru st on the rod is computed from sys tem maximum pre ssure and the diameter of the cy linder.

F=p XA F

= 2000

x 7.07

=

14,143 Ibf

4. The pist on rod diameter is det erm ined from Table 8-3 by first locating the ne xt larger value, 16,000, in the left hand column, and moving ac ross to 35, the corrected rod length. Finally, read ing upward to the value at the top , a rod diameter of P/4 in. is indic ated .

Fluid at 2000 ps i

3 in. bo re - L cylind er

Figure 8·6

Sec. 8-2

Cyl inders

Exa mple 3.

169

5. Finally. stop tube length is computed from Eq . 8-3 Stop Tube Length

=

Corrected Length - 40 in. 10

Stop Tube Length

=

30 - 40 10

=-)

The minus value indicates that a stop tube for a corrected cylinder length under 40 in. is not appropriate in this application.

8-3 CUSHIONING DEVICES

Cushioning devices are provided in the ends of hydraulic cylinders or as separate units when loads must be decelerated. Hydraulic shocks to the cylinder and system are reduced by slowing down the piston just before it contacts the cylinder end caps . The sleeve or spear-type cushioning device is often used as the decelerating mechanism . The energy absorbed by decelerating a body in motion is converted to heat and dissipated to the environment through the fluid and cylinder body. Refer to Fig. 8-7. As the piston completes its extension stroke, the slightly tapered end cushioning sleeve enters the rod end cap blocking the normal flow of fluid. The flow of fluid is then rerouted through the bypass port and needle valve at a controlled rate, decelerating the piston. Cushioning at the head end of the cylinder during the retraction stroke is accomplished in the same manner by the action of the tapered cushioning spear plunger as it enters the end cap . This blocks the normal flow of fluid and causes fluid to be rerouted through the bypass port and metering valve to decelerate the piston. The check valve, shown only on the blank end of the cylinder, is used to direct fluid at system pressure to the full area of the piston during acceleration.

Figure 8-'

170

Cylinder cushioning devices (court esy of Cart er Controls . Inc.).

Hydraulic Cylinders and Cushioning Devices

Chap. 8

The force exerted by a body put into motion is derived from Newton 's law F= Ma

(8-4)

where the mass (M) is equal to the weight of the body (IV) divided by the acceleration (g) due to gravity , and the acceleration (a) is that given to the body. In notation , IV

M=-g

(8-5)

The force then becomes

w Xa F=-g

(8-6)

where the force (F) is in Ibf (N), the weight (w) is in Ibf (N), the acceleration or deceleration of the body (a) is in ft/sec? (m /s-) and the acceleration due to gravity (g) is in ft/sec? (rn/s-). An acceleration force factor (ga) can be defined as

a

s: = g

(8-7)

where the acceleration factor (ga) is without dimension. Values for the acceleration factor (ga) may be read from graphs (Fig. 8-8) or computed from the acceleration (a) given to the body. The acceleration (a) of a body, for example , attached to a hydraulic cylinder, is computed through a stroke (5) to a velocity (v) . That is (8-8)

where the acceleration (a) is in ft /sec-, the velocity (v ) is in ft /sec and the stro ke (5) is in ft. If the velocity is given in ft /min and the stro ke is in in. , the formula becomes v2 3,600

a=--

2 x 5 12

and a

v2

=S x

.00166

(8-9)

If the acceleration due to gravity (g) is given the value 32.2 ft/sec-, the acceleration factor (g a) thus becomes

s, =

v 2 x .00166 5 x 32.2

and 2

s; = Sv Sec. 8-3

Cushioning Devices

x .0000517

(8- I0)

171

300

.....

rTTT

rrrrr

N

Qj

:; c

'E

... '"c. ~ :=.. '"

60

I

'(3

o

E

I V

40I

30

,

V

20I

V

. "/-"./

15

/"

'Vi,.-/V

10

~~

0.010

t,../

v

V

V

V

V

V

91.4

.>

b::

61.0

~

t:: t:: ~ . /

V

k

V

I~ ~ / v . . . i-

v

I

V ~t5"/ ........ v""""V ___ ' 1/V

V

VV~V

V

[.....< p ~

V~t::~t::~~k V vV'

~~r;/VV /

/ . '>

~

I II I I

V V

- (5) F= Wg- W

V'

I

VV V ~ /VVVVV~

- (4)F = Wg + W

~V/VVV

I

I

>

I

._J,~~E:::~VV V

:3

.S X "' E

i

vvt::~V

V

I

I

, - If load friction (II I, 10~I~d. 4+~-' S - 1 ;0./25 mm add or subtract as appl icable . S = 1~ 1n./31 mm ........ 80I '- Cy linder friction need not be S = 11 . 38 ~t::/': considered since it is insignificant . 2 I~.~ ~V ./ ) ~ i n most applications. S = 1% 1n./44 m m /

... >

- (1)F = Wg - (2) F = W(g + f - (3) F = W(g - f

t: 100

>-

a;

I" l

III I

For horizontal motion To accelerate or decelerate To accelerate load and overcome friction 200I '-T o decelerate load and friction For vertical motion Accelerat ion upward or deceleration 150I f-- downward Acceler ation downward or deceleration upward

-..I

t--.

""""

V~

V

.....

f- - -

/'

5= 1 " ~

ceW~'3

Figure 8-8 Acce leration force factor gra ph.

18.3

...>-

II>

'"E

' (3 0

4

. ._

i I I

~\o(\

_ . . 1--

(\ce 6\

. f-

- --

-. I

0.20 0.25 0.30 0.40 0 .50

9 - accelerat ion force factor

24.4

P--t-- 5 ' J " ,

E

--......'" >

II6.1

= = = -

4.6

I~ 3

11 11

0.15

30.5

.S

a;

--

0.0150.02 0.0250.03 0.040.050.06 0.08 0.10 0.07 0.09

:3

VV

~c

V

:

i

~\o(\

--

...

Qj

-

V

5 = 1" I ~e~'3 4 oece , o~

V

45.7

0.70 0.90 0.60 0.80 1.00

1.50

2.00

The force exerted on or by the system that is attributable only to the weight being accelerated or decelerated equals (8-11)

if tabled values for gu are used , and IV X

S

F =

v2

x .0000517

(8- 12)

for computed value s of gu. Example 4 Compute the force required to de celerate a free bod y weighing 8,000 Ibf (35,584 N ) from 120 ft /min (36. 576 m/min ) within a distance of 1.5 in. (0.038 1 m) (See Fig. 8-9. ) Solution

First, compute the acceleration factor ( gil) using Eq, 8- 10 gil

v2

= :f x

gil =

0.0000517

(120)2

IT x

0.0000517

and gil

= 0.496

Finally , compute the force to accelerate bod y using Eq . 8-11

F

=

8,000 x 0.496

=

3,968 Ibf

In SI metric unit s the solutio n is computed using the same method and Eq s. 8-7 and 8-1 I.

v2 2 x S v2 gil = - g- = 2gS

and 1

gil

= (36.576 mlmin x 1/60 min /sec )2 = 0496 (2)(9.8 m/ s2 )(0.038 m) .

Finally , solving for the force to decelerate the body using Eq. 8-11 F = w

X

gil = (35,584 N)(0.496) = 17.649 kN

8000 Ibs.

J II

Decelerat ion

-+i ~ distance 1.5 in. Sec. 8-3

Cus h io n ing Dev ices

Figure 8·9

Example 4.

173

When a friction factor (f) acts to impede the motion of a body, it must be added to the total force when the body is accelerated and subtracted from the total force when the body is decelerated. For tabled values of the acceleration factor (g a), the basic acceleration or deceleration formula becomes F

=w

F

=

w x

X ga ±

f

and (8-13)

w(ga ± f)

For computed values of s, F

= w

[( .0000517

X~2) ± f]

(8-14)

If the coefficient of friction factor (f) is greater than the acceleration factor (g a) and negative, then a negative force (F) results from these two formulas . The additional force being imparted to a vertically moving body being accelerated or decelerated must also be considered and consists of the load resistance of the body itself (w). This situation exists, for example, when a hydraulic elevator must be accelerated or decelerated, and the load must be added or subtracted appropriately. For tabled values of the acceleration factor (ga), the total force during acceleration or deceleration equals F

=

w(ga ± f) ± w

Specifically, the following conditions may exist:

+w w( ga - f) + w w(ga + f) - w

(8-15)

F = w(ga - f) - w

(8-18)

Accelerate upward

F

=

Decelerate downward

F

=

Accelerate downward

F

=

Decelerate upward

w(ga + f)

(8-16)

(8-17)

Similar formulas are obtained if computed values are substituted for the acceleration factor (g a). For bodies moving horizontally, the additional force (F,J imparted to the body by an actuator that continues to have fluid power applied to it during deceleration must be added to compute the total force . For a body actuated by a cylinder, for example, the formula to compute the total force during deceleration , including friction, is

F

=

w(ga - f) + Fa

(8-19)

for tabled values of the acceleration factor (ga), and

F

=

w [(.0000517 x

~)

-

f] + Fa

(8-20)

for computed values of ga' This is the case existing when a cushioning device is used on the rod side of the piston to decelerate the load as fluid continues to be 174

Hydraulic Cylinders and Cushioning Devices

Chap. 8

supplied to the blind side of the piston. A similar condition exists when a separate cushioning device is used to decelerate a load moving under a continued force supplied by a cylinder actuator. Example 5

Compute the total force necessary to decelerate a load of 1,500 Ibf (6672 N) traveling at 50 ftlmin (15.24 m/min) with a coefficient of friction (f) of 0.12 within a distance of 3/4 in. (0.01905 m). The cylinder actuator moving the load resistance has a 3-in. (0.0762 m) bore operating at a maximum relief valve pressure of 750 lbf/in.? (5.172 MPa) (Fig. 8-10). Solution Since the load is moving horizontally and decelerating under an applied force (Fa), the formulas used are Eq. 8-19 F = w(ga - f)

+

Fa

for tabled values of the acceleration factor ( g ll), or Eq. 8-20 F =

II'

[(.0000517

X~2) - f]

+ F lI

for computed values of g ll' The maximum additional force (FlI ) applied by the cylinder to the load resistance during deceleration is F lI =p xA

and

FlI = 750 x 7.065 = 5,299 Ibf Using tabled values for the acceleration factor (gll) (see Fig . 8-8), the velocity of the load resistance, 50 ftlmin, is located first in the left margin of the table. Moving horizontally to the right, locate the line indicating the length of the stroke (5) for deceleration. This is 3/4 in. Then, moving vertically down to the lower margin, locate the acceleration factor , which is approximately 0.172 . Substituting in Eq. 8-19

F = 1,500(0.172 - 0.12) + 5299

and

= 78 + 5299 = 5377 Ibf

F

750 psi relief pressure

1500 load Friction coefficient 0.12

_.. - ~ 50 ft /m in

3 in. bore cyl inder

Figure 8·10

Sec. 8-3

Deceleration distance 0.75 in.

~I ~

/

Cushioning Devices

Example 5.

175

Using Eq. 8-20 for computed values for the acceleration factor ( ga)' the final force (F) would be computed similarly.

F =

IV

[ ( .0000517

~2)

X

F = 1,500 [ ( .0000517 x

- 0.12] + Fa

(~~n

-

0.12] + 5298.75

and F = 1,500(.052) + 5299 F

= 78 + 5299 = 5377 Ibf

which is the same value obt ained using tabled values for ga' In 81 metric unit s, solving for g" using Eq , 8-7,

s,

= ~ = (15.24 m/mi n. x

1160 min/sec) 2 = 0 172 (2)(9.8 m/s 2 )(0.019 m) .

2gS

Finally, solving for the deceleration force using Eq , 8- 19,

F = lV( g" - f ) + F" and F = (6672 N)(O.I72 - 0.12)

+ (23 576 N)

= 23922 N

The pres sure generated on the cushion side of the piston is a function of the force app lied and must be taken into account to prevent rupturing the cylinder. An example illustrates this computation. Example 6 Assuming a total deceleration force of 7,500 Ibf on the stationary end of a cylinder with a 3-in. bore and I liz-in . single end cylinder rod , compute the fluid pre ssure at the rod end of the cylinder (Fig. 8-1 I) . Solution Fluid pre ssure at the rod end of the cylinder is computed from the force applied duri ng deceleration and the effective area. The effective area of the cylinder 7500 Ibf force

Pressure

3 in . bore

/

Rod d iameter 1.5 in .

Mete ring device

Figure 8-11 Ex ample 6.

176

Hydraulic Cylinde rs and Cushio ning Devices

Chap . 8

Shoc k

!! absorbers

(a) Stopping of loads transferring from one convey or to anotherpaper rolls , met al coils , slabs, other large loads.

(bl Un iform decelerat ion of product s in aut omati c assembly op erations .

Shoc k absorber

(c) Hydraulic bumpers for railroad cars; endstops for inp lant mater ial tra nsf er cars.

/

/

/'?\

(e) Cushion ing for roll -over applicati ons o f all types.

Load force

(d) Disappearing stops in material handling transfe r lines.

Shock absorber

(f) Forward and reverse cushion ing

of shuttle devices.

Figure 8-12 Typical deceleration application using shock absorbers.

Sec. 8-3

Cushion ing Dev ices

177

at the rod end is

and AI = 7.065 - 1.766 = 5.299 in.?

Pre ssur e in the cylinder is then computed from

F p =A and _ 7500 _

p - 5.299 - 1414

tbf/ '

tn.

2

Thi s is the pre ssure the cylinder or cu shioning device must withstand. Since it is at the rod end that has the smalles t effective area, given the same deceleration force in both directions, it will generate the highest unit pressure (lbf/in. t) that must be contained by the cy linder. External hydraulic energy absorbing devices also are employed many times to reduce shock loading of machinery that would result in failure. Carriage stops on turntables , doors, tran sfer machines , turnover s, and small cr ane s requiring controlled deceleration use the se devices (Fig. 8-12). In operation, fluid is metered through adj ustable orifice s for a wide range of impact load s and velocities. In Fig. 8-13 , for ex ample , liquid is metered through the orifice plate valve along the tapered metering pin as the piston is forced into the cylinder. Adju stment is accomplished by removing the protective plug and rotating the adjustment screw with a hexagonal socket ke y for more or less damping. The hardened and chrome plated tubular piston re sists buckling and side load s under impact. The metered liquid is stored in the spring-loaded separator chamber which also repositions the piston. Gu ide Bu sh ing s Wiper.Scraper Seal

Ta pe re d Meter ing P in

Ncn-E xtr usion Seals

Cyl inde r P rot ec ti ve P lug

A Mount ing Flange

pa rat or Da s hpot Heo d Orifice Vo lve P late Adjus table Meteri ng Scr e w

Figure 8·13 Typical hydraulic shock abso rber with full-range adjustability (courtesy of Enidi ne Corporation).

178

Hydrau lic Cylinders and Cush ioning Devices

Chap . 8

8-4 SUMMARY

Hydraulic cylinders convert the high energy level of the fluid to a linear motion by applying the fluid to the movable piston. Movement thereby accomplishes the output objective of the system and lowers the energy level of the fluid, after which it is returned to the reservoir at low pressure . Cylinders are sized not only to meet the force and distance requirements of the machine , but also to provide the strength needed to prevent buckling. The most severe example is rear flange mounting with the piston rod unsupported , which requires an increase in the stroke by a column strength factor of four to provide the necessary strength. One means to prevent overstroking is to install stop tubes on the cylinder rod to limit the stroke. Hydraulic cylinders extend and retract to complete a cycle and sometimes include stepped or spear type cushioning devices in the ends to prevent shock loading and noise associated with inertia and overrunning loads. External cushioning devices are another means to prevent shock loading and use a hydraulic shock absorber principle. They are constructed with spring-returned, adjustable pistons and internal metering devices. In both cases , the force to be absorbed by the cylinder or external cushioning device is computed from the magnitude of the load , hydraulic force , acceleration , direction of travel , and the friction factor. S TU D Y QU ESTIONS A ND P R OB LEMS 1. Ignoring friction and fluid losses, how large would the blank end of a cylinder operating at 2000 lbf/in. ? have to be to raise 25 tons? 2. Compute the linear speed of a 2-in. bore cylinder receiving fluid at 5 gal/min. 3. A cylinder has a bore of 4 in. and a rod diameter of 2 in. Compute the velocity of the cylinder as it extends and retracts if 10 gal/min is supplied to the system. 4. A lO-kN car and 5-kN hoist are supported 1.8 m above the floor by a cylinder with a 300-mm bore . What pressure will this develop in the system? S. If the car in Problem 4 is lowered at 0.33 ftlsec, what would be the velocity of the fluid through the return line if the cylinder were plumbed with l-in, Schedule 40 pipe (clue: see Table 8-1 for pipe dimensions). 6. A cylinder with a 7-cm bore and 2.5-cm rod receives fluid at 40 I/min. What is the velocity of the cylinder rod extending and returning? 7. If the cylinder in Problem 6 has a stroke of 40 ern, what is the maximum cycle rate that might be achieved? 8. A hydraulic cylinder with a 2.5-in. bore and a 1.25-in. cylinder rod receives fluid at 12 gal/min through 3/4 -in. Schedule 40 pipe. What is the velocity of the fluid through the return line when the cylinder is retracting (clu e: draw an illustration showing the direction of flow). 9. Calculate the corrected length and diameter size for a piston rod in a rear trunnion mounted hydraulic cylinder that must exert a 5-ton force through a 24-in. stroke. 10. A center trunnion mounted cylinder with a 35-mm diameter cylinder rod and 100-cm stroke is extending under a force of 35 kN. How far can the cylinder safely extend without exceeding the recommended stroke? Chap. 8

Study Questions and Problems

179

11. A rear flange mounted cy linder extends 14 in. under a load of 10,000 lbf. What size cylinder rod would be required ? 12. For the cylinder in Problem II, what stop tube length would be required ? 13. Compute the force required to decelerate a free bod y weighing 2000 lbffrom 150ft/min within a distance of 2 in. 14. A 3-in. bore cylinder operating at 2000 lbf/in .? is driving a load of 3000 Ibf horizontally at 20 ft/min . Compute the force nece ssary to decelerate the load within a distance of 2 in. if the coefficient of friction is 0.15. 15. A hydraulic cu shioning device has a 2-in. bore and a 0.75-in. cylinder rod . If the fluid pressure on the rod end of the cylinder increases as a result of absorbing a total deceleration force of 6000 lbf, what is the pressure generated ? 16. Neglecting friction, what force could be expected from the rod end of a cylinder with a 5-in. bore if the blank end receives fluid at 2500 Ibf/in. 2? 17. A filled drum weighing 2 kN rolls against a cushing device at I m/s. If friction is negligible, what force would be required to decelerate the drum within 2 em? 18. If in Problem 17 the force is absorbed through a hydraulic cushioning device with a cylinder bore of 350 mrn, what pressure could be expected? 19. A 2000 lbf elevator with a 1000 lbf cargo descends at 2 ft/sec. If the friction factor equals 0.08, and the elevator decelerates within a distance of 6 in., what force downward would the cushioning device have to absorb? 20. In Problem 19, what would be the deceleration distance if the elevator were ascending and the cushioning device absorbed the same deceleration force?

180

Hydraulic Cylinders and Cushioning Devices

Chap. 8

_ _ _ _ _ _ _ _ _ _CHAPTfR

9

HYDRAULIC MOTORS

9-1 INTRODUCTION

Rotary actuators convert the energy in the pressurized fluid to the turning motion of a shaft. They exert both limited rotation, as in the case of torque motors , and continuous rotation as with hydraulic motors (Fig. 9-1). Limited rotation actuators use several novel arrangements of pistons and vanes to achieve rotation, from a few degrees to several rotations . Continuous rotation actuators, commonly called hydraulic motors, use internal geometry similar to gear, vane, and piston pumps to achieve continuous angular motion of the shaft. In addition to off-the-shelf units, there are a number of specialized hydraulic motors. Radial piston motors , for example , exert a high torque, low speed characteristic because of the size and geometry of the piston arrangement around a common crankshaft. Two-speed operation of these units is achieved by delivering pressurized fluid to the pistons in sequence or in tandem pairs. Another application of the radial piston motor concept is wheel motors, used for agricultural equipment and off-the-road vehicles. Wheel motors are available as stationary bolt-on units and as steerable yoke units. The machine tool industry and other precision applications make extensive use of electrohydraulic pulse motors to provide precise positioning and velocity control. These units combine a low power digital stepping motor with a high power precision hydraulic piston motor to achieve a position or speed in proportion to the input pulse rate of the control signal. 181

Actuators

I Linear

I

Rotary

I---~

Limited rotation r---

Piston

Continuous rot at io n

I- ---..,

Gear Vane

Vane

Piston HTLS

Wheel motors

EHPM

Figure 9-1

Actuator cla ssificat ion.

9-2 TORQUE, SPEED, POWER, AND EFFICIENCY

Rotary actuators convert system pressurized fluid to torque at the dutput shaft. Theoretical torque at the output shaft is derived by equating the basic fluid horsepower formula to the brake horsepower formul a (as wa s done in Eq. 2-11) and solving for the va lue of T. That is,

p X Q 1,714

=

T xN 5,252

T = p x Q x 5, 252 N x 1,714

in Ibf-ft

It is ob served in the formula that

Q =v, -

-

N

(9-1)

231

where the volume di splacement of the motor ( V m) is in in .! per re volution . If torque in lbf-in . is de sired , then T = p

x

Vm x 5,252 x 12 231 x 1,714

and

or (9-2) where torque ( T) is computed in lbf-in. , pressure is in lbf/in .", and displacement of the motor (Vm) is in in .' per revolution. 182

Hy draulic M otors

Chap. 9

Example 1 Compute the theoretical torque from a rotating hydraulic motor operating at 2000 lbf/in. ? with a displacement of 0.5 in. ! per revolution . Solution

Th eoretical torque is computed fro m Eq . 9-2 T =p x V m x 0.16

a nd T = 2000 x 0.5 x 0.16 = 160 Ibf-in.

It should be noticed that the torque is not affected by changes in speed or flow rate so long as the pre ssure remains constant, since from Eq . 9-1,

v,

Q N

231

Also notice that the theoretical speed (N) of the motor is computed by solving for N N

=

231 x Q

Vm where the speed (N ) of the motor is in rpm. Similarly , theoretical flow through the motor (Q) in gal/min is computed by solving Eq . 9-1 for Q: Q

v, x N = ---'--...,,---231

Example 2 Compute the th eoretical speed of a fluid po wer motor with a 2 in .! (0.0328 I/rev) per re volution displ acement recei ving fluid at the rat e of 7.5 ga l/min (28.39 I/min). Solution

The speed can be computed from Eq . 9-1 N = 231 x Q

Vm and N = 231 x 7.5 = 866 rpm

2

In 81 metric units , where the displacement of the motor is given in liters and the flow rate is in liters/min, Eq. 9-1 simply becomes

N

Q

28.39 I/min . I/rev - 866 rev /min

= Vm = 0.0328

Example 3 Compute the effective flow rate through a fluid power motor with a displacement of 0.55 in. ' per revolution turning at 3400 rpm . Sec. 9-2

Torque, Speed, Power, and Efficiency

183

Solution

Flow rate is computed using Eq . 9-1

Q=

Vm x N 231

Q=

0.55 x 3400 231

and . = 8 .09 ga II rmn

Motor efficiency is measured primarily by how well the fluid horsepower input is converted to useable brake horsepower at the motor output shaft. Overall motor efficiency can be computed by comparing the useable brake horsepower output to the fluid horsepower input. That is,

e = (J

T xN 5252 - - x 100 p x Q 1714

and

T xN eo = p x Q x 3.06 x 100

(9-3)

where the torque (T) is measured in lbf-ft , the speed (N) is in rpm, the pressure (p) is measured across the motor inlet and outlet ports in lbf/in .2 , and the flow rate (Q) is in gal/min. Notice that this formula is the reciprocal of the formu la for overall pump efficiency . Example 4 Compute the overa ll efficiency of a motor operating at a pre ssure of 2,000 Ibf/in .2 , a flow rate of 7.5 gallmin , a speed of 1,200 rpm and exerting a torque of 32 lbf-ft . Solution Ove rall efficienc y is computed fro m Eq . 9-3

e

o

=p

T xN x 100 x Q x 3.06

and 32 x 1,200

eo = 2,000 x 7.5 x 3.06 x 100 = 84% The volumetric effic iency of a fluid power motor can be measured by comparing the effective vo lume flow rate (Q e) ind icated by th e product of the d isplaceme nt of the motor (Vm ) times its speed (N), to the actual vo lume flow rate (Q,,) supplied to the motor. That is, e v

ev

184

=

Qe x 100

Q"

Vm x N 231

=- - -

x 100

Q" Hydraulic Motors

Chap. 9

and ev

v, = Qa

x N x 231 x 100

as a percent

(9-4)

where the displacement of the motor (Vm) is in in.! per re volution , the speed (N) is in rpm , and the actual volume flow rate ( Qa) is in gal/m in. A major factor in volumetric efficiency that gives an indication of motor condition can be mea sured by ob serving the flow rate of fluid returning from the ca se drain (q) to the reservoir and comparing this value to the tot al actual flow rate ( Qa) supplied to the motor. The percent lost to the case drain ca n then be computed using the formula Percent Case Drain Lo ss

=

Jax 100

(9-5)

where the ca se drain loss ( q) and the actual flow rat e ( Qa) are measured in gpm . On small motors with minimum leakage, percent case drain loss may be negligible and difficult to compute . Example 5

Compute the volume tric efficiency of a mot or with a displacement of 0. 6 in.! per revolution (0. 00983 l/ rev) turni ng at 2,400 rpm that is rece iving fluid at the rat e of 6.5 gal/min (24.61 l/ min). How much fluid might be ex pec ted to return to the rese rvo ir from the case drain if 50% of the loss in volume tric efficie ncy ca n be attributed to inte rna l leakage ? Solution

Volumetric efficiency for thi s motor is co mputed from Eq. 9-4 Vm x N ev = Qa x 231 x 100

0.6 x 2,400 e., = 6.5 X 231 x 100 = 95.9% In SI met ric units , where the displacement is give n in liter s/re v, and the del ivery in liters/min , Eq. 9-4 becomes

e., =

(0.0098 l/rev)(2400 rev/min ) (24.6 1 l/ min) x 100 = 95.9%

Th e tot al fluid flow lost from slippage through the high pre ssure ports and chambers of the pump as well as internal leak age that return s through the case drain is Tot al Fluid Loss

= 0.041

x 6.5

= 0.267 gal/min

If half thi s loss ca n be att ributed to interna l leakage Case Drain Loss Sec. 9-2

= 0.50 x 0.267 = 0. 134 gal/ min

Torque, Speed, Power , and Effic iency

185

Mechanical efficienc y (em) for the motor is then computed from related efficiencies the same as for hydraulic pumps. That is,

and as a percent where the overall efficienc y (eo) and volumetric efficiency (e v) are also computed as percent. This is a restatement of Eq s. 7-14 and 7-15. Example 6 Compute the mechanical efficiency of a fluid power motor that has an overall efficiency of 88% and a volumetric efficiency of 97%. Solution

Mech anic al efficiency of the motor is computed from e

m

=

eo x 100

ev

and 88 em = 97 x 100

= 90.72%

9 -3 LIMITED ROTATIO N ACTUATORS

Limited rotation actuators, called torque motors, have a wide variety of applications where a limited specified degree of rotation at the output shaft is required . Rotation is usuall y limited to 720°. The y are used exten sively in indu stry for actuating clamping devices , material handling , rotating cams for braking mechanisms, tumbling and dumping, positioning and turning , and many other situations where an economical application of fluid power for limited rotation is desirable (Fig. 9-2). Vane -type limited actuators apply fluid force to the cro ss section area of single or multiple vane s. Rack and pinion type actuators apply fluid force to the cylindrical chambers which move the rack to drive the pinion gear.

(a)

186

Figure 9-2 (a) Rack and pinion limited rot ation ac tuator (co urtesy of Flo-Tork , Inc. ); (b) Vane type limited rotation actuator (co urtesy of Bird-John son ); (c) Ty pical applications of limited rot ation actuators (co urtesy of Bird-Johnson ).

Hydraul ic Motors

Chap. 9

(e)

Plug or butter f ly valve turn ing or posi ti o ning

A///~_

cy

~

Osci ll at ing harmonic mot ion

"""- - - - ..... ,

tID 1:[

/'

I /

r

--

'-

"-

I I :' I I 1: I

I II

\

"

II , :~ : II

I

\

11'1' --- -11--,

-(g C

\

Roll over or posit ion ing f or weldi ng or machin in g f i xt ur es

Flipover between wo rk stat io ns

All bending oper ati ons



OOWELED CONSTRUCTION INTEGRAL GEAR AND SHAFTDESIGN

DRAIN LINE PORT MAXIMUM PEAK SIDE LOAD 4,000 LBS. MAXIMUM SIDELOAD 2,500 LBS. MAXIMUM END THRUST 2,000 LBS.·EITHER DIRECTION

Symbol Figure 9-19 High-torqu e low-speed gear mot or and redu cti on unit (courtesy of Borg Warn er).

multiplies the displacement , flow rate , and torque of the motor by the same amount. At the same time , output shaft speed is reduced to 1/5.41 of motor speed . 9-7 ELECTROHYDRAULIC PULSE MOTORS (EHPM)

Electrohydraulic pulse motors offer the designer, builder, and user of machines an unusual combination of power and precision for position and velocity control. EHPMs provide precision pos itioning, a high transverse rate, and very simp le control and installation. They are es sentially bolt on and use unit s. An EHPM combines a low power digital electric stepping motor with a high power preci sion hydraulic motor (Fig. 9-20). Con ventional square wave digital pulses are used as input commands to the electric stepping motor. Thi s , in turn , controls the direction , speed and position of the high-torque hydr aulic motor shaft. Precision control of the shaft is accomplished without requiring the use of external feedback transducers . Since clo sed-loop feedback is not used , the EHPM sy stem is defined as open loop. The output shaft speed of the hydr aulic 200

Hydraulic Motors

Chap. 9

Figure 9·20 Electrohydra ulic pulse motor (cou rtesy of Sperry Vickers).

motor is in direct proportion to the input pulse rate to the electric stepping motor. Output shaft position is directly related to the number of input pulse s. The rotational direction of the output shaft is determined by the sequence in which the input pulses are applied to the various phases (or coils) of the electric stepping motor. In operation , low power digital electrical pulses, from a standard digital electronic or NC control , are fed to the electric stepping motor that rotates depending on the sequence, direction , pulse rate , and numbe r of pulses. The stepping motor opens the four- way spool valve in either direction by rotating the spool through a gear set, in or out of the ball nut (Fig. 9-21). HydrauHydrau lic piston motor

Figure 9-21 ers ).

Sec. 9-7

Electro hydra ulic pulse motor operation (courtesy of Sp erry Vick-

Electrohydraul ic Pulse Motors (EHPM)

201

lie fluid is allowed to pass to the hydraulic motor through the spool at a rate determined by the spool opening. The ball nut is coupl ed to the hydraulic motor shaft, and as the hydraulic motor rot ate s, the ball nut turn s with the spool, tending to close it. When electrical impulses stop, the spool stops rotating. The hydraulic motor continues to rotate until the attached ball nut completely closes the spool. The hydraulic motor remain s positioned under load because of the high stiffne ss of the drive . Positive logic level input pulses are fed to the 3-2 pha se electronic control (Fig. 9-22). The electric stepping motor is a five-phase motor and requires that 3 and 2 phases be alterna tely excited . The logic level comm and signal is properly sequenced into 3-2 phase excitation and amplified to power the electric stepping motor. This controls the four-way hydrauli c valve that , in turn , controls the hydrauli c motor output. Command

Output

Logic level co mmand pulses

+

t A C power 115 volt

Figure 9-22 Vick ers ).

Hydraulic pow er

Electrohydraulic pulse motor co ntrol syste m (co urtesy of Sp erry

EHPMs are controlled by digital signals and therefore are well suited for industrial applications where digital controls such as numerical control , process controllers, mini-computers, and other digital logic controls are found . The convenience and availabilit y of digital control can be used in applications requiring 120 hp. These includ e NC position and contouring systems, turret indexing , cut to length sys te ms, synchronized drive systems, grinder feed s, conveyer drive systems, ten sion syste ms, winders, and proc ess control.

9-8 OUTPUT PERFORMANCE AND TESTING Hydraulic motors are performance tested both at the bench and as part of complete systems to determine operation characteristic s. Bench testing of the motor determines such characteristics as torque, speed, brake hor sepower, effective flow rate , volumetric efficienc y, and overall efficiency; whereas testing the motor in a system determines its compatibility with other components. Hydrostatic drive s, for exa mple , that couple variable displacement pump s to fixed displacement motors, add braking loads and other conditions to the motor during the application of overhauling loads which test the appropriateness of the reservoir size, line sizes, oil coolers, and so on , for particular applic ation s. Motor performance data are collected in tabular form and then plotted on graph s. Independent variable s are specified (abscissa) and dependent variables 202

Hydraulic Motors

Chap . 9

(ordinate) are read directl y or computed . Independent variables usually include displacement , pre ssure , and speed. Dependent variables usually include torque , fluid hor sepower, flow rate , and volumetric efficiency . Hor sepower and speed are oftentimes interchanged and specified as independent or dependent variables depending on the purpose and application of the graph. Other variables may also be interchanged to accommodate the purposes of specific manufacturers . Graphs are commonly drawn to represent single independent versus dependent variables. Figures 9-23, 9-24, and 9-25 are typical for a high speed gear motor and illustrate this example. In Figure 9-23, output horsepower is plotted again st speed for four pre ssures. The pressure range helps the fluid power system designer select the appropriate motor for a given hor sepower requirement , since for a given pressure requirement and motor speed, the output hor sepower can be selected directly from the graph. Figure 9-24 illustrates the plot of input flow rate in gpm against motor speed for the same four pressures. As pressure increases , the flow rate increases slightly for any given speed, indicating increased slippage

I I

20

~Si-

16 ~

,/"

;:

8.

~

~

a

:r: 8 4

o

~

~

~ - --v V I--

1000

2000

3000 4000 Speed (rpm)

2000PS~

I

,

~12

;: a ;;:: 8

o

o

5000

6000

Figure 9-23 Fluid horsepo wer vs . motor speed .

I I

16

4

I--

500 psi

::::--- ~

20

'E -.,

L-r-

V ........ ' / . . . . ~1000 psiI /'

12

Q>

?

1500 psi

V

~

./' 1000

Sec. 9-8

~~

~

for

2000

....::

,p~

V

~ "-

1500 psi

"" 1000 psi :

I :

3000 4000 Speed (rpm )

5000

6000

Output Performance and Testing

Figure 9-24 Oil flow vs. motor speed .

203

250

- r-- -

200

c

~

150

-

Ql

&

:--. 2000 psi r i

-r--

100

C5

-r--.J

1500 psi I

I

1000 psi

~

50 500 psi

o

I o

1000

2000

3000

4000

Speed (rpm )

5000

6000

Figure 9-25 speed.

Motor torque

YS.

motor

through the motor with increased load. Figure 9-25 shows the accompanying droop in torque that occurs because of fluid slippage and increased mechanical friction . The tendency for characteristic curves to droop is an indication of the volumetric and mechanical efficiency of the unit, and is largely influenced by the design of the unit. Gear motors exhibit the greatest droop and least efficiency , but are less expensive to manufacture and purchase than equiv alent vane and piston motor counterparts. Given the approximate speed range and pressure requirements, gear motors appropriate for a specific task can be selected. Where more exacting requirements must be met over a wide range of speed and loads , for example in airborne applications, close-fitting piston motors are more appropriate, even though they have a higher initial cost. Another variable affecting motor performance is the viscosity of the fluid. As fluid temperature increases , it has a tendency to thin and increase slippage . Oil that becomes too thin reduces volumetric efficiency below an acceptable minimum. Conversely, selecting a more viscous oil will increase the volumetric efficiency but may also increase mechanical friction and heat associated with excessive fluid shear. Care should be exercised to match the required viscosity of the fluid to the system at the expected operating temperature. While torque is generally considered to be constant with respect to speed for a given pressure in fixed displacement motors , the droop in the torque curves seen in Fig. 9-25 indicates that this is not the case. To produce a flat torque curve the speed range requires that pressure must be increased from the source to make up the fluid lost through slippage. In matching the pump to the motor in the system, this is taken into account by oversizing the pump so that the additional volume can be made up with increased pressure at higher motor speeds. This is also the case if constant speed is to be maintained at increased torque and pressure. As the load is increased on the motor, pressure and slippage increase, with some loss in speed . To maintain constant speed, pressure and fluid volume flow rate must be increased. 204

Hydraulic Motors

Chap. 9

9-9 SUMMARY Hydraulic motors convert the high energy level of the fluid to a more useable form by applying the fluid to the movable internal mechanism of the actuator. Movement thereby accomplishes the output objective of the system and lowers the energy level of the fluid, after which it is returned to the reservoir at low pressure. Rotary actuators include continuous as well as limited rotation devices. Limited rotation devices, sometimes called torque motors, rotate from a few degrees to several turns in both directions , and receive widespread application in the production and material handling industries. Another continuous rotation hydraulic actuator is the high torque, low speed motor, used extensively where rotation speed is low, but where substantial torque must be generated. Gear reduction units are commonly connected to the actuator in a common case to keep the size of the unit small. Applications include windlasses, crane hoists, wheel motors, mixers, and others . Rotary actuators are classified by type as gear, vane, and piston motors . Their displacement may be either fixed, as is the case with most gear motors, or variable, as can be the case with vane and piston motors. Cost and efficiency are typically lower with gear motors and higher for piston motors that represent the highest available standard in the industry with respect to precision engineering and machining. Efficiencies are also higher, although the mechanism of the motor is usually less tolerable of dirt , adverse conditions , and system degradation. Where a high level of system and fluid maintenance is provided, and costs are secondary to system quality , piston motors are preferred . Variable displacement piston motors receive widespread use in industry. Displacement may be varied either manually, or the motor may be sensitive to changes in pressure brought about by changes in the load at the output shaft. If the motor receives fluid at constant volume from the pump , a decrease in displacement is accompanied by an increase in the speed of the output shaft. The torque from the variable displacement motor varies with the change in volume displacement and the horsepower, neglecting losses, is constant since the product of the pressure and volume flow rate remains relatively the same. Variable displacement motors , then , provide the capability for a constant horsepower output. Electrohydraulic pulse motors are used on machinery where an unusual combination of power and precision for position and velocity control are required. Recent development in this technology has made them available in bolton versions suitable to a variety of applications . The output shaft speed of an EHPM is proportional to the impulse rate to the stepping motor, and the position of the output shaft is directly related to the number of input pulses. Rotation direction of the output shaft is determined by the sequence in which the input pulses are applied to the various phases of the electric stepping motor. Since EHPMs do not use a feedback loop, they are defined as open loop systems . Hydraulic motors are performance tested both at the bench and as part of complete hydraulic systems to determine characteristics such as speed, torque, brake horsepower, and efficiency. Other variables, such as fluid temperature rise and pressure drop are considered when particular applications are treated . Motor Sec. 9-9

Summary

205

performance data are typically collected in table form and then transposed to graphs using suitable independent and dependent variables. STUDY QUESTIONS AND PROB LEMS 1. Compute the speed of a hydraulic motor with a fixed displacement of 2.2 in.Vrev receiving fluid at 5 gal/min . 2. What flow rate at 4000 lbf/in .? would it take for a hydraulic motor to deliver 20 FHP? 3. What is the displ acement of a hydraulic motor receiving 18 gal/min at 1500 rev /min? 4. What flow rate would be required to drive a hydraulic motor with a displacement of 20 cmt/rev turning at 2000 rev /min? 5. What FHP would the motor in Problem 4 deliver at a pressure of 25 MPa? 6. What theoretical torque could a 5 hp motor be expected to deliver at a rated speed of 1000 rev/min? 7. If in Problem 6 the pre ssure remains constant at 1750 Ibf/in . 2 , what effect would doubling the speed have on torque? How about halving the speed ? What about at stall? Construct a table listingfiow rate, speed, torqu e, and power to explain what happens . 8. What is the SI metric equivalent of Eq . 9-1? 9. Compute the theoretical torque from a hydraulic motor with a displacement of2 .5 in. 3 / rev operating at 1500 lbf/in ", 10. What theoretical torque could be expected from a hydraulic motor with a displacement of 75 crnvrev operating at 15 MPa? 11. As suming no volumetric los se s, what is the operating speed of a hydraulic motor with a 28 cm l /rev displacement receiving fluid at 45 liters/minute? U. At what pre ssure mu st a hydraulic motor with a displacement of 3.6 in. Vre v operate to deliver a torque of 50 Ibf/ft? 13. A hydraulic motor with a displacement of 1.1 in.t/rev operates at 1500 rev /min. Assuming no volumetric losses, what mu st be the deli very to the motor? 14. Compute the overall efficiency of a hydraulic motor that operates at a pre ssure of 2500 lbf/in. ", a flow rate of 5 gal/min , a speed of 1500 re v/min, a nd a torque of 20 Ibf-ft. 15. What is the overall efficiency of a hydraulic motor with a displacement of 4.2 in. t /rev that delivers 78 Ibf-ft of torque at 1000 rev /min and 1750 lbf/in ."? 16. In a bench test of a gear motor, it was found that a maximum overall efficiency of 84% was achieved at a pressure drop of 3400 lbf/in .? across the ports when the ratio of the flow rate to speed was 0.01. Compute the torque and displacement of the motor. 17. Compute the volumetric efficiency of a hydraulic motor with a displaceme nt of 0.55 in. Vrev turning at 1800 rev /min and receiving fluid at 5 gal /mi n . How much fluid car. be expected to return through the case drain if 75% of the loss in volumetric efficie ncy can be attributed to internal leakage? 18. If the expected wear over the life ofa hydraulic motor with a displacement of 4 in .Vrev results in a ca se drain loss of 3% , how much fluid will be returned through the case drain line at 1200 rev /min? 19. Compute the mechanical efficiency of a motor with a n overall efficiency of 85% and a volumetric efficiency of 95%. 20. Describe the major difference between piston motors and pumps.

206

Hydraulic Motors

Chap. 9

_ _ _ _ _ _ _ _CHAPTER

10

VALVES

10-1 INTRODUCTION

Valve s control the pressure , rate of flow, and direction of fluids in accorda nce with basic principles of flow. Several valves classified by function are listed in Table 10-1. Pressure control valves limit and reduce pre ssure, sequence hydraulic operations by stepping system pressure, and counterbalance such external load s as vertical pre sses so they can 't fall by maintaining back-pressure on the underside of the pre ss cylinders. The y are also used to unload slack cycle circuits at low pre ssure to reduce power consumption and provide switch signals at specified pre ssures to interface hydraulic systems with adjacent electrical controls. Flow control valves vary the fluid flow rate using restrictions in fluid passages which may be fixed, variab le, or flow and pres sure compensated. Directional control valves are used to check, divert , shuttle, proportion , and by other mean s manage the flow of fluid in one , two , three , four , or more ways. Pressure and flow compensation are commonly included in directional control valves . Valve s may be direct acting because of the arrangement of the internal mec hanism design, or pilot operated from an adjacent or remote location. The act uat ing force to operate the valve can be supplied manually by the human operator, directly by the fluid under pressure or from a pilot circuit, or by electrical devices such as solenoids or servo electric drive s. The internal mechanisms which physically accomplish the valving action use a variety of elements including poppets, diaphragms, flat slides , balls, round shear-action plate s, and rotating or sliding spools (Table 10-2). Other considerations related to valves in hydraulic transmission control include appropriate valve sizing, control pane ls, remote circuitry servo valve controls, and valve testing . 207

TABLE 10-1

VALVE FUNCTIONS CLASSIFICATION

Pressure co ntro l

Flow control

Directional co ntrol

Pressur e relief Hydraulic fuse Pressure reducing Sequencing Unloading Counte rbalance Press ure switches

Fixed Variable Compensated Dece leration Flow divider Electrohydraulic Servo

Two way Chec k Shuttle Three way Fou r way Limit switches Proportional electro hydra ulic Servo

TABLE 10·2

VALVE PHYSICAL CHARACTERISTICS CLASSIFICATION

Pressure cont rol

Flow co ntro l

Direct ion contro l

Direct acting

j

j

j

Pilot operated

j

j

j

j

j

j

j

j

j

MODE OF OPERATION

METHOD OF ACTUATION Manu al Flu id

j

Electrica l ACTUATING ELEMENT Popp et

j

j

j

Diaphragm Plug

j

j

Ball

j

j

Rotating spool Sliding spool

j

j

j

j

j

10-2 PRESSURE CONTROL VAL VES Relief valves are the most common of the pre ssure control valves . They are located near the pump outlet to protect the pump and provide the system plumb ing and components with protection again st pre ssure overload s. The y also limit the output force exerted by cylinders and rotary motors. Circuits using positive fixed displacement pumps must have pre ssure relief valve s. Figure 10-1 illustrates the operation of a direct-acting spring-type ball relief valve . Spring pressure act s to keep the ball element again st the seat and the valve in the closed position. System pre ssure at the port marked " inlet" acts again st 208

Valves

Chap. 10

Out let

Sy m bo l

Figure 10-1 Direct-acting spring-type ball relief valve (co urtesy of Sperry Vickers ).

the exposed area of the ball. When the force of the fluid (press ure x area) becomes greater than the opposing resi stance offered by the spring, the ball is forced from its seat, the valve opens, and fluid is directed to the reser voir at low pre ssure through the port marked "outlet. " The pre ssure at which the valve opens is called the cracking pressure. The pressure at which the rated flow passes through the valve is termed full fl ow pressure. The pre ssure at which the valve ceases to pa ss fluid after being opened is called the closing pressure. Adjustment within the pre ssure range of the valve is made with the adju stment screw which acts to compress the check valve spring. Compound pilot drained pressure relief valves increase pre ssure sensitivity by reducing the pre ssure override usually encountered with valve s using onl y the direct-acting forc e of sys tem pre ssure against the valve element. The valve in Fig. 10-2 illustrates the principle of operation. In schematic (a) , fluid pre ssure act s on both sides of piston I , which is held closed on its seat by a relatively light bias spring 2. In schematic (b), when the pre ssure increases sufficiently to move the pilot poppet valve 4 from its seat , fluid behind the piston goes to drain , and the resulting pre ssure imbalance on the valve piston causes it to move in the direction of lower pre ssure . Th is action compresses the piston spring 2 and opens the discharge port , pre venting a further rise in pre ssure. The pre ssure setting is adjusted with the adju stment screw that increases or decreases the pre ssure required to unseat the pilot drain poppet valve . Pre ssure reducing valve s are used to supply branch circuits with fluid at a pre ssure lower than sys tem pres sure . Essentially , the y step the pre ssure down to the requirements of the branch circuit by restricting fluid flow when branch circuit pre ssure reaches the pre-set limit determined by the rating and adjustment of the spool positioning spring. A pressure reducing valve is seen in Fig. 10-3. Fluid passes unob structed from C to D in illustration (a). The pre ssure reducing spool valve is held open by spring 2, and leakage around the free floating spool passes at low pre ssure to drain. As the pressure of the system at the outlet of the valve Sec. 10-2

Pressure Control Valv es

209

3

Drain

\

/

4

D

c ' .v/yh,.L-~r//j . -

1

Discharge

ral

2

C -- r/.