About Pearson FLUID MECHANICS Pearson is the world’s learning company, with presence across 70 countries worldwide Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world We believe learning opens up opportunities, creates fulfilling careers and inhence SI Units better lives We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum Superior learning experience and improved outcomes are at the heart of we This product is the result of one such effort R.everything C Hibbeler Your feedback plays a critical role in the evolution of our products and you can reachus@pearson.com We look forward to it SI Conversion by Kai Beng Yap Boston Columbus Indianapolis New York San Francisco Hoboken Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo FLUID MECHANICS in SI Units R C Hibbeler SI Conversion by Kai Beng Yap Boston Columbus Indianapolis New York San Francisco Hoboken Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo FLUID MECHANICS in SI Units R C Hibbeler SI Conversion by Kai Beng Yap Boston Columbus Indianapolis New York San Francisco Hoboken Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo To the Student With the hope that this work will stimulate an interest in Fluid Mechanics and provide an acceptable guide to its understanding Copyright © 201 Pearson India Education Services Pvt Ltd Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128, formerlyknown as TutorVista Global Pvt Ltd, licensee of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN eISBN 978-93-325-8700-7 Head Office: Tower-B, Trade Tower, Plot No 1, Block-C, Sector201 16,309, Noida 201 301, Head Office:15th A-8Floor, (A), 7th Floor,World Knowledge Boulevard, Sector 62, Noida Uttar Pradesh, India Uttar Pradesh, India Registered Office: UI'MPPS, 4PGUXBSF#MPDL, Elnet Software City, TS-140, Block & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email: companysecretary.india@pearson.com PREFACE PREFACE This book has been written and revised several times over a period of nine years, in order to further improve its contentsThis andbook account for the many suggestions and times comments my has been written and revised several over afrom period of students, nine university colleagues,years, and reviewers It is hoped that this effort will provide those who use this in order to further improve its contents and account for the many work with a clear andsuggestions thorough presentation of both the theory and application of fluid mechanics and comments from my students, university colleagues, and To achieve this objective, I haveItincorporated of the pedagogic features have used in reviewers is hoped thatmany this effort will provide those whothat useIthis work my other books These include the following: with a clear and thorough presentation of both the theory and application of fluid mechanics.Each To achieve this have well-defined incorporatedsections many of that Organization and Approach chapter is objective, organizedI into the of pedagogic that I haveexample used in my other books contain an explanation specific features topics, illustrative problems, and These at the include end of the the following: chapter, a set of relevant homework problems The topics within each section are placed into subgroups defined byOrganization boldface titles The purpose of this organization is toispresent a structured and Approach Each chapter organized into method for introducing each new definition or concept, and to make the book a well-defined sections that contain an explanation of specific convenient topics, resource for later reference and review illustrative example problems, and at the end of the chapter, a set of relevant homework problems The topics each section areand placed Procedures for Analysis This unique feature provides thewithin student with a logical orderly intoapplying subgroups defined boldface titles.discussed The purpose this organization method to follow when the theorybythat has been in a of particular section The is to present a structured introducing each definition or example problems are then solved using this method outlinedfor method in order tonew clarify its numerical make the book aprinciples convenienthave resource later reference application Realize, concept, however,and thattoonce the relevant been for mastered, and enough and review confidence and judgment has been obtained, the student can then develop his or her own procedures for solving problems Procedures for Analysis This unique feature provides the student with a feature logical and orderly method to follow of when applying the theory Important Points This provides a review or summary the most important concepts that has been discussed in apoints particular The example problems are in a section, and highlights the most significant that section should be remembered when applying then solved using this outlined clarify the theory to solve problems A further review of the method materialinis order given to at the endits of numerical the chapter application Realize, however, that once the relevant principles have been Photos The relevance of knowing the subject matterand is reflected realistic applications mastered, and enough confidence judgmentby hasthe been obtained, the depicted in the manystudent photoscan placed throughout the book These photos are often used to show then develop his or her own procedures for solving problems how the principles of fluid mechanics apply to real-world situations Important Points This feature provides a review or summary Fundamental Problems These problem sets areinselectively just afterthe the most example of the most important concepts a section,located and highlights problems They offer significant students simple applications of the concepts and therefore provide them points that should be remembered when applying the theorywith the chance to develop their problem-solving skillsreview beforeof attempting to solve anyatofthe theend standard to solve problems A further the material is given of problems that follow.the Students may consider these problems as extended examples, since they all chapter have complete solutions and answers given in the back of the book Additionally, the fundamental Photos Themeans relevance of knowing subject matter reflected problems offer students an excellent of preparing for the exams, and they canisbe used at by a later the realistic applications depicted in the many photos placed throughout time to prepare for the Fundamentals in Engineering Exam the book These photos are often used to show how the principles of fluid Homework Problems The majority of problems in the book depict realistic situations mechanics apply to real-world situations encountered in engineering practice It is hoped that this realism will both stimulate interest in Fundamental Problems Thesetoproblem sets problem are selectively the subject, and provide a means for developing the skills reduce any from located its physical just after the example problems They offer students simple description to a model or symbolic representation to which the principles of fluidapplications mechanics may of the concepts and therefore provide them with the chance to develop then be applied their problem-solving skills before attempting solveing anydifficulty of the standard An attempt has been made to arrange the problems in order of to increas Except for problems that follow Students may consider these problems as extended every fourth problem, indicated by an asterisk (*), the answers to all the other problems are given examples, since they all have complete solutions and answers given in the in the back of the book back of the book Additionally, the fundamental problems offer students Accuracy Apart from my work, the accuracy of the andand problem solutions all been an excellent means of preparing fortext exams, they can be usedhave at a later thoroughly checked time by other parties Most importantly, Kai Beng Yap, Kurt Norlin along with to prepare for the Fundamentals in Engineering Exam Bittner Development Group, as well as James Liburdy, Jason Wexler, Maha Haji, and Brad Saund VI P REFACE PREFACE Homework Contents Problems The majority of problems in the book depict realistic situations encountered in engineering practice It is hoped The that bookthis is divided intoboth 14 chapters with an introduction realism will stimulateChapter interest1inbegins the subject, and provide a to fluid mechanics, followed by a discussion of units and some important fluid properties The concepts of fluid statics, means for developing the skills to reduce any problem from its physical including constant accelerated translation of a liquid and its constant rotation, are covered in description to a model or symbolic representation to which the principles Chapter In Chapter 3, the basic principles of fluid kinematics are covered This is followed by of fluid mechanics may then be applied the continuity equation in Chapter 4, the Bernoulli and energy equations in Chapter 5, and fluid An attempt has been made to arrange the problems in order of increasmomentum in Chapter In Chapter 7, differential fluid flow of an ideal fluid is discussed Chapter ing difficulty Except for every fourth problem, indicated by an asterisk covers dimensional analysis and similitude Then the viscous flow between parallel plates and (*), the answers to all the other problems are given in the back of the within pipes is treated in Chapter The analysis is extended to Chapter 10 where the design of book pipe systems is discussed Boundary layer theory, including topics related to pressure drag and lift, Accuracy Apart from my 12 work, the accuracy of the text and problem is covered in Chapter 11 Chapter discusses open channel flow, and Chapter 13 covers a variety solutions have all been thoroughly checked by other parties Most of topics in compressible flow Finally, turbomachines, such as axial and radial flow pumps and importantly, Kai Beng Yap, Kurt Norlin along with Bittner Development turbines are treated in Chapter 14 Group, as well as James Liburdy, Jason Wexler, Maha Haji, and Brad Alternative Coverage After covering the basic principles of Chapters through 6, at the Saund discretion of the instructor, the remaining chapters may be presented in any sequence, without the loss of continuity If time permits, sections involving more advanced topics, may be included in the course Most of these topics are placed in the later chapters of the book In addition, this material Contents also provides a suitable reference for basic principles when it is discussed in more advanced courses The book is divided into 14 chapters Chapter begins with an introduction to fluid mechanics, followed by a discussion of units and some important fluid properties The concepts of fluid statics, including Acknowledgments constant accelerated translation of a liquid and its constant rotation, areendeavored covered in to Chapter In Chapter 3, itthe basic principles fluid I have write this book so that will appeal to bothofthe student and instructor kinematics aremany covered Thishave is followed continuity equation Through the years people helped inbyitsthe development, and I willinalways be grateful for 4, the Bernoulli and energy equations in years, Chapter 5, and theirChapter valued suggestions and comments During the past I have hadfluid the privilege to teach my momentum in Chapter Chapter 7, differential fluidand flow an idealI would like to thank students during the summer6 atIn several German universities, in of particular is discussed Chapter covers dimensional and similitude Prof.fluid H Zimmermann at the University of Hanover, analysis Prof F Zunic of the Technical University in Thenand theProf viscous flow between parallel of plates and within pipes is treated for their assistance Munich, M Raffel at the Institute Fluid Mechanics in Goettingen, in Chapter The analysis is extended to of Chapter 10 where of theMecklenburg design In addition, I Vogelsang and Prof M Geyh the University have provided of pipe systems is discussed Boundary Ilayer including topics me with logistic support in these endeavors wouldtheory, also like to thank Prof K.Cassel at Illinois related to pressure Prof drag A and lift, at is the covered in Chapter 11 Chapter 12 Institute of Technology, Yarin University of Illinois-Chicago, and Dr J Gotelieb for open flow,In and Chapterthe 13following covers a variety of topics theirdiscusses comments andchannel suggestions addition, individuals havein contributed important compressible flow Finally, turbomachines, such as axial and radial flow reviewer comments relative to preparing this work: pumps and turbinesSchool are treated in Chapter 14 S Venayagamoorthy, Colorado State University S Kumpaty, Milwaukee of Engineering N Kaye, Clemson University Knight, University Alternative Coverage After coveringD.the basicRutgers principles of J Crockett, Brigham Young University B Hodge, Mississippi State University Chapters through 6, at the discretion of the instructor, the remaining B Wadzuk, Universityin any sequence, without L Grega, The of New Jersey chaptersVillanova may be presented the loss of College continuity K Sarkar, of Delaware R.topics, Chen,may University of Central Florida If timeUniversity permits, sections involving more advanced be included E Petersen, Texas A&M Poly Institute in the course Most University of these topics are placedR inMullisen, the laterCal chapters of J Liburdy, Oregon State University C.aPascual, Poly Institute the book In addition, this material also provides suitableCal reference for B Abedian, Tufts University basic principles when it is discussed in more advanced courses There are a few people that I feel deserve particular recognition A longtime friend and associate, Kai Beng Yap, was of great help in checking the entire manuscript, and helping to further check all the problems And a special note of thanks also goes to Kurt Norlin for his diligence and support in this regard During the production process I am also thankful for the support of my long time Production Editor, Rose Kernan, and my Managing Editor, Scott Disanno My wife, Conny, and V www.freebookslides.com PREFACE P REFACE V I I VII daughter, Mary Ann, have been a big help with the proofreading and typing needed to prepare the Indian Adaptation manuscript for publication The publishers would likemany to thank theare following for to their Lastly, thanks extended all contribution my students who have given me their suggestions and to the Global Edition: comments Since this list is too long to mention, it is hoped that those who have helped in this manner will accept this anonymous recognition Contributor I value your judgment as well, and would greatly appreciate hearing from you if at any time you Kai Beng Yap have any comments or suggestions that may help to improve the contents of this book Kai is currently a registered Professional Engineer who works in Malaysia He has BS and MS degrees in Civil Engineering from the University of Louisiana, Lafayette, Louisiana; and he has done further graduate work at Virginia Polytechnic Institute in Blacksberg, Virginia His professional experience has involved teaching at the University of Louisiana, and doing engineering consulting work related to structural analysis and design and its associated infrastructure Russell Charles Hibbeler hibbeler@bellsouth.net Indian Adaptation Reviewers The publishers would of like to thank Dr Pravin Kumar, Jitendra Singh Rathore, Department Mechanical Engineering, Birla Department of Mechanical Engineering, Delhi Technological Institute of Technology and Science University, Delhi for his valuable suggestions and inputs for enhancing the content of this of book to suit the requirementMiddle of Indian M Haluk Aksel, Department Mechanical Engineering, Eastuniversities Technical University Suresh Babu, Centre for Nano Sciences and Technology, Pondicherry University Resources for Instructors r Instructor’s Manual instructor’s solutions manual s Instructor’s Solutions Solutions Manual AnAn instructor’s solutions manual was was prepared by the author The manual includes homework assignment lists and was also checked prepared by the author The manual includes homework assignment lists as part of the accuracy checking program The Instructor Manual is available at www.pearsoned.co.in/rchibbeler and was also checked as part ofSolutions the accuracy checking program The Instructor Solutions Manual is available at www.pearsoned.co.in/rchibbeler r Presentation Resource All art from the text is available in PowerPoint slide and JPEG format TheseResource files are available download the Instructor Resource Center at s Presentation All art for from the textfrom is available in www.pearsoned.co.in/rchibbeler PowerPoint slide and JPEG format These files are available for download from the Instructor Resource Center at www.pearsoned.co.in/rchibbeler www.freebookslides.com VIII P REFACE PREFACE ABOUT THE AUTHOR Indian Adaptation The publishers would like to thank the following for their contribution to the Global Edition: ContributorR.C Hibbeler graduated from the University of Illinois at Urbana with a BS in Civil Engineering Kai Beng Yap (majoring in Structures) and an MS in Nuclear Engineering He obtained his PhD in Theoretical and Applied Mechanics from Northwestern University Professor Hibbeler’s professional experiKai is currently a registered Professional Engineer who works in ence includes postdoctoral work in reactor safety and analysis at Argonne National Laboratory, Malaysia He has BS and MS degrees in Civil Engineering from the and structural and stress analysis work at Chicago Bridge and Iron, as well as at Sargent and Lundy University of Louisiana, Lafayette, Louisiana; and he has done further in Chicago He has practiced engineering in Ohio, New York, and Louisiana graduate work at Virginia Polytechnic Institute in Blacksberg, Virginia His professional experience hascurrently involvedteaches teachingboth at the University of Professor Hibbeler civil and mechanical engineering courses at the UniLouisiana, and doing engineering consulting work related to structural versity of Louisiana– Lafayette In the past, he has taught at the University of Illinois at Urbana, analysis andYoungstown design and its associated infrastructure State University, Illinois Institute of Technology, and Union College Reviewers Jitendra Singh Rathore, Department of Mechanical Engineering, Birla Institute of Technology and Science M Haluk Aksel, Department of Mechanical Engineering, Middle East Technical University Suresh Babu, Centre for Nano Sciences and Technology, Pondicherry University VII www.freebookslides.com CONTENTS Fundamental Concepts Chapter Objectives 3 1.1 Introduction 1.2 Characteristics of Matter 1.3 The International System of Units 1.4 Calculations 1.5 Problem Solving 1.6 Basic Fluid Properties 1.7 Viscosity 1.8 Viscosity Measurement 1.9 Vapor Pressure 10 12 17 22 26 1.10 Surface Tension and Capillarity 27 Fluid Statics 45 Chapter Objectives 45 2.1 Pressure 45 2.2 Absolute and Gage Pressure 2.3 Static Pressure Variation 2.4 Pressure Variation for Incompressible Fluids 51 2.5 Pressure Variation for Compressible Fluids 53 2.6 Measurement of Static Pressure 48 50 56 www.freebookslides.com 2.7 EXAMPLE HYDROSTATIC FORCE ON A PLANE SURFACE—FORMULA METHOD 2.10 Determine the magnitude and location of the resultant force acting on the triangular end plate of the settling tank in Fig 2–28a The tank contains kerosene 1m 0.5 m B x yP ϭ m 2m A P 2m FR ϭ 5.32 kN y (a) (b) Fig 2–28 SOLUTION Fluid Description The kerosene is considered an incompressible fluid for which gk = (814 kg/m3)(9.81 m/s2) = 7985 N>m3 (Appendix A) Analysis From the inside back cover, for a triangle, y = h = (2 m) = 0.6667 m 1 Ix = ba3 = (1 m)(2 m)3 = 0.2222 m4 36 36 Therefore, FR = gk h A = 7985 N>m3 (0.6667) c = 5323.56 N = 5.32 kN yP = y + 69 Ix = 0.6667 m + yA (1 m)(2 m) d Ans 0.2222 m4 = 1.00 m Ans (0.6667 m)c (1 m)(2 m) d The triangle is symmetrical about the y axis, so Ixy = Thus, xP = x + Ix = + = yA These results are shown in Fig 2–28b Ans www.freebookslides.com 70 CHAPTER F L U I D S TAT I C S 2.8 x p1 ϭ gh1 x dF u p y y Rather than using the equations of the previous section, the resultant force and its location on a flat submerged plate can also be determined using a geometrical method To show how this is done, consider the flat plate shown in Fig 2–29a Resultant Force If an element dA of the plate is at a depth h, where the pressure is p, then the force on this element is dF = p dA As shown in the figure, this force geometrically represents a differential volume element d V of the pressure distribution It has a height p and base dA, and so dF = d V The resultant force can be obtained by integrating these elements over the entire volume enclosed by the pressure distribution, we have h dV dA Hydrostatic Force on a Plane Surface—Geometrical Method p2 ϭ gh2 (a) FR = ⌺F ; p dA = dV = V (2–12) LA LV Therefore, the magnitude of the resultant force is equal to the total volume of the “pressure prism.” The base of this prism is the area of the plate, and the height varies linearly from p1 = gh1 to p2 = gh2, Fig 2–29a Centroid of volume x FR yP CV y P xP Center of pressure FR equals the volume of the pressure diagram, and it passes through the centroid CV of this volume FR = Location To locate the resultant force on the plate, we require the moment of the resultant force about the y axis and about the x axis, Fig 2–29b to equal the moment created by the entire pressure distribution about these axes, Fig 2–29a; that is, 1M R2y = ⌺M y; 1M R2x = ⌺M x; xP FR = L x dF yP FR = y dF L Since FR = V and dF = d V , we have (b) Fig 2–29 xP = LA x p dA = p dA L A yP = LA (2–13) y p dA LA y dV = p dA x dV LV V LV V These equations locate the x and y coordinates of the centroid CV of the pressure prism volume In other words, the line of action of the resultant force will pass through both the centroid CV of the volume of the pressure prism and the center of pressure P on the plate, Fig 2–29b www.freebookslides.com 2.8 Centroid of pressure volume HYDROSTATIC FORCE ON A PLANE SURFACE—GEOMETRICAL METHOD p1 h1 FR u CV p2 71 FR h2 w2 P Centroid of pressure area w1 h u CA P h2 b FR equals the volume of the pressure diagram, and it passes through the centroid of this volume FR equals the area of the w diagram, and it passes through the centroid of this area (b) (a) Fig 2–30 Plate Having Constant Width As a special case, if the plate has a constant width b, as in the case of a rectangle, Fig 2–30a, then the pressure loading along the width at depth h1 and at depth h2 is constant As a result, the loading may be viewed along the side of the plate, in two dimensions, Fig 2–30b The intensity w of this distributed load is measured as a force > length, and varies linearly from w1 = p1b = (g h1)b to w2 = p2b = (g h2)b The magnitude of FR is then equivalent to the trapezoidal area defining the distributed loading, and FR has a line of action that passes through both the centroid CA of this area and the center of pressure P on the plate Of course, these results are equivalent to finding the trapezoidal volume of the pressure prism, FR, and its centroidal location CV , as shown in Fig 2–30a Important Points s The resultant force on a plane surface can be determined graphically by finding the volume V of the pressure prism, FR = V The line of action of the resultant force passes through the centroid of this volume It intersects the surface at the center of pressure P s If the submerged surface has a constant width, then the pressure prism can be viewed from the side and represented as a planar distributed loading w The resultant force equals the area of this loading diagram, and it acts through the centroid of this area www.freebookslides.com 72 CHAPTER EXAMPLE F L U I D S TAT I C S 2.11 The tank shown in Fig 2–31a contains water to a depth of m Determine the resultant force, and its location, that the water pressure creates both on side ABCD of the tank and on its bottom B A C 3m 2m D 1.5 m (a) Fig 2–31 SOLUTION Fluid Description The water is considered to be incompressible, with rw = 1000 kg>m3 Analysis I Loading The pressure at the bottom of the tank is p = rwgh = 11000 kg>m3219.81 m>s22(3 m) = 29.43 kPa Using this value, the pressure distribution along the side and bottom of the tank is shown in Fig 2–31b Resultant Forces The magnitudes of the resultant forces are equal to the volumes of the pressure prisms 1FR2s = (3 m)129.43 kN>m22(2 m) = 88.3 kN 1FR2b = 129.43 kN>m22(2 m)(1.5 m) = 88.3 kN Ans Ans These resultants act through the centroids of their respective volumes, and define the location of the center of pressure P for each plate, Fig 2–31 www.freebookslides.com 2.8 73 HYDROSTATIC FORCE ON A PLANE SURFACE—GEOMETRICAL METHOD z (a) 2m (FR)b P 3m zP xP CV (FR)s CV yP 29.43 kPa P 2m 1.5 m 29.43 kPa x y xP x (b) Location Using the inside back cover, for the side plate, zP in Fig 2–31b is determined for a triangle to be 13 a, so that xP = m z P = (3 m) = m Ans xP = m yP = 0.75 m Ans Ans Ans For the bottom plate, 3m P CV zP ϭ m 58.86 kN/m (FR)b 58.86 kN/m CV P yP = 0.75 m 1.5 m (FR)s Analysis II Loading Since the side and bottom plates in Fig 2–31a both have a constant width of b = m, the pressure loading can also be viewed in two dimensions The intensity of the loading at the bottom of the tank is w = 1rwgh2b = 11000 kg>m3219.81 m>s22(3 m)(2 m) = 58.86 kN>m The distributions are shown in Fig 2–31c Resultant Forces Here the resultant forces are equal to the areas of the loading diagrams (3 m)(58.86 kN>m) = 88.3 kN (FR)b = (1.5 m)(58.86 kN>m) = 88.3 kN (FR)s = (c) Fig 2–31 (cont.) Ans Ans Location These results act through the centroids of their respective areas as shown in Fig 2–31c www.freebookslides.com 74 CHAPTER EXAMPLE F L U I D S TAT I C S 2.12 A The storage tank contains oil and water at the depths shown in Fig 2–32a Determine the resultant force that both of these liquids together exert on the side ABC of the tank if the side has a width of b = 1.25 m Also, determine the location of this resultant, measured from the top of the tank Take ro = 900 kg>m3, rw = 1000 kg>m3 0.75 m B 1.5 m SOLUTION Fluid Description incompressible C Both the water and the oil are assumed to be (a) Loading Since the side of the tank has a constant width, the intensities of the distributed loading at B and C, Fig 2–32b, are A 0.75 m B 8.277 kN/m w B = roghA Bb = 1900 kg>m3219.81 m>s22(0.75 m)(1.25 m) = 8.277 kN>m w C = w B + rwghBCb = 8.277 kN>m + 11000 kg>m3219.81 m>s22(1.5 m)(1.25 m) = 26.67 kN>m 1.5 m C 26.67 kN/m (b) A 0.75 m B Resultant Force The resultant force can be determined by adding the three shaded triangular and rectangular areas shown in Fig 2–32c FR = F1 + F2 + F3 1 = (0.75 m)(8.277 kN>m) + (1.5 m)(8.277 kN>m) + (1.5 m)(18.39 kN>m) 2 = 3.104 kN + 12.42 kN + 13.80 kN = 29.32 kN = 29.3 kN Ans F1 ϭ 3.104 kN 8.277 kN/m F2 ϭ 12.42 kN 1.5 m F3 ϭ 13.80 kN C 8.277 kN/m A = 26.67 kN/m Ϫ 8.277 kN/mϭ 18.39 kN/m (c) Location As shown, each of these three parallel resultants acts through the centroid of its respective area y = (0.75 m) = 0.5 m y = 0.75 m + (1.5 m) = 1.5 m 2 y = 0.75 m + (1.5 m) = 1.75 m The location of the resultant force is determined by equating the moment of the resultant about A, Fig 2–32d, to the sum of the moments of all the component forces about A, Fig 2–32c We have y P FR = ⌺ ෂ y F; yP FR ϭ 29.32 kN (d) Fig 2–32 y P (29.32 kN) = (0.5 m)(3.104 kN) + (1.5 m)(12.42 kN) + (1.75 m)(13.80 kN) y P = 1.51 m Ans www.freebookslides.com 75 2.9 HYDROSTATIC FORCE ON A PLANE SURFACE—INTEGRATION METHOD 2.9 Hydrostatic Force on a Plane Surface—Integration Method If the boundary of the flat plate in Fig 2–33a can be defined in terms of its x and y coordinates as y = f(x), then the resultant force FR and its location P on the plate can be determined by direct integration Resultant Force If we consider a differential area strip dA of the plate that is located at a depth h, where the pressure is p, then the force acting on this strip is dF = p dA, Fig 2–33a The resultant force on the entire area is therefore FR = ⌺F; FR = A L (2–14) p dA Location Here we require the moment of FR about the y and x axes to equal the moment of the pressure distribution about these axes Provided dF passes through the center (centroid) of dA, having ෂ,y ෂ), then from Fig 2–33a and Fig 2–33b, coordinates (x 1M R2y = ⌺M y ; x P FR = 1M R2x = ⌺M x ; LA y P FR = LA The hydrostatic force acting on the elliptical back plate of this water truck can be determined by integration ෂ x dF ෂ y dF Or, written in terms of p and dA, we have xP = L A ෂ x p dA L A yP = p dA A L ෂ y p dA A L (2–15) p dA Application of these equations is given in the following examples x dF x u p FR y yP h y ˜˜ dA (x,y) P dy y ϭ f(x) y (a) xP (b) Fig 2–33 www.freebookslides.com 76 CHAPTER EXAMPLE F L U I D S TAT I C S 2.13 The bin in Fig 2–34a contains water Determine the resultant force, and its location, that the water pressure exerts on the circular plate 0.75 m y 3m SOLUTION Fluid Description We assume the water to be incompressible For water, rw = 1000 kg > m3 x + y2= Resultant Force We can determine the resultant force on the plate by using integration, since the circular boundary can be defined from the center of the plate in terms of the x, y coordinates shown in Fig 2–34a The equation is x2 + y2 = The rectangular horizontal strip shown in Fig 2–34b has an area dA = 2x dy = 211 - y 221>2dy It is at a depth h = - y, where the pressure is p = gwh = gw(3 - y) Applying Eq 2–14, to obtain the resultant force, we have x 1m (a) y x yP F= x h 3m LA p dy y L- c (1000 kg)19.81 m>s22(3 - y) d (2)11 - y 221>2 dy = 19 620 x P p dA = L-1 c 3(1 - y 2)1>2 - y(1 - y 2)1>2 d dy = 92.46 kN Ans (b) Location The location of the center of pressure P, Fig 2–34b, can be determined by applying Eq 2–15 Here dF = p dA is located at ෂ x = and ෂ y = - y and so Fig 2–34 xP = Ans yP = LA (3 - y)p dA = LA p dA (3 - y)(1000 kg) 9.81 m>s2 (3 - y)(2) 1 - y 2 1>2dy L-1 (1000 kg) 9.81 m>s (3 - y)(2) 1 - y L-1 2 = 3.08 m 1>2 dy For comparison, this problem was also worked in Example 2.9 Ans www.freebookslides.com 77 2.9 HYDROSTATIC FORCE ON A PLANE SURFACE—INTEGRATION METHOD EXAMPLE 2.14 Determine the magnitude and location of the resultant force acting on the triangular end plate of the settling tank in Fig 2–35a The tank contains kerosene SOLUTION Fluid Description The kerosene is considered an incompressible fluid for which gk = (814 kg>m3)(9.81 m>s2) = 7985 N>m3 (Appendix A) 1m B 2m A Resultant Force The pressure distribution acting on the end plate is shown in Fig 2–32b Using the x and y coordinates, and choosing the differential area strip shown in the figure, we have (a) 0.5 m dF = p dA = (gk y)(2x dy) = 15.97(103)yx dy x B The equation of line AB can be determined using similar triangles: x 0.5 m = 2m - y 2m x y dy 2m (2 m Ϫ y) x = 0.25(2 - y) dF A Hence, applying Eq 2–14 and integrating with respect to y, from y = to y = m, yields y (b) 2m F = p dA = LA L0 15.97(103)y[0.25(2 - y)] dy = 5.323(103) N = 5.32 kN Location x Ans Because of symmetry along the y axis, ෂ x = and so xP = yP ϭ 1.00 m P Ans Applying Eq 2–15, with ෂ y = y , we have yP = LA ෂ y p dA LA = p dA y 2m L0 y[15.97(103)y][0.25(2- y)]dy 5.323(10 ) (c) = 1.00 m Ans The results are shown in Fig 2–35c They were also obtained in Example 2.10 using the formula method Fig 2–35 CV FR ϭ 5.32 kN www.freebookslides.com 78 CHAPTER F L U I D S TAT I C S 2.10 Hydrostatic Force on an Inclined Plane or Curved Surface Determined by Projection dF FR p dA CV u h P If a submerged surface is curved, then the pressure acting on the surface will change not only its magnitude but also its direction, since it must always act normal to the surface For this case, it is generally best to determine the horizontal and vertical components of the resultant force caused by the pressure, and then use vector addition to find the resultant The method for doing this will now be described with reference to the submerged curved plate in Fig 2–36a Realize that this method will also work for an inclined flat plate, as in Fig 2–30 Horizontal Component The force shown acting on the differential element dA in Fig 2–36a is dF = p dA, and so its horizontal component is dFh = ( p dA) sin u, Fig 2–36b If we integrate this result over the entire area of the plate, we will obtain the resultant’s horizontal component (a) Fh = dF ϭ p dA dFv u dFh dA u dA cos u Horizontal projected area dA sin u Vertical projected area LA p sin u dA Since dA sin u is the projected differential area onto the vertical plane, Fig 2–36b, and the pressure p at a point must be the same in all directions, then the above integration over the entire area of the plate can be interpreted as follows: The resultant horizontal force component acting on the plate is equal to the resultant force of the pressure loading acting on the area of the vertical projection of the plate, Fig 2–36c Since this vertical area is flat or “planar,” any of the methods of the previous three sections can be used to determine Fh and its location on this projected area (b) Fv Projected area FR Fh ϭ the resultant pressure loading on the vertical projected area Fv ϭ the weight of the volume of liquid above the plate P Centroid of pressure volume Fh CV CV (c) Fig 2–36 www.freebookslides.com 79 2.10 HYDROSTATIC FORCE ON AN INCLINED PLANE OR CURVED SURFACE DETERMINED BY PROJECTION Vertical Component The vertical component of the resultant force acting on the element dA in Fig 2–36b is dFv = (p dA) cos u This same result can also be obtained by noting that the horizontal projection of dA is dA cos u, and so dFv = p(dA cos u) = gh(dA cos u) Fv C B Since a vertical column of liquid above dA has a volume of d V = h(dA cos u), then dFv = gd V Therefore, the resultant’s vertical component is CV D P Fv = LV g dV = gV In other words, the resultant vertical force acting on the plate is equivalent to the weight of the volume of the liquid acting above the plate, Fig 2–36c This force acts through the centroid CV of the volume, which has the same location as the center of gravity for the weight of the liquid, since the specific weight of the liquid is constant Once the horizontal and vertical components of force are known, the magnitude of the resultant force, its direction, and its line of action can be established As shown in Fig 2–36c, this force will act through the center of pressure P on the plate’s surface This same type of analysis can also be applied in cases where the liquid is below the plate, rather than above it For example, consider the curved plate AD of constant width shown in Fig 2–37 The horizontal component of FR is determined by finding the force Fh acting on the projected area, DE The vertical component of FR, however, will act upward To see why, imagine that liquid is also present within the volume ABCD If this were the case, then the net vertical force caused by pressure on the top and bottom of AD would be zero In other words, the vertical pressure components on the top and bottom surfaces of the plate will be equal but opposite and will have the same lines of action Therefore, if we determine the weight of imaginary liquid contained within the volume ABCD, and reverse the direction of this weight, we can then establish Fv acting upward on AD Gas If the fluid is a gas, then its weight can generally be neglected, and so the pressure within the gas is constant The horizontal and vertical components of the resultant force are then determined by projecting the curved surface area onto vertical and horizontal planes and determining the components as shown in Fig 2–38 Fh FR A E Fh ϭ the resultant pressure loading on the vertical projected area DE Fv ϭ the weight of the volume of imaginary liquid ADCBA above the plate Fig 2–37 Fv ϭ pAh p Centroid of pressure volume Ah p Av Fh ϭ pAv Centroid of pressure volume Weight of gas is negligible Gas pressure is constant Fig 2–38 www.freebookslides.com 80 CHAPTER F L U I D S TAT I C S Important Points s The horizontal component of the resultant force acting on a submerged flat inclined or curved surface is equivalent to the force acting on the projection of the area of the surface onto a vertical plane The magnitude of this component and the location of its point of application can be determined using the methods outlined in Sec 2.7 through 2.9 s The vertical component of the resultant force acting on a submerged flat inclined or curved surface is equivalent to the weight of the volume of liquid acting above the surface This component passes through the centroid of this volume If the liquid is confined below the flat inclined or curved surface, then the vertical component is equal but opposite to the weight of imaginary liquid located within the volume extending above the surface to the liquid level s Pressure due to a gas is uniform in all directions since the weight of the gas is generally neglected As a result, the horizontal and vertical components of the resultant force of pressure acting on a flat inclined or curved surface can be determined by multiplying the pressure by its associated vertical and horizontal projected areas, respectively These components act through the centroids of these projected areas The formula, geometrical, or integration method can be used to determine the resultant pressure force acting on a surface, such as the endplates of this water trough or oil tank www.freebookslides.com 81 2.10 HYDROSTATIC FORCE ON AN INCLINED PLANE OR CURVED SURFACE DETERMINED BY PROJECTION EXAMPLE 2.15 The sea wall in Fig 2–39a is in the form of a semiparabola Determine the resultant force acting on m of its length Where does this force act on the wall? Take rw = 1050 kg>m3 SOLUTION Fluid Description y 2m We treat the water as an incompressible fluid Horizontal Force Component The vertical projection of the wall is AB, Fig 2–39b The intensity of the distributed load caused by water pressure at point A is 8m w A = (rwgh)(1 m) = 11050 kg>m3219.81 m>s22 (8 m)(1 m) = 82.40 kN>m Thus, Fx = x (8 m)(82.40 kN>m) = 329.62 kN (a) Using the table on the inside back cover for a triangle, from the surface of the water, this component acts at y = (8 m) = 5.33 m B x C y 8m Fx 82.40 kN/m Fy = 1rwg2AABC (1 m) x A 2m 1050 kg>m3 9.81 m>s2 c (2 m)(8 m) d (1 m) = 54.94 kN This force acts through the centroid of the volume (area); that is, from the inside back cover, x = Resultant Force Fy Ans Vertical Force Component The vertical force is equivalent to the weight of the water contained within the volume of the exparabolic segment ABC, Fig 2–39b From the inside back cover, the area of this segment is A A BC = ba Thus, = y (2 m) = 1.5 m Ans The resultant force is therefore FR = 2(329.62 kN)2 + (54.94 kN)2 = 334 kN Ans (b) Fig 2–39 www.freebookslides.com 82 CHAPTER EXAMPLE F L U I D S TAT I C S 2.16 The semicircular plate in Fig 2–40a is m long and acts as a gate in a channel Determine the resultant force the water pressure exerts on the plate, and then find the components of reaction at the hinge (pin) B and at the smooth support A Neglect the weight of the plate A SOLUTION Fluid Description Water is assumed to be an incompressible fluid for which gw = (1000 kg/m3)(9.81 m/s2) = 9.81(103) N/m3 3m B Analysis I We will first determine the horizontal and vertical components of the resultant force acting on the plate (a) D Horizontal Force Component The vertical projected area of AB is shown in Fig 2–40b The intensity of the distributed loading at B (or E) is A Fy wB = gwhBb = [9.81(103) N/m3](6 m)(4 m) = 235.44(103) N/m d Therefore, the horizontal force component is C [235.44(103) N/m](6 m) = 706.32(103) N = 706.32 kN This force acts at h = (6 m) = m Fx Fx = h B E (b) D A D A + C C Force acting on segment CA (d) B Force acting on segment CB Vertical Force Component From Fig 2–40b, note that the force pushing up on segment BC is due to the water pressure under this segment It is equal to the imaginary weight of water contained within BCDAB, Fig 2–40c And the vertical force pushing down on segment AC in Fig 2–40b is due to the weight of water contained within CDAC, Fig 2–40d The net vertical force acting on the entire plate is therefore the difference in these two weights, namely an upward force equivalent to the volume of water contained within the semicircular region BCAB in Fig 2–40b Thus, Fy = gwVBCAB = [9.81(103) N/m3]e [p(3 m)2] f (4 m) = 176.58p(103) N = 176.58p kN The centroid of this semicircular volume of water can be found on the inside back cover (c) Fig 2–40 d = 4(3 m) 4r = = m p 3p 3p www.freebookslides.com 83 2.10 HYDROSTATIC FORCE ON AN INCLINED PLANE OR CURVED SURFACE DETERMINED BY PROJECTION Resultant Force The magnitude of the resultant force is therefore FR = 2F 2x + F 2y = 2(706.32 kN)2 + (176.58p kN)2 = 898 kN Ans FA Reactions The free-body diagram of the plate is shown in Fig 2–40e Applying the equations of equilibrium, we have + c ⌺Fy = 0; 176.58 -By + 176.58p kN = By = 176.58p kN = 555 kN 4m Ans a + ⌺MB = 0;FA(6 m) - (706.32 kN) (2 m) - (176.58p kN) a m b = p FA = 353.16 kN = 353 kN + g Fx = S Ans kN 6m 706.32 kN 2m Bx 706.32 kN - 353.16 kN - Bx = B x = 353.16 kN = 353 kN By Ans (e) Analysis II We can also determine the resultant force components directly using integration In Fig 2–40f, notice how the pressure varies over the cross section To simplify the analysis, we will use polar coordinates because of the circular shape The elemental strip of width b has an area of dA = b ds = (4 m)(3 du m) = 12 du m2 Therefore, the pressure acting on it is p = gwh = [9.81(103) N>m3](3 - cos u) m dF = pdA = 29.43(10 )(1 - cos u) N>m h 3m d dA O For the horizontal component, dFx = p dA sin u, and so 3m p Fx = p sin u dA = 29.43(103) (1 - cos u)(sin u)(12 du) = 706.32 kN LA LO In a similar manner, the y component can be found from dFy = p dA cos u You may wish to evaluate this to verify our previous result for Fy.* (f) Fig 2–40 (cont.) *Be aware that this method can only be used to determine the components of the resultant force The resultant force cannot be found from FR = because it does not account for the changing direction of the force LA p dA ... Compressible Flow 704 13 .9 Normal Shock Waves 710 13 .10 Shock Waves in Nozzles 713 13 .11 Oblique Shock Waves 13 .12 Compression and Expansion Waves 723 13 .13 Compressible Flow Measurement 718 728 14 ... 3 1. 1 Introduction 1. 2 Characteristics of Matter 1. 3 The International System of Units 1. 4 Calculations 1. 5 Problem Solving 1. 6 Basic Fluid Properties 1. 7 Viscosity 1. 8 Viscosity Measurement 1. 9... Boundary Layer 525 11 .2 Laminar Boundary Layers 11 .3 The Momentum Integral Equation 11 .4 Turbulent Boundary Layers 11 .5 Laminar and Turbulent Boundary Layers 546 11 .6 Drag and Lift 11 .7 Pressure Gradient