•1 like•341 views

Report

Share

Download to read offline

Fluid Mechanics and Hydraulics

- 1. id Mechanics & Hydraulics lea }i AL Y AY oS PNL4 4 DIEGO INOCENCIO T. GILLESANIA Civil Engineer BSCE, LIT — Magna Cum Laude 5th Place, PICE National Students’ Quiz, 1989 Awardee, Most Outstanding Student, 1989 3rd Place, CE Board November 1989 Review Director & Reviewer in all Subjects Gillesania Engineering Review Center Reviewer in Mathematics and General Engineering Sciences MERIT Philippines Review, Manila Author of Various Engineering Books
- 2. Fluid Mechanics and Hydraulics Revised Edition The cardinal objective of this book is to provide reference to ngineering students taking-up Fluid Mechanics and Hydraulics. his may also serve as a guide to engineering students who will be taking the licensure examination given by the PRC. The book has 9 chapters. Each chapter presents the principles Copyright © 1997, 1999, 2003 and formulas involved, followed by solved problems and by Diego Inocencio Tapang Gillesania supplementary problems. Each step in the solution is carefully explained to ensure that it will be readily understood. Some All rights reserved. No part of this book may be problems are even solved in several methods to give the reader a reproduced, stored in a retrieval system, or choice on the type of solution he may adopt. transferred, in any form or by any means, without the prior permission of the author To provide the reader easy access to the different topics, the book includes index. Most of the materials in this book have been used in my review classes. The choice of these materials was guided by their effectiveness as tested in my classes. I wish to thank all my friends and relatives who inspired me ISBN 971-8614-28-1 in writing my books and especially to my children and beloved wife Imelda who is very supportive to me. Printed By: ; I will appreciate any errors pointed out and will welcome any GPP Gillesania Printing Press Feott for furthess t eo Ormoc City, Leyte suggestion for further improvement. Philippines bs Cover design by the author DIEGO INOCENCIO T. GILLESANIA Cebu City, Philippines *
- 3. To my mother Iuminada, my wife Imelda, and our Children Kim Deunice, Ken Dainiel, and Karla Denise TABLE OF CONTENTS Preface 4.22 ath eee pe Ue a Sd te ee Vii DISCICA LON sicdac vse cons del cP ks wap en ech Ladd po beve ego ne nad Vili CHAPTER 1 Properties Of Fluid .........cccsseees estes teeeeseessnessenessenenseneceneeceneenens 1 Types of Fluid......escesssecssesnecseessennecsessnsesressesssesnnsesncensesvensennneadicsseness 1 Mass Density.......sesssccsesseseeeeserneccnsersneeeneseesesssveneasscensanrusnesenseneneneeengss 2 Specific VOIUMe.......esccssessesreeeereesesesseeeeneeseessessesecneensenenenseneeseeneancsess 3 Unit Weight or Specific Weight .........cceesessessssstestesseneeneneeneeneaneenns 3 Specific Gravity ....cccccessesecseececeeeeesssseeneeseesseseessenseseensesennensensanssnsoess 4 ViSCOSItY vessesseseseesescseessssnenesneenseneenenscntenssseensanecsecnessensensenecseenenusenseees 4 Kimematic VisCOSity .......cssesecsesceeceseeseeeeeetseseseseeseeereeneseseseanenens 5 GTFACE PERGION:. ocsscepce sass cantiencssstestaastasessdeaealnacers Feedba saa sehen -stesseie 6 Capillary ccsatedenscnsles cs ecsuservenscvanaeisd setsctoresssonoensnaactoonnensygbentens dipeys i Comp ressibility .....c.ssccsessesseeseeseeneneereensenesserssnessrissenesneenennserenseanenys 8 Pressure Disturbances ..:...:.s...cscsconcssseoversetoesescevansseneonsersencesnoneonennens 9 Property Changes in Ideal Gas ........scssecsessssssessseeeerteeneceneetienneneyy 9. Vapor Pressure ...c.cssssessecsesosesnesseesrennstrcossonsesenesensecnnennennennentenasanennays 10 SOLVED PROBLEMS SUPPLEMENTARY PROBLEMS .......:....:ccceeeeeereseeeene 24 to 26 CHAPTER 2 Principles of Hydrostatics ........cceessecsssessrees ees eceeeseenecneennenens 27 Unit Pressure ........ povechessesccensnnsesesonscnsanescsossnsscscsusessessanseceanansccnsnanescs 27 Pascal’s Law...secessessesssesseesssensensesersersenenneessecsscseenseenesenenvensansessasse 27 Absolute and Gage Pressures.......csscsesesessrsseseesensecereesssesnenerseneny 29 Variations in Pressure Pressure below Layers of Different Liquids..........cccee ae Prassiate Fea snsscasdiscsitscthesceshecs gts woes devarn snes coraashbtnnghinsunntcertaodens 33 Manometers.iiinie).:.cagecao swiss capes epee aa 34 SOLVED PROBLEMG................... ba We cadasibehe inesdigedee exes 35 to 68 SUPPLEMENTARY PROBLEMS ..............ccec:sseseseusteessiens 69 to 72
- 4. TABLE OF CONTENTS TABLE OF CONTENTS iii CHAPTER 3 otal Hy drostatie Force on.Surfaces 2.0 iaccissssccscescis eee 79 Total Hydrostatic Force on Plane Surface Properties of Common Geometric ShapeS......ccscsceecssssesesseees 76 Total Hydrostatic Force on Curved Surface ...cccccccecccsssseseseeeee: 78 CHAPTER 5 Fundamentals of Fluid FLOW .........000::ccsscecseseresereteeseeseseesereenes 241 Discharge........--ssscecssesesnserernsenenrenceavesncerseretesenennenvenssccucansareseeseners Dieinittiony OF Terrrts feces ores cts cecgenteesenenesenonneonnenatrebonsedanrie 241 Energy and Head .....sssesesssensssssneereeereenseneens ee Power and Efficiency .....sesscesssereseecneseecesenerenensnerensseneeeenenes vee 245 Bernoulli’s Energy Theorem... 246 Energy and Hydraulic Grade Lines .......cssesesetereieenerieseeesten 248 SOLVED PROBLEMS i..i.0j25.0.socinsessscssessenees 14200 t0:273 SUPPLEMENTARY PROBLEMS .........:cesseseeeeee 274 to 276 PVAEG, diss cicdk an caoveleracrasec eosthoee GAR leek auido lane ee 81 TY pes OF Deis i. tas otaku dice ane ie eee 81 Analysis ot Gravity Das c...45.00:isiauleniatalen ene 84 Buoyancy Archimedes Principles 4s s.c<sissecessunos cack aeeeaeee eal 88 Statical Stability of Floating Bodies...) sti nacntel ene 90 Stress on Thin-Walled Pressure Vessels o..ccccccccsccccssessessecsessscesece 96 Cylindrical Taik cc dncuecedeusscauienuc me nee 96 Spherical Shell........ VNOOC: Steve PH tay. tiacertes casa vii dnnsinhioncidevl ee ila WIR RIN 98 CHAPTER 6 Biitidl RiGwvieastirCinenie |i eteseaaviraenctsre aps oe Device CORP MGICIIES Son tcrecaceissscscesveawes eee as aptewerecaeesuesenenteeree= mens 275 Head lost in Measuring Devices .......sscceseceereeeereeseneeteenrererene GRC ale ee nee al irene el eae Ne. be suche pseannseseepensensiaty 281 Values of H for Various Conditions ........:..scccseceesecenseeeteeees 283, Copttaction Of the Jet .c..:..cccieuseresensacnes CcleSieeeiien ner nee 284 Once UO Or OW FLCAGSiiiteuiite ct atsacasearessaceescaessccusamasitenrs 285 STA EULTE IN LOLOL. oie area ates rcne ey bal ou tadocaentarennevnens pesmitnedens 285 IN ee eva aiaass AUCH eas canteens rs tuteaa nes eetisavaamnee sets Prtotlubensd...0tias. URE aN eee ING celts aeen CireeR Eater cen nes CHAPTER 4 Relative Rquilibriuan Of Liquide (25 cose. casccesenisoiomteneletes 201 Rectilinear Translation Horizontal Motion ERCINC ELIS HON wi gptinl blu decaccecd ei RE a es 202 Vertical Motion..........0...... Rapides sacticeativ es See ae be he a RTs 203 Rotation Voluine-of Parabolotd:. s1:.ccubueeete eee Peed Bee Sess, 205 Liquid Surface Conditions jc... Meera ateotcicde eh 206 SOLVED PROBLEMS Rectangular Wei0.....:ceceseersseseereseeseenentenetneneens Contracted Rectangular Weirs Triangular WEILS .....scsercecerersereesoseereess Trapezoidal WeirS.......sereresesene nae Cipolletti Weir ........ oe eee Suttro Weils.......ccstgemmeeaeentessessssuceseessssnasugeivanaeonssonosesalsesoatenee
- 5. iv TABLE OF CONTENTS Submerged TW Git ics odicvecieic ae ae IL Eel 305 OTIS VY BLOWic oscctdiovethcx.actasdesteges Sr RM nanos eae agee 306 SOLVED PROBLEMS. «. .1<cereicresetearundnoermr tla eiucesivenseds 307 to 371 SM hana 372 to 374 CHAPTER 7 MUTI BLOW TRA A OS ar can scocrnot eon cht ace byaventa Wavedactscadgreuenen es idenonesen 37/9 TRATES <casascanss ccevaver buat vaca once aiataetausessmar pt eeenccsvcae ds iene a70 Revol dis NANOOM 5: cesccast ectersvantedosl acters annotate puis 376 Viewty Distmbuportin Pipes... diaiidearctsotnmnnuaarare 377 Shearing Stress in Pipes ........... Dich eeMNl PN ta tn do cai AWNG ehalydeten Sea eye Head Losses in Pipe Flow Major Head Loss... sssisastteers he tater Seana. siepcciaetrdes 381 Datey-Wets bach Porntla s.iss:.stvosjecs eh coattstaniugaramersnelinteers 381 MINDS OE, Fs a. 5 Spaces enus cai angn vane Arado Sele usa ch teois oe Ma ced 382 WOOdy Dig STATI .s.uc,sseueiveerecces aesimeatshes sacs ounuG cases 384 Manning Formula ci.c3.5.s.¢hiianteats Riri nes enka Weal aan deh 385 Hazen Willams Formula iis.) diniiotdswuungeaeires 386 MEmOt ECA LOSSahs ana. traconncicoyeires tances ea enmmeeemen 387 DUCILET) BIUAVECHIOINE 114.0... acceacvrssebive Mera cheet eter aeteies 388 Gradual Enlargement Bizdiden Contract Gt ic.cisivetiesscteteeyvnctetees oeesereceeier ean eke 388 Bends and Stancard VIN 6. scpestierloadaratiset dpa staan 390 Pipe Discharaing froin ReSGrVOlt :)., cesar tiserremosssnteadanaracy 390 Pipe Commecting I Wo Reser viOins sito dices var ieeisninsatcccmnnrateen 391] Prpes tm series-and Parallels: .ic.c.yscu seat ere a eevvns een peritg 392 BGPP GUOE PPI ot ceurtdiyeensdncdbansdoinns Wapato sisi cahinkensenves 394 Reservoir Problems PEPS NCEW OPIS saceaasasstges caeevaveteaiaus o)anges canarias ces cctvsus nearieosed 398 SOLVED PROBLEMS TABLE OF CONTENTS Vv CHAPTER 8 Open Channel .........sseteccssecseestantecrsssteonssrsseveesvenssssnennesene se Terns 481 Specific Energy ..:ssessessecrvecseetecreessessesseeseesessessscencensennsenenncencnnseseene 481 Chézy Formula .....sccccessecssssesesnnessessesssrescearsssensensensensaeneenennenenseness 482 Kutter and Gunguillet Formula............... cosneeessnenecenversouneessen 483 Manning Formula ......:ccccecsesseeseeeeeneseesecseeseeseeneensensnsenenneenenes 483 BeeZaen FONE ELG cy cases Sepeceved mh vnadv al eee so ves) cn trign bade nnpeiten> erat 483 Powell Equation .......c:scssessessesessesseeeceeseeseeneesesnsessensseeneesenaneeny 484 TROP POW csvnteicercasrcoidersttiw ctapratonians linea eigenen omen 485 Boundary Shear Stress.......ccscccssessessessesseseesecssesssnsersennenneeeeneaneenns 485 Normal Depth.....seccccecessecessseeeerereesseeeenesessesesrsaneneeeeneasenenrencones 486 Most Efficient S@Ctions .......::cse0ssscessessesessescscnssareneonencensssanesenrenes 486 Proportions for Most Efficient Sections .....cssssseeeeee 487 Rectangular Section........ccccseseeeeeeeneesesesssensensenecnnenreneeny 487 Trapezoidal Section ...secsscsesseennresesssecsnsessennnecsnecanennnens 487 Triangular Section........csssessecereereersesreseesersesseasenneensenenneenns 489 Circular SOCHONS sesdscccsiVi.vsedcrsenuteasedecenstpnostnotoniensapensnesnenyaeeyons 490 Velocity Distribution in Open Channel ..........ssesessereeeieeiees 491 Alternate Stages Of FlOW .....scscsecsssesessesesseeseenesnseeseesseneeneerensenn 491 PEGIES DBE Yoshio av cs eases ng enacvngtins hats opennenrecastoasualseeieeers 492 Criticall Depthicasedar av iebdistea stb tagtannsvngeinadlecdomeonenbee tiie 492 Non-Uniform or Varied FIOW ..:...:.ccssconseessescerscstecerserenepensensonees 495 Hydraulic Jump.....scccssecssssssissesrersneensesnedesnsssnsennneceneccnnseteennneensensess 497 Flow around Channel BendS.....ccccccscscsecesereesssssetenlonesseensenevees 500 SOL VED. PROBLEMS .icssjcncpitiissscss sccssnspevacerscssonsentaoterenen 501 to 547° SUPPLEMENTARY PROBLEMS ...........:-:cesererrneeees 547 to 550
- 6. FLUID MECHANICS & HYDRAULICS Chapter 1 Properties of Fluids CHAPTER ONE 1 Properties of Fluids vi TABLE OF CONTENTS CHAPTER 9 Hydrodynamics............ Piraeus: RITE Redon enre: 551 Force against Fixed Flat Plates......... snes as ee 551 Force against Fixed Curved Vanes........ Bocuisrecctes eae fetal DOS Force against Moving Vanes.......... Leet rts a kalat added 554 Work Done on Moving Vane ..ecssccscseseeneceeese ashliciee 555 Force Developed on Closed Conduit ......cccccscesee: ue hemearOO0 Drag and Litt wei ia.e eer Gents ani abil dieachats eich ; 557 terminal V Glocity...-.discacacden Libjaedoraouse acetone eels A 559 Water Hamme ticcicarccciec dessus hadanvaccaieee eee ee f “an 560 SOLVED PROBLEMB............... fesiiasevelaid ‘diac arene’ 2568.16 597 SUPPLEMENTARY PROBLEMS ......... Ei tlegsilles Gutcaeced 597 to 598 FLUID MECHANICS & HYDRAULICS Fluid Mechanics is a physical science dealing with the action of fluids at rest or in motion, and with applications and devices in engineering using fluids. Fluid mechanics can be subdivided into two major areas, fluid statics, which deals with fluids at rest, and fluid dynamics, concerned with fluids in motion. The term hydrodynamics is applied to the flow of liquids or to low-velocity gas flows where the gas can be considered as being essentially incompressible. Hydraulics deals with the application of fluid mechanics to engineering devices involving liquids, usually water or oil. Hydraulics deals with such problems as the flow of fluids through pipes or in open channels, the design of storage dams, pumps, and water turbines, and with other devices for the control or use of liquids, such as nozzles, valves, jets, and flowmeters. APPENDIX Properties of Fluids and Conversion Factors ....... eee 2 O99 Table A - 1: Viscosity and Density of Water at 1 atm............ 599 Table A - 2: Viscosity and Density of Air at 1 atm............. she 600 Table A - 3: Properties of Common Liquids at 1 atm & 20°C.. 601 Table A - 4: Properties of Common Gases at 1 atm & 20°C...... 601 Table A - 5: Surface Tension, Vapor Pressure, pe and Sound Speed of Water ............. Etc L alone a 602 Table A - 6: Properties of Standard Atmosphere........... vid 603 Table A - 7: Conversion Factors from BG to SI Units le te 604 Table A - 8: Other Conversion Factors .........scscsssesssssesseesee ein 605 , TYPES OF FLUID Fluids are generally divided into two categories: ideal fluids and real fluids. Ideal fluids e Assumed to have no viscosity (and hence, no resistance to shear) e Incompressible e Have uniform velocity when flowing e No friction between moving layers of fluid INDEX e No eddy currents or turbulence I-IV Real fluids e Exhibit infinite viscosities ‘ e Non-uniform velocity distribution when flowing e Compressible e Experience friction and turbulence in flow
- 7. 2 CHAPTER ONE Properties of Fluids FLUID MECHANICS & HYDRAULICS Real fluids are further divided into Newtonian fluids and non-Newtonian fluids. Most fluid problems assume real fluids with Newtonian characteristics for convenience. This assumption is appropriate for water, air, gases, steam, and other simple fluids like alcohol, gasoline, acid solutions, etc. However, slurries, pastes, gels, suspensions may not behave according to simple fluid relationships. Ideal Fluids | Real Fluids ! Newtonian Fluids ¥ ¥ y Pseudoplastic Fluids} [ Delatant Fluids} [Bingham Fluids} Non-Newtonian Fluids |i Figure 1 - 1: Types of fluid MASS DENSITY, p (RHO) The density of a fluid is its mass per unit of volume. oe mass of fluid, M Eq. 1-1 volume, V Units: English slugs/ ft? Metric gram/cm? SI : kg/més Note: Pslugs = Pibm/ For an ideal gas, its density can be found from the specific gas constant and ideal gas law: Eq. 1-2 FLUID MECHANICS & HYDRAULICS Properties of Fluids f se > where: p= absolute pressure of gas in Pa R = gas constant Joule / kg-°K For air: R = 287 J/kg- °K R= 1,716 lb-ft/slug-°R T = absolute temperature in °Kelvin °K ="C+ 273 oR = °F + 460 CHAPTER ONE Table 1 - 1: Approximate Room-Temperature Densities of Common Fluids Fluid p in kg/m? Air (STP) 129 Air (21°F, a 1tm) 1.20 Alcohol 790 Ammonia 602 Gasoline 720 Glycerin 1,260 Mercury 13,600 Water 1,000 SPECIFIC VOLUME, V; | 7 Specific volume, Vs, is the volume occupied by a unit mass of fluid. 3 1 Wee P UNIT WEIGHT OR SPECIFI€ WEIGHT, y aN Specific weight or unit weight, y, is the weight of a unit volume of a fluid. Eq, 1-3 weight of fluid, W v= volume, V VS PS Eq. 1-4 Eq.1-5
- 8. FLUID MECHANICS 4 CHAPTER ONE & HYDRAULICS Properties of Fluids Units: English :. lb/ft Metric : dyne/cm? SI : N/m or kN/m3 SPECIFIC GRAVITY Specific gravity, s, is a dimensionless ratio of a fluid’s density to some standard reference density. For liquids and solids, the reference density is water at 4° C (39.2° F). _ Pliquid $ Eq.1-6 P water In gases, the standard reference to calculate the specific gravity is the derisity of air. Peas Pair $5 Eq. 1-7 For water at 4°C: y = 62.4 Ib/ ft? = 9.81 kN/m? p= 1.94 slugs/ft? = 1000 kg/m3 s=1.0 VISCOSITY, 1: (MU) The property of a fluid which determines the amount of its resistance to shearing forces. A perfect fluid would have no viscosity. Consider two large, parallel Raat plates at a small distance y z - apart, the space between them being filled with a fluid. Consider the upper plate to y be subject to a force F so as to move with a constant velocity U. The fluid in contact with ~ A moving plate fixed plate CHAPTER ONE 5 FLUID MECHANICS Properties of Fluids & HYDRAULICS the upper plate will adhere to it and will move with the same velocity LU while the fluid in contact with the fixed plate will have a zero velocity. For small values of U and y, the velocity gradient can be assumed to be a straight line and F varies as A, U and yas: ee AU F U Fe —— or —« y A y3 bit Doig - (from the figure) t ay = Shearing stress, t |>4|< dV dV te = or t= Kk dy dy where the constant of proportionality k is called the dynamic of absolute viscosity denoted as i. : Eq. 1-8 MS dV / dy where: * t = shear stress in lb/ft or Pa u = absolute viscosity in lb sec/ ft? (poises) or Pa-sec. y = distance between the plates in ft or m U = velocity in ft/s or m/s KINEMATIC VISCOSITY v (NU) Kinematic viscosity is the ratio of the dynamic viscosity of the fluid, 1, to its mass density, p. ¥ vet Eq. 1-9 where: u = absolute viscosity in Pa-sec p = density in kg/m?
- 9. 6 CHAPTER ONE FLUID MECHANICS Properties of Fluids & HYDRAULICS Table 1 - 2: Common Units of Viscosity System Absolute, Kinematic, v lb-sec/ ft? English FLUID MECHANICS & HYDRAULICS Capillarity CHAPTER ONE 7 Properties of Fluids d ae (slug/ft-sec) fe pec dyne-s/cm? cm2/s (poise) (stoke) Pa-s Si. (N-s/m?) Bere Metric Note: 1 poise = 1 dyne-s/cm? = 0.1 Pa-séc (1 dyne = 105 N) 1 stoke = 0.0001 m?2/s ; * SURFACE TENSION o (SIGMA) The membrane of “skin” that seems to form on the free surface of a fluid is due to the intermolecular cohesive forces, and is known as surface tension. Surface tension is the reason that insects are able to sit on water and a needle is able to float on it. Surface tension also causes bubbles and droplets to take on a spherical shape, since any other shape would have more surface area per unit volume. Pressure inside ‘a Droplet of Liquid: ’ p= Eq. 1- 10 where: o = surface tension in N/m d = diameter of the droplet in m p = gage pressure in Pa (a) Adhesion: > Cohesion (b) Cohesion > adhesion Capillarity (Capillary action) is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall or a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liquid to the walls of the containing vessel. Liquids rise in tubes they wet (adhesion > cohesion) and fall in tubes they do not wet (cohesion > adhesion). Capillary is important when using tubes smaller than about 3/8 inch (9.5 mm) in diameter. - 2 4ocos8 yd h Eq. 1-11 For complete wetting, as with water on clean glass, the angle 0 is 0°. Hence 8 8 the formula becomes Eq: d= 12 where: h = capillary rise or depression in m y = unit weight in N/m? d = diameter of the ttube inm o = surface tension iin Pa
- 10. 8 CHAPTER ONE FLUID MECHANICS Properties of Fluids & HYDRAULICS Table 1 - 3: Contact Angles, 6 Materials Angle, 0 mercury-glass 140° water-paraffin 107° water-silver 90° kerosene-glass 26° lycerin-glass 19° water-glass 0° ethyl alcohol-glass 0° COMPRESSIBILITY, B Compressibility (also known as the coefficient of conipressibilit) is the fractional change in the volume of a fluid per unit change in pressure in a constant- Temperature process. Fo Meare eer Eq. 1-18 Ap q Or & ae Eq. 1-14 where: : AV = change in volume V = original volume ‘ Ap = change in pressure dV/V = change in volume (usually in percent) BULK MODULUS OF ELASTICITY, E, The bulk modulus of elasticity of the fluid expresses the compressibility of the fluid. It is the ratio of the change in unit pressure to the corresponding volume change per unit of volume. FLUID MECHANICS CHAPTER ONE 9 & HYDRAULICS Properties of Fluids a stress _ Ap Eq. 1-15 strain AV a or E; =-——— Eq. 1-16 PRESSURE DISTURBANCES Pressure disturbances imposed on a fluid move in waves. The velocity or celerity of pressure wave (also known as acoustical or sonic velocity) is expressed as: pe ee [a Eq. 1-17 p Bp PROPERTY CHANGES IN IDEAL GAS Hor any ideal gas experiencing any process, the equation of state is given by: - Bea Eq. 1-18 T; a When temperature is held constant, Eq. 1 - 18 reduces to (Boyle’s Law) fa Vi = p2 V2 Eq. 1- 19 | When temperature is held constant (isothermal condition), Eq. 1 - 18 reduces to (Charle’s Law) " 2 Eq.1 -20 Mkt Ty
- 11. CHAPTER ONE Properties of Fluids FLUID MECHANICS & HYDRAULICS For Adiabatic or Isentropic Conditions (no heat exchanged) pr Vik = po Vek Eq. 1- 21 wy as Va Pi or = Constant Eq. 1 - 22 ka pe Goa Ty Pi and Eq. 1-23 where: pi = initial absolute pressure of gas p2 = final absolute pressure of gas V, = initial volume of gas V2 = final volume of gas T, = initial absolute temperature of gas in °K (°K = °C + 273) T2 = final absolute temperature of gas in °K k = ratio of the specific heat at constant pressure to the specific heat at constant volume. Also known as adiabatic exponent. VAPOR PRESSURE Molecular activity in a liquid will allow some of the molecules to escape the liquid surface. Molecules of the vapor also condense back into the liquid. The vaporization and condensation at constant temperature are equilibrium processes. The equilibrium pressure exerted by these free molecules is known as the vapor pressure or saturation pressure. Some liquids, such as propane, butane, ammonia, and Freon, have significant vapor pressure at normal temperatures. Liquids near their boiling point or that vaporizes easily are said to volatile liquids. Other liquids such as mercury, have insignificant vapor pressures at the same temperature. Liquids with low vapor pressure are used in accurate barometers. The tendency toward vaporization is dependent on the temperature of the liquid. Boiling occurs when the liquid temperature is increased to the point that the vapor pressure is equal to the local ambient (surrounding) pressure. Thus, a liquid’s boiling temperature depends on the local ambient pressure, as well as the liquid’s tendency to vaporize. PLUID MECHANICS CHAPTER ONE 1 1 & HYDRAULICS Properties of Fluids Table 1 - 4: Typical Vapor Pressures Fluid kPa, 20°C mercury 0.000173 turpentine 0.0534 water 2.34 ethyl alcohol 5.86: ether 58.9 butane 218 Freon-12 584 propane 855 ammonia 888 Solved Problems Problem 1-1 A reservoir of glycerin has a.mass of 1,200 kg and a volume of 0.952 cu. m Pind its (a) weight, W, (b) unit weight, y, (c) mass density, p, and (d) specific ravity (s). Solution ) (b) (¢) Weight, V=M eg = (1,200)(9.81) Weight, W= 11,772 N or 11.772 kN WwW Unit weight, y= — nit weight, 7 . T1772 0.952 Unit weight, y = 42.366 kN/m$ | M Density, p = — nsity, p 7 1200 0,952 Density, p = 1,260.5 kg/m? Density, p =
- 12. CHAPTER ONE 12 FLUID MECHANICS Properties of Fluids FLUID MECHANICS CHAPTER ONE 1 3 & HYDRAULICS & HYDRAULICS Properties of Fluids Solution (a) W=mg=22(9.75) W= 214.5 N (d) Specific gravity, s = fey P water _ 1,260.5 1,000 1.26 Specific gravity, s il (b) Since the mass of an object is absolute, its mass will still be 22 kg Specific gravity, s Problem 1-5 What is the weight of a 45-kg boulder if it is brought to a place where the acceleration due to gravity is 395 m/s per minute? Problem 1 - 2 The specific gravity of certain oil is 0.82. Calculate its (a) specific weight, in Ib/ft3 and kKN/m3, and (b) mass density in slugs/ft? and kg/m’. Solution W= Mg Solution (a) Specific weight, y = Ywater X $ Specific weight, y = 62.4 x 0.82 = 51.168 Ib/ft3 Specific weight, y = 9.81 x 0.82 = 8,044 kN/m$3 m/s lmin Seas min 60sec g = 6.583 m/s? W = 45(6.583) W = 296.25 N g = 395 (b) Density, p = pwater x Density, p = 1.94 x 0.82 = 1.59 slugs/ft? Density, p = 1000 x 0.82 = 820 kg/m3 Problem 1 - 6 A lf the specific volume of a certain gas is 0.7848 m°/kg, what is its specific Weight? Problem 1 - 3 A liter of water weighs about 9.75 N, Compute its mass in kilograms, Solution Solution Mass = Yes Mass =:=—— a p Nee V, 0.7848 Mass = 0. 1.2742 kg/m bo x a oT Oo i i} p * Specific weight, y = p x g = 1.2742 x 9.81 Specific weight, y = 12.5 N/m3 Problem 1 - 4 If an object has a mass of 22 kg at sea level, (2) what will be its weight at a point where the acceleration due to gravity g = 9.75 m/s? (b) What will be its mass at that point?
- 13. FLUID MECHANICS 1 4 CHAPTER ONE & HYDRAULICS Properties of Fluids Problem 1 - 7 What is the specific weight of air at 480 kPa absolute and 21°C? Solution Y= pRig o= a where R = 287 J/kg-°K 480 x 10° 287 (21 + 273) 5.689 kg Pp y = 5.689 x 9.81 y = 55.81 N/m3 FLUID MECHANICS CHAPTER ONE 1 5 & HYDRAULICS Properties of Fluids Solution Density, p = is 8 18:7 9. 1.397 kg/m? P RT _ (205+ 101.325) x 10° R(32 + 273) i Gas constant, R = 718.87 J/kg - °K Density, p 1.397 Note: Patm = 101.325 kPa a Problem 1-8 Find the mass density of helium at a temperature of 4 °C and a pressure of 184 kPa gage, if atmospheric pressure is 101.92 kPa. (R = 2079 J/kg * °K) Solution P D Ss * Se esas ensity, p RT P = Peage ot Patm = 184 + 101.92 p = 285.92 kPa T =4 +273 =277°K 285.92 x 10° 2,079(277) Density, p = 0. 4965 kg/m$ Density, p = Problem 1 - 10 Air is kept at a pressure of 200 kPa absolute and a temperature of 30°C in a 500-liter container. What is the mass of air? Solution eh PO RT 200 x 10° 287(30 + 273) 2.3 kg/m? p Mass =p x V a 5( 2.3 x 1000 Mass = 1.15 kg Problem 1 - 9 t 32°C and 205 kPa gage, the specific weight of a certain gas was 13.7 N/m‘%. | etermine the gas constant of this gas. Problem 1-11 A cylindrical tank 80 cm in diameter and 90 cm high is filled with a liquid. The tank and the liquid weighed 420 kg. The weight of the empty tank is 40 kg. What is the unit weight of the liquid in kN/m’.
- 14. FLUID MECHANICS 1 CHAPTER ONE & HYDRAULICS Properties of Fluids Solution j= MV i) = ee = 840 kg/m? £ (0.8)" (0.90) Y=pPg = 840(9,81) = 8240.4 N/m3 y = 8.24 kN/m? Problem 1 - 12 A lead cube has a total mass of 80 kg. What is the length of its side? Sp. gr. of lead = 11.3. Solution Let L be the length of side of the cube: M=pV 80 = (1000 x 11.3) L3 : L=0.192 m=19.2 cm Problem 1 - 13 A liquid compressed in a container has a volume of 1 liter at a pressure of 1 MPa and a volume of 0.995 liter at a pressure of 2 MPa. The bulk modulus of elasticity (Eg) of the liquid is: Solution dP 2-1 Eat dV/V (0.995—1)/1 Es = 200 MPa Problem 1 - 14 What pressure is required to reduce the volume of water by 0.6 percent? Bulk modulus of elasticity of water, Eg = 2.2 GPa. FLUID MECHANICS CHAPTER ONE 1 7 & HYDRAULICS Properties of Fluids Solution . dP dV /V dp = pr- pr m=09 dp = pr dV = V2 = V1 dV =-0.6% V = -0.006V Eg= . P2 Eg = -—————_ = 2.2 0.006V /V p2 = 0.0132 GPa * p2= 13.2 MPa Problem 1 - 15 Water in a hydraulic press, initially at 137 kPa absolute, is subjected to a pressure of 116,280 kPa absolute. Using Ex = 2.5 GPa, determine the percentage decrease in the volume of water. Solution dV /V (116,280 — 137) x 10° dV /V 2.5.x 10? = - Oe -0.0465 V dv = 4.65°% decrease V Problem 1 - 16 * If 9 m3 of an ideal gas at 24 °C and 150 kPaa is compressed to 2 m°, (a) what is the resulting pressure assuming isothermal conditions. (b) What would have been the pressure and temperature if the process is isentropic. Use k= 1.3
- 15. 8 CHAPTER ONE : F Properties of Fluids Wee & HYDRAULICS — Solution (a2) For isothermal condition: nV= p2 V2 150(9) = po (2) p2 = 675 kPa abs (b) For isentropic process: Pi Vik = po Vok 150(9)!3 = p2 (2)!8 p2 = 1,060 kPa abs (k-1)/k Ty a Pa Ty Pi 244273 ( 150 Tz = 466.4°K or 193.4°C Problem 1 - 17 If the viscosity of water at 70 °C is 0.00402 poise and its specific gravity is 0.978 determine its absolute viscosity in Pa - s and its kinematic viscosity in m2/s and in stokes. Solution Absolute viscosity: u = 0.00402 pewex 2 a8 lpoise u = 0.000402 Pa-s Kinematic viscosity: 0.000402 (1000 x 0.978) 4.11 x 107 m4/s vets p Vv lstoke v=4.11 x 10” m2/s x 0.0001 m2/s v = 4,11 x 103 stoke CHAPTER ONE 19 FLUID MECHANICS Properties of Fluids & HYDRAULICS LA Problem 1 - 18 lwo large plane surfaces are 25 mm apart and the space between them is filled with a liquid of viscosity ,1 = 0.958 Pa-s. Assuming the velocity gradient to be ght line, what force is required to pull a very thin plate of 0.37 m? area at stant speed of 0.3 m/s if the plate is 8.4 mm from one of the surfaces? Solution F=F,+ Fo T usy 16.6 ELA 25mm U/y 8.4 i e wllA eae y 0.958(0.3)(0.37) F,= =6.4N iz 0.0166 _ 0.958(0.3 OI) A eae P= 0.0084 = 12.66 N Lt w= F=6.4 + 12.66 F=19.06 N Problem 1 - 19 A cylinder of 125 mm radius rotates concentrically inside a fixed cylinder of 140 mm radius. Both cylinders are 300 mm long. Determine the viscosity of the liquid which fills the space between the cylinders if a torque of 0.88 N-m is red to maintain an angular velocity of 27 radians/sec Assume the elocity gradient to be a straight line
- 16. FLUID MECHANICS 20 CHAPTER ONE & HYDRAULICS Properties of Fluids Solution T U/y U=ra U = 0.125(2n) rotating U = 0.785 m/s eee y = 0.005 m Torque = F(0.125) - Deer realli | Torque = tA (0.125) ae ce fixed cylinder 0.88 = t [27(0.125)(0.3)] (0.125) t = 29.88 Pa L=0.3m <— liquid : 29.88 | SS Sanna 0.785/0.005 : / 0.005 47° NL 09.125 u = 0.19 Pa-s eI 0.13 m Problem 1 - 20 An 18-kg slab slides down a 15° inclined plane on a 3-mm-thick film of oil with viscosity }. = 0.0814 Pa-sec. If the contact area is 0.3 m2, find the terminal velocity of the slab. Neglect air resistance. Solution W = 18(9.81) = 176.58 N slab a ¥ / Terminal velocity is attained when the sum of all forces in the direction of motion is zero. FLUID MECHANICS & HYDRAULICS CHAPTER ONE 2 1 Properties of Fluids [DF = 0] Wsin 6 - F;=0 F,=W sin 0 F, = 176.58 sin 15° ll WW L (Fs=tTA= goa y babe, Ble Ul re 176.58 sin 15° = 0.0814 —— (0.3) 0.003 Ll = 5.614 m/s , = 5.614 m/s ¢ Problem 1 - 21 Estimate the height to which water will rise in a capillary tube of diameter 3 mm. Use o = 0.0728 N/m and y = 9810 N/m? for water. Solution Note: 6 = 90° for water in clean tube ‘Capillary rise, = ae yd 4(0.0728) 0.0099 m = 9.9 mm Capillary rise, h Capillary rise, Ii Problem 1 - 22 Estimate the capillary depression for mercury in a glass capillary tube 2 mm in diameter. Use o = 0.514 N/m and 8 = 140° Solution 4ocos@ _ 4(0.514)(cos 140°) ya (9810 x 13.6)(0.002) Capillary rise, h = -0.0059 m (the negative sign indicates capillary depression) Capillary rise, h = Capillary depression, li = 5.9 mm
- 17. 2 CHAPTER ONE CHAPTER ONE FLUID MECHANICS FLUID MECHANICS 23 Properties of Fluids i f Fluids & HYDRAULICS "& HYDRAULICS Properties u — Problem 1 - 26 2S de if S d to A sonar transmitter operates at 2 impulses per second. If the device is tes Hie surface of fresh water (Ez = 2.04 x 10° Pa) and the echo is received midway 8s c 5 helween impulses, how deep is the water? Problem 1 - 23 What is the value of the surface tension of a small drop of water 0.3 mm in pressure within the droplet is 561 Pa? Solution Solution The velocity of the pressure wave (sound wave) is 4o 0.0003 o = 0.042 N/m 561 = ee Sonar i — =1,428 m/s eludig stp te Mneisek pk Aa 1000 lam 1 ‘ - > 0 Problem 1 - 24 Ninice the echo is received 3 iiltlway between impulses, then ihe total time of travel of sound, O }» ¥4(0.5) = % sec and the total distance covered is 2/1, then; h An atomizer forms water droplets 45 um in diameter. Determine the excess pressure within these droplets using o = 0.0712 N/m. Solution pa oe e d 4(0.0712) 45x 107% 2h=ct 2h = 1,428(%) = 178.5 m P= = 6,329 Pa . Problem 1 - 27 Ai what pressure will 80 °C water boil? | (Vapor pressure of water at 80°C = 47 4 kPa) Problem 1 - 25 Distilled water stands in a glass tube of 9 mm diameter at a height of 24 mm. hat is the true static height? Use o = 0.0742 N/m. Solution Solution Water will boil if the atmospheric pressure equals the vapor p.essure _ 40cosé yd where 0 = 0° for water in glass tube ‘ re water at 80 °C will boil at 47.4 kPa 4(0.0742) 9810(0.009) h= = 0.00336 m = 3.36 mm True static height = 24 - 3.36 True static height = 20.64 mm
- 18. 24 CHAPTER ONE ~ Properties of Fluids FLUID MECHANICS & HYDRAULICS [Supplementary Problems Problem 1 - 28 What would be the weight of 1 3. -ke mass $ ; due to gravity is 10 m/s2? & mass on a planet where the acceleration Ans: 30 N Problem 1 - 29 A vertical cylindrical tank with a diame with water to the top with water at 20° much water will spill over? Unit wei kN/m$3 and 9.69 kN/m3, respectively. ter of 12 m and a depth of 4 m is filled C. If the water is heated to 50°C, how ght of water at 20°C and 50°C is 9.79 Ans: 4.7 m3 Problem 1 - 30 A rigid steel container is partially filled with a li the liquid is 1.23200 L. Ata pressure of 30 a 1.23100 L. Find the average bulk modulus of given range of pressure if the temperature a return to its initial value. What is the coefficie quid at 15 atm. The volume of tm, the volume of the liquid ‘is elasticity of the liquid over the fter compression is allowed to nt of compressibility? Ans: Eg = 1.872 GPa; B = 0.534 GPa") Problem 1 - 31 Calculate the density of water va is 0.462 kPa-m3/kg-°K. poratoot-kFaabs and 20°C i ity gas constant Ans: 2.59 kg/m Problem 1 - 32 Air is kept at a pressure of 200 kPa and a : t oC container. What is the mass of the air? Peer ory OLS jek S008 Ans: 1.15 kg CHAPTER ONE 25 FLUID MECHANICS Properties of Fluids & HYDRAULICS Problem 1 - 33 (a) If 12 m? of nitrogen at 30°C and 125 kPa abs is permitted to expand jsothermally to 30 m3, what is the resulting pressure? (b) What would the pressure and temperature have been if the process had been isentropic? Ans: (a) 50 kPa abs (b) 34.7 kPa abs; -63°C Problem 1 - 34 A square block weighing 1.1 kN and 250 mm on an edge slides down an incline on a film of oil 6.0 ym thick. Assuming a linear velocity profile in the vil and neglecting air resistance, what is the terminal velocity of the block? The viscosity of oil is 7 mPa-s. Angle of inclination is 20°. Ans: 5.16 m/s Problem 1 - 35 flenzene at 20°C has a viscosity of 0.000651 Pa-s. What shear stress is required to deform this fluid at a strain rate of 4900 s?? Ans: tT =3.19 Pa Problem 1 - 36 A shaft 70 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250 mm long. The clearance, assumed uniform, is filled with oil at 20°C with v = 0.005 m2/s and sp. gr. = 0.9. Find the force exerted by the oil in the shaft. Ans: 987 N Problem 1 - 37 Two clean parallel glass plates, separated by a distance d = 1.5 mm, are dipped in a bath of water. How far does the water rise due to capillary action, if o = 0.0730 N/m? ‘ Ans: 9.94 mm
- 19. 2 CHAPTER ONE Properties of Fluids FLUID MECHANICS & HYDRAULICS Problem 1 - 38 Find the angle the surface tension film leaves the glass for a vertical tube immersed in: water if the diameter is 0.25 inch : re inch. Use o = 0.005 lb/ft. .25 inch and the capillary rise is 0.08 Ans: 64,3° Problem 1 - 39 What force is required to lift a thin wire ring 6 cm in diameter from a water ae (o of water at 20°C = 0.0728 N/m). Neglect the weight of the Ans: 0.0274 N FLUID MECHANICS CHAPTER TWO 2 7 & HYDRAULICS Principles of Hydrostatics Chapter 2 Principles of Hydrostatics UNIT PRESSURE OR PRESSURE, p Pressure is the force per unit area exerted by a liquid or gas on a body or surface, with the force_acting at right angles to the surface uniformly in all directions. ‘ Force, F = = Eq. 2-1 P Area, A 4 In the English system, pressure is usually measured in pounds per square inch (psi); in international usage, in kilograms per square centimeters (kg/cm?), or in atmospheres; and in the international metric system (SI), in Newtons per square meter (Pascal). The unit atmosphere (atm) is defined as a pressure of 1.03323 kg/cm? (14.696 lb/in’), which, in terms of the conventional mercury barometer, corresponds to 760 mm (29.921 in) of mercury. The unit kilopascal (kPa) is defined as a pressure of 0.0102 kg/cm? (0.145 Ib/sq in). PASCAL’S LAW Pascal’s law, developed by French mathematician Blaise Pascal, states that the pressure on a fluid is equal in all directions and in all parts of the container. In Figure 2 - 1, as liquid flows into the large container at the bottom, pressure pushes the liquid equally up into the tubes above the container. The liquid rises to the same level in all of the tubes, regardless of the shape or angle of the tube.
- 20. 28 CHAPTER TWO : FLUID MECHANICS . aug : FLUID MECHANICS CHAPTER TWO 29 Principles of Hydrostatics & HYDRAULICS & HYDRAULICS Principles of Hydrostatics ABSOLUTE AND GAGE PRESSURES Gage Pressure (Relative Pressure) Gage pressures are pressures above or below the atmosphere and can be Measured by pressure gauges or manometers., For small pressure differences, a U- tube manometer is used. It consists of a U-shaped tube with one end connected to the container and the other open to the atmosphere. Filled with a liquid, such as water, oil, or mercury, the difference in the liquid surface levels in the two manometer legs indicates the pressure difference from local atmospheric tonditions. For higher pressure differences, a Bourdon gauge, named after the french inventor Eugéne Bourdon, is used. This consists of a hollow metal tube with an oval cross section, bent in the shape of a hook. One end of the tube is tlosed, the other open and connected to the measurement region. Figure 2 ~ 1: Illustration of Pascal’s Law The laws of fluid mechanics are observable in many everyday situations. For example, the pressure exerted by water at the bottom of a pond will be the same as the pressure exerted by water at the bottom of a much narrower pipe, provided depth remains constant. If a longer pipe filled with water is tilted so that it reaches a maximum height of 15 m, its water will exert the same pressure as the other examples (left of Figure 2 - 2). Fluids can flow up as well as down in devices such as siphons (right of Figure 2 - 2). Hydrostatic force causes water in the siphon to flow up and over the edge until the bucket is empty or the suction is broken. A siphon is particularly useful for emptying containers that should not be tipped. Atmospheric Pressure & Vacuum Atmospheric Pressure is the pressure at any one point on the earth's surface from the Weight of the air above it. A vacuum is a space that has all matter removed from it. lt is impossible to create a perfect vacuum in the laboratory; no matter how advanced a vacuum system is, some molecules are always present in the vacuum area. Even remote regions of outer space have a small amount of gas. A vacuum ran also be described as a region of space where the pressure is less than the formal atmospheric pressure of 760 mm (29.9 in) of mercury. Under Normal conditions at sea level: Patm = 2166 lb/ft? = 14.7 psi = 29.9 inches of mercury (hg) = 760 mm Hg = 101.325 kPa Absolute Pressure Absolute pressure is the pressure above absolute zero (vacuunt) Py = Pa = Ps Figure 2 - 2: Illustration of Pascal’s Law Pabs = Pgage + Patm Eq: 2-2 Note: * Absolute zero is attained if all air is removed. It is the lowest possible pressure attainable. * Absolute pressure can never be negative. * The smallest gage pressure is equal to the negative of the ambient atmospheric pressure.
- 21. CHAPTER TW/O 3 1 eaten Principles of Hydrostatics & HYDRAULICS — _ VARIATIONS IN PRESSURE FLUID MECHANICS 3 0 CHAPTER TWO & HYDRAULICS Principles of Hydrostatics 6 Standard 0 gage 58.675 gage atmosphere-= 101.325 abs re i | Nae Current atmosphere = 100 abs -40 gage ~41.325 gage 4-9 160 abs 60 abs : Absolute zero = -101.325 gage / or -100 gage All pressure units in kPa Figure 2 - 3: Relationship between absolute and gage pressures Note: Unless otherwise specified in this book, the term pressure signifies gage pressure. MERCURY BAROMETER 760 mm nanan A mercury barometer. is an accurate and relatively simple way to measure changes in atmospheric pressure. At sea level, the weight of the atmosphere forces mercury 760 mm (29.9 in) up acalibrated atmospheric glass tube. Higher elevations yield lower readings pressure because the atmosphere is less dense there, and the thinner air exerts less pressure on the mercury. Mercury yr at Sea Level Pointers ANEROID BAROMETER In an aneroid barometer, a partially evacuated metal drum expands or contracts in response to changes in air pressure. A series of levers and springs translates the up and down movement of the drum top into the circular motion of the pointers along the aneroid barometer's face, Metal drum (partial vacuum) Hairspring Consider any two points (1 & 2), ends of an elementary prism having a cross whose difference in elevation is li, to lie in the -sectional area a and a length of L i i i ilibrium. Since this prism is at rest, all forces acting upon it must be in equili Free liquid surface eS ao : Pi & pe are gage pressures O69 N oa Figure 2 - 4: Forces acting on elementary prism j r with Note: Free Liqufd Surface refers to liquid surface subject to zero gage pressure oO atmospheric pressure only. With reference to Figure 2- 4: W=yV W= y (aL) {=F = 0] Fy - Fy =W sin 9 pra-pia=y (aL) sin 8 po-pi=yLsin® but Lsin9 =h e po-m=yh Eq.2-3 | | : ! ints in a homogeneous fluid Therefore; the difference in pressure between any tivo points tr oe at Goat is nal to the product of the unit weight of the fluid (y) to the vertica (h) between the points.
- 22. CHAPTER TWO 32 FLUID MECHANICS Principles of Hydrostatics & HYDRAULICS Also: L . p2= pit wh Eq. 2-4 | This means that any change in pressure at point 1 would cause an equal change at point 2. Therefore; a pressure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point in the liquid. Let us assume that point © in Figure 2 - 4 lie on the free liquid surface, then the gage pressure p, is zero and Eq. 2 - 4 becomes: p=wh Eq.2-5 | This'means that the pressure at any point “h” below a free liquid surface is equal to the product of the unit weight of the fluid (y) and h. Consider that points © and @ in Figure 2 - 4 lie on the same elevation, such that h = 0; then Eq. 2 - 4 becomes: L Pi = pe Eq. 2-6 | This means that the pressure along the same horizontal plane in a homogeneous fluid at rest are equal. Pressure below Layers of Different Liquids Air, pressure = pa FLUID MECHANICS CHAPTER TWO 3 3 & HYDRAULICS Principles of Hydrostatics Consider the tank shown to be filled with liquids of different densities a with air at the top under a gage pressure of pa, the pressure at the bottom o the tank is: | Proton = LY H+ p= yi In+ yatta + ys ha + pa Eq.2-7| PRESSURE HEAD Pressure head is the height “h” of a column of homogeneous liquid of unit weight y that will produce an intensity of pressure p. h= : E Eq. 2-8 Y To Convert Pressure head (height) of liquid A to liquid B YA hg = Ia SA or hg = Ita PA or hg = ha—— Eq. 2-9 5B PB YB 7 To convert pressure head (height) of any liquid to water, just multiply its height by its specific gravity | hiwater = Miiquid X Sliquid Eq:2-10 |
- 23. FLUID MECHANICS & HYDRAULICS 34 CHAPTER TWO Principles of Hydrostatics MANOMETER A manometer is a tube, usually bent in a form of a U, containing a liquid of known specific gravity, the surface of which moves proportionally to changes of pressure. It is used to measure pressure. Types of Manometer Open Type - has an atmospheric surface in one leg and is capable of measuring gage pressures. Differential Type - without an atmospheric surface and capable of measuring only differences of pressure. Piezometer - The simplest form of open manometer. It is a tube tapped into a wall of a container or conduit for the purpose of measuring pressure. The fluid in the container or conduit rises in this tube to form a free surface Limitations of Piezometer: * Large pressures in the lighter liquids require long tubes * Gas pressures can not be measured because gas can not form a free surface tae (a) Open manometer (b) Differential manometer (c) Piezometer ID MECHANICS HWYDRAULICS — — _ Bleps in Solving Manometer Problems: 1. Decide on the fluid in feet or meter, of which the heads are to bc expressed, (water is most advisable). 2 Starting from an end point, number in order, the interface of different fluids. 4, Identify points of equal pressure (taking into account that for a homogeneous fluid at rest, the pressure along the same horizontal plane are equal). Label these points with the same number. - 4. Proceed from level to level, adding (if going down) or subtracting (if CHAPTER TW/O 3 5 Principles of Hydrostatics ; going up) pressure heads as the elevation decreases or increases, respectively with due regard for the specific gravity of the fluids. ‘|Soived Problems Problem 2-1 If a depth of liquid of 1 m causes a pressure of 7 kPa, what is the specific Pavity of the liquid? i - Solution Pressure, p = yh : 7 = (9.81 x s) (1) ’ s=0.714 > Specific Gravity ’ F Se Problem 2 - 2 What is the pressure 12.5 m below the ocean? Use sp. gr. = 1.03 for salt water. Solution ] p=yh p = (9.81 x1.03)(12.5) p =126.3 kPa i ie
- 24. CHAPTER TWO Principles of Hydrostatics FLUID MECHANICS — & HYDRAULICS Problem 2 - 3 If the pressure 23 meter below a liquid is 338.445 kPa, datenmune te unit weight y, mass density p , and specific gravity s. Solution (a) Unit weight, y p=yh 338.445 = y (23) y = 14.715 kN/m3 «°) Mass density, p p a g 14.715x 10° 9.81 p = 1,500 kg/m’ p= (c) Specific gravity, s g = Pefluid P water 1,500 1,000 s=15 Problem 2 - 4 If the pressure at a point in the ocean is 60 kPa, what is the pressure 27 meters below this point? Solution The difference in pressure between any two points in a liquid is po - pi =yh p2 = pit yh = 60 + (9.81x1.03)(27) pr = 332.82 kPa FLUID MECHANICS CHAPTER TW/O 3 7 & HYDRAULICS Principles of Hydrostatics Problem 2-5 if the pressure in the air space above an oil (s = 0.75) surface in a closed tank is 115 kPa absolute, what is the gage pressure 2 m below the surface? Solution P = Psurface +y h Psurface = 115 - 101.325 Psurface = 13.675 kPa gage p = 13.675 + (9.81x0.75) (2) p = 28.39 kPa Note: Patm = 101.325 kPa Problem 2 - 6 Find the absolute pressure in kPa at a depth of 10 m below the free surface of oil of sp. gr. 0.75 if the barometric reading is 752 mmHg. Solution Pats = Patm i Peage Patm = Ym hy = (9.81 x 13.6)(0.752) Patm = 100.329 kPa Pabs =: 100.329 ae (9.81 x« 0.75)(1 0) Pabs = 173.9 kPa Problem 2 - 7 A pressure gage 6 m above the bottom of the tank containing a liquid reads 90 kPa. Another gage height 4 m reads 103 kPa. Determine the specific weight of the liquid. Solution po-pi=yh 103 - 90 = y(2) * y = 6.5 kN/m3
- 25. 33 CHAPTER TW/O FLUID MECHANICS FLUID MECHANICS CHAPTER TW/O 39 Principles of Hydrostatics & HYDRAULICS & HYDRAULICS Principles of Hydrostatics Problem 2 - 8 An open tank contains 5.8 m of water covered with 3.2 m of kerosene (y = 8 Since the density of the mud varies with depth, the pressure kN/m3). Find the pressure at the interface and at the bottom of the tank. should be solved by integration Solution Solution dp =y dh a (10 + e h)dh [u ~ foososna : 5 (a) Pressure at the interface pa = Yen = (8)(3.2) pa = 25.6 kPa Kerosene pa 3 (b) Pressure at the bottom re EEN => yh Be i Water = Yo lw + Yk hy y = 9.81 kN/m? = 9,81(5.8) + 8(3.2) pr = 82.498 kPa I iS 10h + 0.25h7 | 0 [10(5) + 0.25(5)2] - 0 p = 56.25 kPa WW 4 Problem 2 - 11 Ii the figure shown, if the atmospheric If atmospheric pressure is 95.7 kPa and the gage attached to the tank reads 188 pressure is 101.03 kPa and the absolute mmHg vacuum, find the absolute pressure within the tank. pressure at the bottom of the tank is Problem 2 - 9 Pa etl et el et et ee SAE Oil, s = 0.89 E oh i, 1 iy : 11.3 kPa, what is the specific gravity { Solution of olive oil? Water 2.5m Pubs = Patm + Pgage i . | Pgage = Ymercury Hmercury | = (9,81 x 13.6)(0.188) : Olive, s =? om 25.08 kPa vacuum -25.08 kPa Pavs = 95.7 + (-25.08) Pabs = 70.62 kPa abs Pgage Mercury, s = 13.6 4m Problem 2 - 10 Solution Gage pressure at the bottom of the tank, p = 231.3- 101.03 TI ht d f di by y = 10 + 0.5h, wh kN/m? and ne Welgt deri aba Hiugsie giver Dy Y iW te tn SIN ay a Gage pressure at the bottom of the tank, p = 130.27 kPa his in meters. Determine the pressure, in kPa, at a depth of 5m [p = Zyh] Pp = Yu Bis a Yo ho 25 Vio hyp a Yoil hot 130.27 = (9.81 x 18.6)(0.4) + (9.81 ~’ s)(2.9) + 9.81(2.5) + (9.81 x 0.89)(1.5) s = 1.38
- 26. FLUID MECHANICS 4 CHAPTER TWO & HYDRAULICS— Principles of Hydrostatics Problem 2 - 12 If air had a constant specific weight of 12.2 N/m? and were incompressible, what would be the height of the atmosphere if the atmospheric pressure (sea level) is 102 kPa? Solution Height of atmosphere, h £ Y 102x 10° 12.2 = 8,360.66 m Height of atmosphere, h Problem 2 - 13 (CE Board May 1994) Assuming specific weight of air to be constant at 12 N/m, what is the approximate height of Mount Banahaw if a mercury barometer at the base of the mountain reads 654 mm and at the same instant, another barometer at the top of the mountain reads 480 mm. Solution hm = 654 mm Poot ~ Prop = yh (Ym Hm) bottom in (Ym lim) top (y h)air (9,810 x 13.6)(0.654) - (9,810 x 13.6)(0.48) = 12h h = 1,934.53 m FLUID MECHANICS & HYDRAULICS Problem 2 - 14 i i Ititude o the barometric pressure in kPa at an a of p ae oe el is 101.3 kPa. Assume isothermal conditions a 21°C. Use pressure at sea lev R = 287 Joule /kg-°K. Solution For gases: dp = -pg dl 287(21 + 273) 0,00001185 p il Pp dp = -(0.00001185 p)(9.81) dh ae 0.0001163 dh P 1200 BR ls =-0,0001163 dh P 101.3x10* 9 P inp | = = 0.0001163% | 101.3x10* CHAPTER TWO 4 1 Principles of Hydrostatics f 1,200 m if the 1200 0 In p - In (101.3 x 10°) = - 0.0001163(1200 - 0) In p = 11.386 ex 311386 Pp = € p = 88,080 Pa
- 27. 42 CHAPTER TWO FLUID MECHANICS FLUID MECHANICS. CHAPTER TWO 43 Principles of Hydrostatics & HYDRAULICS & HYDRAULICS Principles of Hydrostatics — Problem 2 - 18 (CE November 1998) i hile that of piston B is 950 sq. cm. Convert 760 mm of mercury to (a) oil of sp. gr. 0.82 and (b) water. Miston A has a cross-section of 1,200 sq. cm w - "a z with the latter higher than piston A by 1.75 m. If the intervening passages are filled with oil whose specific gravity is 0.8, what is the difference in pressure felween A and B. s niercury Problem 2 - 15 Solution (a) Hoit = Nmercury Soil “hae = 0.76 9 39 Not = 12.605 m of oil Solution PA A ps = V6 No = ( 9,810 x 0.8)(1.75) pa - ps = 13,734 Pa (b) Nwater = Hmereury Smercury = 0.76(13.6) water = 10.34 m of water 1200 cm? 950 cm? Problem 2-16 (CE Board May 1994) A barometer reads 760 mmHg and a pressure gage attached to a tank reads 850 cm of oil (sp. gr. 0.80). What is the absolute pressure in the tank in kPa? Problem 2-19 fi the figure shown, Hetermine the weight W that can be carried by the 15 KN force acting on the piston. 300 mm @ Solution Pats = Pam - Pgage = (9.81 x 13.6)(0.76) + (9.81 x 0.8)(8.5) Pats = 168.1 kPa abs 1.5 kN Oil, s = 0.82 Problem 2 - 17 A hydraulic press is used to raise an 80-kN cargo truck. If oil of sp. gr. 0.82 acts on the piston under a pressure of 10 MPa, what diameter of piston is required? 30 mm @ Solution Since points 1 and 2 lie on the Since the pressure under the piston is uniform: ; same elevation, pr = p2 Force = pressure x Area 80,000 = (10 x 10%) = D? 15 al Ae D=0.1 m=100 mm : f (0.03) 7 (0.3) W=150 kN Solution . BE ic 2 Oil, s = 0.82 Nae mm @
- 28. CHAPTER TW/O Principles of Hydrostatics 45 FLUID MECHANICS & HYDRAULICS 4 CHAPTER TWO FLUID MECHANIC: Principles of Hydrostatics & HYDRAULI Problem 2 - 20 Solution. A drum 700 mm in diameter and filled with water has a vertical pipe, 20 in diameter, attached to the top. How many Newtons of water must poured into the pipe to exert a force of 6500 N on the top of the drum? Solution Force on the top: F=px Area 6500 = p x £ (700? - 20?) p = 0.016904 MPa p = 16,904 Pa Plunger, a = 0.00323 m Oil, s = 0.78 [p=yh] 16,904 = 9810 h h=1,723m [po - pr = yh Ba 0.00823 m = 309.6F (kPa) W 44 BRE ar A 0.323 136.22 kPa Area Weight = y x Volume = 9810 x 4 (0.02)2(1.723) Weight = 5.31 N Area on top — 700 mm @ p2 Problem 2 - 21 The figure shown shows a setup with a vessel containing a plunger and a cylinder. What force F is required to balance the weight of the cylinder if the weight of the plunger is negligible? 136.22 - 309.6 F = (9.81 = 0.78)(4.6) F = 0.326 kN = 326 N Problem 2 - 22 ok Che hydraulic press shown is filled with oil with sp. gr. 0.82. Neglecting . weight of the two pistons, what force F on the handle is required to suppor the 10 kN weight? Cylinder W = 44 kN F=? A = 0.323 m2 4.6m Plunger, a = 0.00323 m? 75mm@ | Hinge 25mm @ Oil, s = 0.78 Oil, s = 0.78 oil
- 29. FLUID MECHANICS & HYDRAULICS 46 CHAPTER TW/O Principles of Hydrostatics Solution Since points | and 2 lie on the same elevation, then; Pi = p2 75mm @ apts VES £(0.075)> = (0.025)? [= Mo = 0] F(0.425) = F3(0.025) F(0.425) = 1.11(0.025) F = 0.0654 kN F=65.4N 400 mm | FBD of the lever arm Problem 2 - 23 The fuel gage for a gasoline (sp. gr. = 0.68) tank in a car reads proportional to its bottom gage. If the tank is 30 cm deep an accidentally contaminated with 2 cm of water, how many centimeters of gasoline does the tank actually contain when the gage erroneously reads “FULL”? Solution @ “Full” “Full” CHAPTER TWO 4 7 FLUID MECHANICS Principles of Hydrostatics & HYDRAULICS ” Since the gage reads “FULL” then the reading is equivalent to 30 cm of gasoline Reading (pressure head) when the tank contain water = (y + 2 gig) em of gasoline Then; y+ 25h =30 0.68 y = 27.06 cm Problem 2 - 24 (CE Board November 2000) " for the tank shown in the Figure, hy = 3m and In =4m_ Determine the value of In ha hy Solution Summing-up pressure head from 1 to 3 in meters of water PA 4 jy(0.84)-x= 7% Y y 0 + 0.84 hn - (4-3) =0 hy =119m
- 30. FLUIDIMEGHANICS of Hydrostatics 48 Principles of Hydrostatics & HYDRAULICS HYDRAULICS Principles of Fry dros Problem 2 - 25 (CE Board May 1992) Problem 2 - 26 CHAPTER TWO: qo In the figure shown, what is the static pressure in kPa in the air chamber? manometer shown): aworle TS MfOMs My OM) TNS & P / For the 4 determine the pressure at the center of the pipe. Pee eenteeeee et sticae? il, s = 0.80 Solution The pressure in the air space equals the pressure on the surface of oil, py Solution | ete 14 A i | mi Sum-up pressure head” fro mt lto3in meters a6" wateal Dic P2 = Vw ha 2 = 9.81(2)¢ p2 = 19.62 kPa P2 - P3 = Yo Ita 19.62 - ps = (9.81 x 0.80)(4) px = -11.77 kPa 14175(9:81) py = 144.7 kPa Another solution. Sum-up pressure head from 1 to 3 in meters of water A +2-4(0.80)= 23 y ja2U39 se 9.81 ps =-11.77 kPa = 13.55
- 31. 5 O CHAPTER TW/O Principles of Hydrostatics Problem 2 - 27 (CE Board November 2001) Determine the value of y in the manometer shown in the Figure Air, 5 KPa = Oil S=0.8 & o g sey = E mi ql Solution FLUID MECHANICS & HYDRAULICS” a Summing-up pressure head from A to B in meters of water: Air, 5 KPa E cal A PA +3(0.8) +15 - (13.6) = 2B oil y y S'3018 5 ~~ +3.9.-13.6y= £2 e ” 9.81 y & Water 1 where pz = 0 0.5m FLUID MECHANICS & HYDRAULICS Problem 2 - 28 (CE May 1993) In the figure shown, when the funnel is empty the water surface is at point A and the mercury of sp. gt. 13.55 shows a deflection of 15° cm Determine the new deflection of mercury when the 1p funnel is filled with water to B. Solution fs 30cm © i y = 0.324 m Mercury Figure (a): Level at A Solve for y in Figure (a): CHAPTER TW/O 51 Principles of Hydrostatics a 30cm @ | 80 cm Figure (b): Level at B Sum-up pressure head from A to 2 in meters of water: PA 4 y.0.15(13.55) = ¥ 0+y-2.03=0 y = 2.03 m
- 32. 1D MECHANICS CHAPTER TWO 53 HYDRAULICS Principles of Hydrostatics 52 CHAPTER TW/O FLUID MECHANICS Principles of Hydrostatics & HYDRAULICS In Figure (0): Sum-up pressure head from 2 to m in meters of water: When the funnel is filled with water to B, point 1 will move down tol with the same value as point 2 moving up to 2’ P2 + (13.6) -x= 2 Y Y 13.6y-x= 4 Eg. (1) Sum-up pressure head from B to 2’: In Figure (b) PB +08+y+x- (x +015 + x)(13,55) = 22 7 ¥ 0+ 0.80 + 2.03 + x - 27.1x - 2.03 =0 26.1 x = 0,80 x=0.031 m=3.1cm Sum-up pressure head from 2’ to m’ in meters of water: P2’ + (0.2sin6 + y+ 0.2)(13.6) - (x + 0.2 oe Y 0 + 2.72 sin © +.13.6y,+ 2.72+ x- 0.22 98 .6y — x = 8,183 - 2.72 sin 8 New reading, R = 15 + 2x = 15 + 2(3.1) 13.6y - x q. (2) New reading, R = 21.2 cm [13.6y - x = 13.6y - x] 8.183 - 2.72 sin @ = 22 sin 6 = 0.3852 0 = 22.66° Problem 2 - 29) The pressure at point m in the figure shown was increased from 70 kPa to 105 kPa. This causes the top level of Water i Problem 2-30 "A tlosed cylindrical tank contains 2 m of water, 3 m of oil (s = 0.82) and the air _ wheve oil has a pressure of 30 kPa. If an open mercury manometer at the yitom of the tank has 1 m of water, determine the deflection of mercury. mercury to move 20 mm in the sloping tube, What is the inclination, 0? Mercury Air, 30 kPa Solution Sum-up pressure head from 1 - Solution 1 to 4 in meters of water: Pair 430.82) +2+1- y(13.6) = F 3m ie Y 4 30 Zo, +246 +3-13.6y=0 on y = 0.626 m 4+-@ Figure (a) Figure (b) In Figure (a):
- 33. 5 CHAPTER TWO Principles of Hydrostatics Problem 2 - 31 The U-tube shown is 10 mm in diameter and contains mercury. If 12 ml of water is poured into the right-hand leg, what are the ultimate heights in the two legs? Eee Solution Solving for h, (see figure b): Volume of water = 4 (20)? h = 12 cm3 Note: 1 ml = 1 cm? 4 “10 h = 15.28 em = 152.8 mm Since the quantity of mercury before and after water is poured remain the same, then; 120(3)=R+x+120+. R + 2x = 240 > Eq. (1) FLUID MECHANICS & HYDRAULICS 120 mm ey Figure (b) ie ei a I Figure (a) -UID MECHANICS WYDRAULICS - In Figure (b): Pl 4 152.8 - R(13.6 ‘ R=11.24 mm In Eq. (2): 11.24 + 2x = 240 x= 114.38 mm Ultimate heights in each leg: Right-hand leg, hr =h +x Left-hand leg, hy =R+x CHAPTER TWO 5 5 Principles of Hydrostatics Summing-up pressure head from 1 to 3 in mm of water: P2 y = 152.8 + 114.38 Right-hand leg, hg = 267.18 mm = 11.24 + 114.38 Left-hand leg, h, = 125.62 mm ;Problem 2-32 Vor a gage reading of -17.1 kPa, _ iletermine the (#) elevations of the liquids in the open piezometer columns E, F, and (; and (b) the deflection of the Wercury in the U-tube Manometer neglecting the Weight of air. J El. 15m || |
- 34. 56 CHAPTER TWO Principles of Hydrostatics Solution Gage Air 2m s = 0.70 f 4m Water beams He it Se Mercury, s = 13.6 Column E Sum-up pressure head from 1 to e in metes of water Be tein = Let y Y ae hi(O.7) = hy = 25 m Surface elevation = 15 - h, Surface elevation = 15 - 2.5=12.5m Column F Sum-up pressure head from 1 to fin meters of water; A + 3(0.7) - h(t) = 24 y Bt +21-m=0 hz = 0.357 m Surface elevation = 12 + /1 Surface elevation = 12 + 0.357 = 12.357 m FLUID MECHANICS & HYDRAULICS hy he cs h3 5 Pi = par = -17.1 kPa FLUID MECHANICS & HYDRAULICS CHAPTER TWO 5 7 Principles of Hydrostatics Column G Sum-up pressure head from 1 to g in meters of water; PL 430.7) + 4(1) - In(1.6) = Las ¥ Y =17.1 ue All +2.1+4-1.6hs=0 hy = 2.72 m Surface elevation = 8 + Is Surfacé elevation = 8 + 2.72 = 10.72 m Deflection of mercury Sum-up pressure head from 1 to 5 in meters of water; Pr 4.30.7) +444 -hy(13.6) = Me ie ZL + 10.1 - 13.6 ‘hy = 0.614 m : Problem 2 - 33 An open manometer attached to a pipe shows a deflection of 150 mmHg with the lower level of mercury 450 mm below the centerline of the pipe carrying water. Calculate the pressure at the centerline of the pipe. Solution Sum-up pressure head from 1 to 4 in meters of water; Y : Y Pi_ +.0.45-2.04=0 981 Pi = 15.6 kPa
- 35. FLUID MECHANICS & HYDRAULICS: FLUID MECHANICS 58 CHAPTER TW/O _& HYDRAULICS Principles of Hydrostatics Solution CHAPTER TWO ‘ Principles of Hydrostatics Problem 2 - 34 if im@ >| For the configuration shown, calculate the weight of the piston if the pressure gage reading is 70 kPa. (a) Gage liquid = mercury, h=01m Sum-up pressure head from a 1 to 4 in meters of water; Pr gy +h-h(13.6)-x-15= Y Ye Pa | Pa ~45-01+0.1(13.6) fy Pi _ P4 ~ 2.76 mof water Y Y Oil Ss = 0.86 Solution Gage liquid = carbon tetrachloride reading, h =? Sum-up pressure head from A to B in meters of water; Ps = 70 kPa PA _1(0.86) = P8 Y x PAA 0366 — 9.81 981 pa = 78.44 kPa Pl 4 y+h-n(1.59)-x-15= = i Pr _ P4 ~454+0.59h iy ¥ where Pi _ P4 2276m > from (a) Py 2.76 =1.5+ 0.59 h= 2.136 m Fy = Pa x Area Weight =F, = pax Area = 78.44 x £(1 Weight = 61.61 kN V2 Sum-up pressure head from 1 to 4 in meters of water; Problem 2-36 .; In the figure shown, determine ihe height 1 of water and the Problem 2 - 35 Two vessels are connected to a differential manometer using mercury, the connecting tubing being filled with water. The higher pressure vessel is 1.5 m age reading at A when the lower in elevation than the other. (a) If the mercury reading is 100 mm, what, Absolute pressure at B is 290 is the pressure head difference in meters of water? (b) If carbon tetrachloride’ kPa. (s = 1.59) were used instead of mercury, what would be the manometer reading for the same pressure difference? Air, p = 175 kPa abs
- 36. CHAPTER TW/O 6 0 CHAPTER TWO FLUID MECHANIC§ § FLUID MECHANICS Principles of Hydrostatics Principles of Hydrostatics & HYDRAULICS | & HYDRAULICS Sum-up pressure (gage) head from 1 to 4 in meters of water; Sum-up absolute pressure head Air, p = 175 kPa abs ieee x(0.9) + 1.3(0.9) - 1303.6) == from B to 2 in meters of water; : Y — 2 40° PB 0.7(13.6) -h= 22 4 + 0,9x- 16.51 = 0 Y Y h 9 Water 981 290 3475 x=13.81m O81 - 9.52 - h = O8r Haat a 4 h= 2.203 m 700 mmm Mercury Then, x + y = 28.42 m Solution Problem 2 - 38 Pe DA For the manometer - 0.7(13.6) + 0.7= —& determine the difference in i i between A and B. 20 .952+07= a 3 1700 mm pa = 203.5 kPa abs Sum-up absolute pressure head from Bto A in meters ofwater; setup shown, pressure Problem 2 - 37 In the figure shown, the atmospheric pressure is 101 kPa, the gage BNR reading at A is 40 kPa, and the vapor pressure of alcohol is 12 kPa alr absolute. Compute x + y. ay Alcohol vapor K Solution x+0.68=y+17 x-y=1.02m * > Eq. (1) Sum-up pressure head from A to B in meters of water; x - 0.68(0.85) + y = Alcohol s = 0.90 PB Y Pa | PB ~y_y+0.578 > Eq. (2) Y i Solution Sum-up absolute pressure head from 1 to 2 in meters of water; Substitute x - y = 1.02 in Eq. (1) to Eq. (2): p =12kPa abs] . | PA _ PB ~102+0.578 ~)A mM Y PA~PB —1 598 9.81 pa ~ ps = 15.68 kPa Air 1 1 - y(0.9) = =2 Y Y Alcohol 40+ 40 +101 s = 0.90 ~0:9y=, 2. Oat aie - y=14.61m > inal : 1.3m im Mercury
- 37. 6 CHAPTER TWO Principles of Hydrostatics Problem 2 - 39 A differential manometer is attached to a pipe as shown. Calculate the difference between points A and B. pressure Mercury Solution Mercury 100 mm Sete es ey beet Oil, s = 0.90 Sum-up pressure head from A to B in meters of water; DA : PA _ W(0.9y-0.1(13.6) + 0.1(0.9) + y(0.9) = 2B ay ’ PA Pp FAS. £8" 210.1(13.6)- 0.1(0.9) he ¥ PA —Ps 9.81 pa - pa = 12.46 kPa =1.27m ‘FLUID MECHANICS & HYDRAULICS CHAPTER TWO UID MECHANICS Principles of Hydrostatics A HYDRAULICS Problem 2 - 40 "ih the figure shown, the deflection of mercury is initially 140 mm. If the pressure at A is Wiereased by 40 kPa, while Maintaining the pressure at B / vonstant, what will be the new Solution Lf L/ Figure (a) Figure (b) In Figure a, sum-up pressure head from A to B in meters of water; PA. _ 9.6 -0.25(13.6) + 0.25 + 21 = 28 y ¥ Pa PR = 1.65 m of water Y y
- 38. FLUID MECHANIC 64 CHAPTER Two & HYDRAULICS Principles of Hydrostatics In Figure b, pa’ = pa + 40 Sum-up pressure head from A’ to B in meters of water; PA’| (0.6- x) - (0.25 + 20)13.6 + (235 +x) = 2 +40 EAT N64 4-34027 de hoes eve 22 Y iF BA, ie + ee RS 25.25 y 981 ee Ay Is Db» 4 ‘ > fa _ FB 959% 2.493 But 2A _ 2B.= 765 a ye oy 1.65 = 25.2 x - 2.423 x = 0.162 m= 162 mm New mercury deflection = 250 + 2x = 250 + 2(162) New mercury deflection = 574 mm Problem 2 - 41 In the figure shown, determine the difference in pressure between points and B. Kerosene, s = 0.82 Air s = 0.0012 Benzene s = 0.88 100 mm 200 mm x iSO ram Solution FLUID MECHANICS CHAPTER TW/O 6 5 ‘& HYDRAULICS ; Principles of Hydrostatics Solution ae Kerosene, s = 0.82 s = 0.0012 Benzene s = 0.88 250 mm 200 mm EN : 150 mm Mercury} Sum-up pressure head from A to B in meters of water, PA + 0,2(0.88) - 0.09(13.6) - 0.31 (0.82) + 0,25 - 0.1(0.0012) = 8 Y y Pa PBs 27.0503 of water yey pa ~ pa = 9.81(1.0523) = 10.32 kPa Problem 2 - 42 (CE Board) Assuming normal barometric pressure, how deep in the ocean is the point where an air bubble, upon reaching the surface, has six times its volume than it had at the bottom? Seer Applying Boyle’s Law (assuming isothermal condition) [pi Vi = p2 V2) : pi = 101.3 + 9.81(1.03)h pi = 101.3 + 10.104 h Vv=V pr = 101.3 + 0= 101.3 V2. =6V (101.3 + 10.104/) V = 101.3 (6 V) 10.104 h = 101.3(6) - 101.3 h= 50.13 m
- 39. IYMECHANICS. CHAPTER TWO & CHAPTER TWO FLUID MECHANIC DRAULICS Principles of Hydrostatics Principles of Hydrostatics & HYDRAULIC Problem 2 - 43 Since the pressure in air inside the tube is uniform, then p. = pr = 20.0124 kPa Pe = Yw lt 20.0124 = 9.81h; h = 2.04 m A vertical tube, 3 m long, with one end closed is inserted vertically, with th open end down, into a tank of water to such a depth that an open manomet connected to the upper end of the tube reads 150 mm of mercury. Neglectin vapor pressure and assuming normal conditions, how far is the lower end ¢ the tube below the water surface in the tank? Then; x=h+y=2.04+ 0.495 x = 2.535 m Solution Aah iblam 2 - 44 nile consisting of a cylinder 15 cm in diameter and 25 cm high, has a neck } is 5 cm diameter and 25 cm long. The bottle is inserted vertically in ‘ f, with the open end down, such that the neck is completely filled with +. Find the depth to which the open end is submerged. Assume normal yometric pressure and neglect vapor pressure. 3mi(C > y; 5s Kee Bea lution plying Boyle’s Law pi Vi = p2 V2 > Applying Boyle’s Law: ‘pi Vi=p2 Vo wre the bottle was inserted Volume of air: V; 5 Sy (25) + 4 (5)? (25) V, = 4,908.74 cm* Before the tube was inserted; Absolute pressure of air inside, p; = 101.3 Volume of air inside, V; =3A Absolute pressure in air: When the tube was inserted; pr = 101.325 Absolute pressure of air inside, p2 = 101.3 + 9.81(13.6)(0.15) Absolute pressure of air inside, p) = 121.31 kPa Yhien the bottle is inserted: Volume of air inside the tube, V2 = (3 - y)A Volume of air: V2= % (15)? (25) V2 = 4,417.9 cm* "vessure in air: po = 101.325 + 9.81h [p1 Y= p2 V2] 101.3 (3 A) = 121.31 [3- yA] 3-y = 2.505 y =0.495 m [pi Vi = pr Val 101.325(4,908.74) = (101.325 + 9.81 h)(4,417 9) 101.325 + 9.81 h = 112.58 h=1.15 cm , y=h-+25= 26.15 cm From the manometer shown; Pb = Vn hy, = (9.81 x 13.6)(0.15) Pr = 20.0124 kPa.
- 40. 68 Proble temper Solutio CHAPTER TWO Principles of Hydrostatics m 2-45 A bicycle tire is inflated at sea level, where the atmospheric pressure is 10] kPaa and the temperature is 21 °C, to 445 kPa. Assuming the tire does expand, what is the gage pressure within the tire on the top of a mountal where the altitude is 6,000 m, atmospheric pressure is 47,22 kPaa, ature is 5 °C; Vi _ P2V2 T2 At sea level: Absolute pressure of air, pi = 101.3 + 445 Absolute pressure, pi = = 546.3 kPaa Volume of air, V; = V Absolute temperature of air. T; = 21 + 273 = 294 °K On the top of the mountain: [ Absolute pressure of air, p2 = 47.22 + p Since the tire did not expand, volume of air, V) = V Absolute temperature of air, T; = 5 + 273 = 278 °K PiVy _ Ne T, 546.3(V) _- (47.22 + p)V 94 . 978 47.22 + p = 516.57 p = 469.35 kPa FLUID MECHANIC & HYDRAULIC it) MECHANICS CHAPTER TWO VO RAULICS Principles of Hydrostatics 69 pplementary Problems blem 2 - 46 Weather report indicates the barometric pressure is 28.54 inches of mercury. » atmospheric pressure in pounds per square inch? -° i 4 Ans: 14.02 psi lem 2 - 47 if tube shown is filled with oil. Determine the pressure heads at B and C in ra of water. B Rare dl 2.2m Po = 05im Air Cc vy Ans: Be 2 -2.38 m x 0.6m ales Oil, s = 0.85 | Oil, s = 0.85 blem 2 - 48 » the tank shown in the figure, compute the pressure at points B, C, D, and E } ee a la, Neglect the unit weight of air. . r F Ans: pg = 4.9; pe = po = 4.9; pe = 21.64
- 41. 70 CHAPTER TWO FLUID MECHANICS UID MECHANICS ____ CHAPTER TWO 69 Principles of Hydrostatics & HYDRAULICS WORAULICS — Principles of Hydrostatics Problem 2 - 49 A glass U-tube open to the atmosphere at both ends is shown. If the U-tub contains oi] and water, determine the specific gravity of the oil Water ere Ans: Pe _ -2.38 m y 2.2m FC. 2 051m Air 2 a) Problem 2 - 50 A glass 12 cm tall filled with water is inverted. The bottom is open. What i the pressure at the closed end? Barometric pressure is 101.325 kPa. Ans: 100.15 kPa 0.6m ee Oil, s = 0.85} Problem 2 - 51 In Figure 13, in which fluid will a pressure of 700 kPa first be achieved? Ans: glycerif Oil, s = 0.85 Hoblem 2 - 48 the tank shown in the figure, compute the pressure at points B, C, D, and E 1) 1a. Neglect the unit weight of air. Po = 90 kPa ethyl alcohol Bon Ans: pe = 4.9; pc = po = 4.9; pe = 21.64 p = 773.3 kg/m? oil 3 m p = 899.6 kg/m? os water 5 p = 979 kg/m? am glycerin © - ee p = 1236 ka/m? |
- 42. 70 CHAPTER TWO FLUID MECHANIC Principles of Hydrostatics & HYDRAULIC 1D MECHANICS CHAPTER TWO 7 1 YVDRAULICS Principles of Hydrostatics Problem 2 - 49 A glass U-tube open to the atmosphere at both ends is shown. If the U-tul contains oil and water, determine the specific gravity of the oil Wblem 2 - 52 Wylindrical tank contains water at a height of 55 mm, as shown. Inside is a fall open cylindrical tank containing cleaning fluid (s.g. = 0.8) at a height ht. pressure ps = 13.4 kPa gage and pc = 13.42 kPa gage. Assume the cleaning Wil is prevented from moving to the top of the tank. Use unit weight of Wier = 9.79 KN/m®. (a) Determine the pressure p in kPa, (b) the value of Ir in _ and (c) the value of y in millimeters. Ans: 04 Ans: (a) 12.88; (b) 10.2; (c) 101 eC?) Pa Air . Water 55mm O202 Problem 2 - 50 A glass 12 cm tall filled with water is inverted. The bottom is open. What the pressure at the closed end? Barometric pressure is 101.325 kPa. Ans: 100.15 kPai "Kerosene (SD paiiiiri AS) Mercury (s.g. =-13.6) kK—_— << ———> yy a Problem 2 - 51 In Figure 13, in which fluid will a pressure of 700 kPa first be achieved? Ans: j ne Bye Problem 2-53 differential manometer shown is measuring the difference in pressure two water pipes. The indicating liquid is mercury (specific gravity = 13.6), I is 675 WM, Jin is 225 mm, and Iz is 300 mm. What is the pressure differential Tetween the two pipes. Po = 90 kPa BAe AAR ethyl alcohol 60m » = 773.3 kg/m? Ans: 89.32 kPa oil p = 899.6 kg/m? 10m water p = 979 kg/m? Sm glycerin 9 = aes ; p = 1236 kg/m?
- 43. 4 7 CHAPTER TWO FLUID MECHANI } MECHANICS - CHAPTER THREE Principles of Hydrostatics & HYDRAULI Pp M 7 3 ULICS Total Hydrostatic Force on Surfaces hapter 3 Otal Hydrostatic Force Surfaces . Problem 2 - 54 A force of 460 N is exerted on lever AB as shown. The end B is connected ‘ piston which fits into a cylinder having a diameter of 60 mm. What force acts on the larger piston, if the volume between C and D is filled with water? Ans: 15.83k 460 N | Water 220 mm AL HYDROSTATIC FORCE ON PLANE SURFACES wu 997 pressure over a plane area is uniform, as in the case of a horizontal « submerged in a liquid or a plane surface inside a gas chamber, the il drostatic force (or total pressure) is given by: | F=pA Eq.3-1 | 120 mm Cc 160 mm @ Problem 2-55 An open tube open tube ts attached to a tank as shown. If water rises to height of 800 mm in the tube, what are the pressures p, and pe of the air aboy water? Neglect capillary effects in the tube. pis the uniform pressure andA is the area. i ‘the case of an inclined or vertical plane submerged in a liquid, the total #ssure can be found by the following formula: Ans: pa = 3.92 kPa; pp = 4.90 kl , Liquid surface NL B E 100 mm E ° - 3S ; = c) 300 mm Water | ‘ Water 4 Center of gravity, cg Center of pressure, cp 3 ~ 1: Forces on an inclined plane
- 44. CHAPTER THREE 74 Consider the plane surface shown inclined at an angle ® with the horizonté To get the total force F, consider a differential element of area dA. Since th element is horizontal the pressure is uniform over this area, then; dF = pda where p= yt p=yysin6@ dF =yysin0 dA dF =ysin0@ ose From calculus, [ree = AY, F=ysin® Ay F=y(y sin®) A From the figure, 7 sin @ = |i Then, FLUID MECHANIC Total Hydrostatic Force on Surfaces & HYDRAULIC MECHANICS CHAPTER THREE AULICS Total Hydrostatic Force on Surfaces 75 ON OF F (yp): re 3 - 1, taking moment of force about S, (the intersection of the iipation of the plane area and the liquid surface), Piyp= |y4F where dF=yysinOdA F=ysin0 Ay I ysin® Ay y= Vy(yysin@dA) yain0 Ay yp =ysin® |y? dA From calculus, y? dA =Is (moment of inertia about S) AY Vp = Is is ype Eq. 3-4 [ss BeynA Eq. 34 Since yh is the unit pressure at the centroid of the plane area, Peg, the formu may also be expressed as: Lo transfer forrhula of moment of inertia: Is=1,+ AY? L ES Peg A Eq. 3-3) Eq. 3 - 2 is convenient to use if the plane is submerged ina single liquid a 1 without gage pressure at the surface of the liquid. However, if the plane j submerged under layers of different liquids or if the gage pressure at th liquid surface is not zero, Eq. 3 - 3 is easier to apply. See Problem 3 - 15 yo) ay vp ™ AY yoy? Eq.3-5 Bince Yp = Y +e, from Figure 3 - 1, then oie , T ARE g tricity, e= — Eq. 3-6 Eccentricity, e 7 q Table 3 - 1 in Page 76 for the propérties of common plane sections. —_—— ee
- 45. 76 CHAPTER THREE Total Hydrostatic Force on Surfaces FLUID MECHANI & HYDRAULI TABLE 3 ~ 1: Properties of Common plane sections Triangle Rectangle ke b/2—ke—by2—a] * %= — y =h/3 Area = Y2bh _ OW bh Sans i 12 pieauecy Ix vy | UD ° i MECHANICS Circle Quarter circle AY Area = ar? = Y% 7 12 zr ar’ _ «D4 4 64 ley Area= % rr*; x.=yc= ae 3x ar’ 16 Ie a ley = 0.0557 L=l= Semicircle Ellipse iy ! 4r Area = Yenrt; y, = — 3n Area = zab mab? ke ; mba? ley i DRAULICS Total Hydrostatic Force on Surfaces CHAPTER THREE 77 Half ellipse Quarter ellipse Area = ¥% mb ee ye 3a 3 ee port ae 8 aba? ley ; ly Area = “% ab ae 2 p= — Stat 32 mb? _ nba? lye 16 16 = 0,055ab* —Iey = 0,055ba° X= = Sector of a circle Parabolic segment Area = ¥2 r? (20) = 170 _ 2 rsin@ 3° 7A I= ae (0 - ¥ sin 20) ro y= = (0+ Yasin 20) Xe Spandrel Area = =e bhi n+1 sf n+1 i, x= b; y.= —— n+2 y 4n+2 Length of arc = r(20) = 270 rsin@ @ When @ = 90° (semicircle) 2r x= — 0 X=
- 46. } MECHANICS 7 CHAPTER THREE FLUID MECHANIC MRAULICS Total Hydrostatic Force on Surfaces & HYDRAULI ; CHAPTER THREE Total Hydrostatic Force on Surfaces 79 TOTAL HYDROSTATIC FORCE ON CURVED SURFACES CASEI: FLUID IS ABOVE THE CURVED SURFACE. F h ere: Curved surface cg of volume Vertical projection. of the curved surfaé CASE II: FLUID IS BELOW THE CURVED SURFACE Volume = V cg of volume ° Vertical projection’ Curved surface of the curved surfaet Fy Fu = prog A Eq. 3-7 ChE eVEA Eq. 3-8 Fy=yV Eq. 3-9 tan 0 = Fy/Fu Eq. 3 - 10 A © vertical projection of submerged curve (plane area) py, ™ pressure at the centroid of A Nie The procedure used in solving Fy is the same are that presented in Page 73. WSF TIT; FLUID BELOW AND ABOVE THE CURVED SURFACE cg of volume lc e oa. 7 ‘ Net vertical eee { Fy oa projection of area
- 47. CHAP 8 O TER THREE Total Hydrostatic Force on Surfaces DAMS Dams are structures that block the flow of a river, stream, or other waterwa' Some dams divert the flow of river water into a pipeline, canal, or channel Others raise the level of inland waterways to make them navigable by shi ) and barges. Many dams harness the energy of falling water to generate elect FLUID MECHANIC & HYDRAULIC! power. Dams also hold water for drinking and crop irrigation, and provi flood control. PURPOSE OF A DAM Dams are built for the following purposes: Irrigation and drinking water Power supply (hydroelectric) Navigation Flood control Multi purposes ai od st is Powerhouse To transmission tines Tail water, Reservoir Bedrock Figure 3 - 2: Section of a dam used for hydroelectric JD MECHANICS CHAPTER THREE 8 1 HYDRAULICS Total Hydrostatic Force on Surfaces (py ea <5, wy if “Equalized water 4 levels Open y 7 sluices 3 ee leaves figure 3 - 3: Boat Passing through Canal Lock. Canal locks are a series of gates designed to allow a boat or ship to pass from one level of water to another. Here, after a boat has eritered the lock and all gates are secured, the downstream sluices open and water flows through them. When the water level is equal on either side of the downstream gate, water stops flowing through the sluices; the downstream gate opens, and the boat continues on at the new water level. - TYPES OF DAMS 1. Gravity dams use only the force of gravity to resist water pressure — that is, they hold back the water by the sheer force of their weight pushing downward. To do this, gravity dams must consist of a mass so heavy that the water in a reservoir cannot push the dam downstream or tip it over. They are much thicker at the base than the top—a shape that reflects the distribution of the forces of the water against the dam. As water becomes deeper, it exerts more horizontal pressure on the dam. Gravity dams are relatively thin near the surface of the reservoir, where the water pressure is light. A thick base enables the. dam to 44withstand the more intense water pressure at the bottom of the reservoir.
- 48. JID MECHANICS CHAPTER THREE 83 CHAPTER THREE FLUID MECHANI 82 WYDRAULICS Total Hydrostatic Force on Surfaces Total Hydrostatic Force on Surfaces 4. A buttress dam consists of a wall, or face, supported by several buttresses on the downstream side. The vast majority of buttress dams are made of concrete that is reinforced with steel. Buttresses are typically spaced across the dam site every 6 to 30 m (20 to 100 ft), depending upon the size and design of the dam. Buttress dams are sometimes called hollow dams because the buttresses do not form a solid wall stretching across a river valley. Conerete Figure 3 - 4: Gravity dam _2. An embankment dam is a gravity dam formed out of loose rock, earth, or a combination of these materials. The upstream and downstream slopes of embankment dams are flatter than those of concrete gravity dams. In essence, they more closely match the natural slope of a pile of rocks or earth 3. Arch dams are concrete or masonry structures that curve upstream into a reservoir, stretching from one wall of a river canyon to the other. This design, based on the same principles as the architectural arch and vault, transfers some water pressure onto the walls of the canyon. Arch dams require a relatively narrow river canyon with solid rock walls capable of withstanding a significant amount of horizontal thrust. These dams do not need to be as massive as gravity dams because the canyon walls carry part of the pressure exerted by the reservoir Figure 3 - 7: Multiple arch dam Figure 3 - 5: Arch dam
- 49. CHAPTER THREE Total Hydrostatic Force on Surfaces 84 ANALYSIS OF GRAVITY DAM A dam is subjected to hydrostatic forces due to water which is raised on its These forces cause the dam to slide horizontally on its These tendencies are resisted by friction on the base of the dam and gravitational forces which causes a moment opposite to the overturning moment. The dam upstream side. foundation and overturn it about its downstream edge or toe. may also be prevented from sliding by keying its base. Upstream Side Downstream Side (Tailwater) Xq Headwater = wg. 1 1 Vertical i Projection of __, | | | ' ' | W2 XQ the submerged face of dam h Heel ie . if | poet lene) Me Uplift Pressure ee eA Diagram — 7? U; a ! 2a Ra Figure 3 - 8: Typical section of a gravity dam showing the possible forces acting Steps ofSolution With reference to Figure 3 - 8, for purposes of illustration, anassumption was made in the shape of the uplift pressure diagram. 1. Consider 1 unit (1 m) length of dam (perpendicular to the sketch) {l. Determine all the forces acting: FLUID MECHANICS & HYDRAULICS CHAPTER THREE FLUID MECHANICS ' Total Hydrostatic Force on Surfaces & HYDRAULICS 85 A. Vertical forces 1. Weight of the dam Wi=yeViz We= 0 Vor Was =e V3 2. Weight of water in the upstream side (if any) W, = Vs 3. Weight or permanent structures on the dam 4. Hydrostatic Uplift Uy =y7 Via Up sy Vi Horizontal Force rt 1. Total Hydrostatic Force acting at the vertical projection of the submerged portion of the dam, F=yhA Wind Pressure Wave Action Floating Bodies Earthquake Load Ao Hi Ill. Solve for the Reaction A. Vertical Reaction, R, R, = 2Fy Ry = Wi + Wa + Ws + We- Ur- Up Horizontal Reaction, R, Ry cf =F), Ry =P IV. Moment about the Toe A. Righting Moment, RM (rotation towards the upstream side) RM = W, x1 + W2 x2 + W3 x3 + We x4 B. Overturning Moment, OM (rotation towards the downstream side) OM =P y+ U1 21 + U2 22 V. Location of Ry (x)
- 50. FLUID MECHANICS CHAPTER THREE 87 86. Cree FLUID MECHANICS az & HYDRAULICS Total Hydrostatic Force on Surfaces Total Hydrostatic Force on Surfaces & HYDRAULICS r where: y = unit weight of water = 9.81 kN/mé (or 1000 kg/m?) Ye = unit weight of concrete Ye = 2.Ay (usually taken as 23.5 kN/m3) 6Rye B2 R a i] R p= (142, where e < 8/6 Eq..3 - 14 Factors of Safety Factor of safety against sliding, FS<: Note: Use (+) to get the stress at point where R, is nearest. In the diagram shown above, use (+) to get qr and (-) to get qu. A negative stress indicates compressive stress and a positive stress indicates tensile stress. uR R FSs* —4 >1 Eq. 3 - 12 x Gince soil cannot carry any tensile stress, the result of Eq. 3 - 14 is invalid if the stress is positive. This will happen if e > B/6. Should this happen, Eq. 3 - 15 will be used. - . Factor of safety against overturning, Fso: FSo= 2M 34 Eq.3-13 |e Bie OM et er where: Lt = coefficient of friction between the base of the dam and the foundation when e > B/6 Foundation Pressure . & X = a/% | Middle Third ' nese For e< B/6 Bt VBi8. ee BIB) : a/3 | | Ry = ¥a(a)(qe)(1) Ry = ¥2(3X )qe | a From combined axial and bending stress formula: PMc Wei on tie | t u =R, Heel P Toe 2R A=B(1)=B nh M y Eq. 3-15 qr Soil Pressure__/7 1D Diagram u — Ry S I S ! ! Ry, (Ry e(B/2) ff ass B? 12 ‘ Me0e B/2 :
- 51. EE CHAPTER THR 89 MECHANICS a Total Hydrostatic Force on Surfaces FLUID MECHANIC! q & HYDRAULICS 8 CHAPTER THREE & HYDRAULICS Total Hydrostatic Force on Surfaces : where: BUOYANCY . q y = unit weight of the fluid Vp = volume displaced. Volume of the body below the liquid surface ARCHIMEDES’ PRINCIPLE A principle discovered by the Greek scientist Archimedes that states that “an body immersed ina fluid is acted upon by an upward force (buoyant force) equal to # weight of the displaced fluid”. To solve problems in buoyancy, identify the forces acting and apply conditions of static equilibrium: = Fy =0 : : 2 Fy=0 This principle, also known as the law of hydrostatics, applies to both floating 5M =0 and submerged bodies, and to all fluids. lor homogeneous solid body of volume V “floating” in a homogeneous fluid at Consider the body shown in Figure 3 - 9 immersed ina fluid of unit weight . rest: The horizontal components of the force acting on the body are all in equilibrium, since the vertical projection of the body in opposite sides is the sp.gr.of body Y body Ea. 3 - 17 same. The upper face of the body is subject to a vertical downward force Vo= aa M a sp. gr. of liquid Viiquid { the body of height H has a constant horizontal cross-sectional area such as vertical cylinders, blocks, etc.: Cross-sectional area, A p= sp. gr. of body H= Y body y Fy2 BF = Fy2 — Fur _ sp.gr. of liquid Yiiquid Figure 3 - 9: Forces acting on a submerged body Fe= Fy2 - Fy = ¥(Volz) - y(Vol:) BF= y(Vole > Vol) If the body is of uniform vertical cross-sectional area A, the area submerged A, is: BE=yVp sp. gr.of body _ A Teody Eq. 3 - 16 | ef sp. gr. of liquid Yiiquid a es Eq. 3-19
- 52. FLUID MECHANICS 90 CHAPTER THREE & HYDRAULICS Total Hydrostatic Force on Surfaces STATICAL STABILITY OF FLOATING BODIES A floating body is acted upon by two equal opposing forces. These are, the body’s weight W (acting at its center of gravity) and its buoyant force BF (acting at the center of buoyancy that is located at the center of gravity of t displaced liquid) When these forces are collinear as shown in Figure 3 - 10 (a), it floats in am upright position. However, when the body tilts due to wind or wave action, the center of buoyancy shifts to its new position as shown in Figure 3 - 10 (b) and the two forces, which are no longer collinear, produces a couple equal to W(x). The body will not overturn if this couple makes the body rotate towards its original position as shown in Figure 3 - 10 (b), and will overturn if the situation is as shown in Figure 3 - 10 (c). lhe point of intersection between the axis of the body and the line of action of the buoyant force is called the metacenter. The distance from the metacente (M) to the center of gravity (G) of the body is called the metacentric height (MG). It can be seen that a body is stable if M is above G as shown in Figure 3 - 10 (b), and unstable if M is below G as shown in Figure 3 - 10 (c) If M coincides with G, the body is said to be just stable Wedge of Immersion RM Metacenter ! : M ! Wedge of emersion § Figure 3 - 10 (a): Upright position Figure 3 - 10 (b): Stable position FLUID MECHANICS 4& HYDRAULICS CHAPTER THREE 91 Total Hydrostatic Force on Surfaces Figure 3 - 10 (c): Unstable position Figure 3 - 10: Forces on a floating body RIGHTING MOMENT AND OVERTURNING MOMENT Eq.3-20 | gy RM or OM = W(x) ¥ ELEMENTS OF A FLOATING BODY: W = weight of the body BF = buoyant force (always equal to W for a floating body) G = center of gravity of the body Bo’= center of buoyancy in the upright position (centroid of the displaced liquid) Bo’ = center of buoyancy in the tilted position Vp = volume displaced al M = metacenter, the point of intersection between the line of action of the buoyant force and the axis of the body c = center of gravity of the wedges (immersion and emersion) s = horizontal distance between the cg’s of the wedges v = volume of the wedge of immersion 8 = angle of tilting MBo = distance from M to Bo GBo = distance from G to Bo zg MG = metacentric height, distance from M to G
- 53. FLUID MECHANICS 9 CHAPTER THREE & HYDRAULICS Total Hydrostatic Force on Surfaces Metacentric height, MG = MB, + GB, Eq. 3 - 21 Use (-) if Gis above Bo Use (+) if G is below Ba / Note: M Is always above B, . VALUE OF MB, lhe stability of the body depends on the amount of the righting moment which in turn is dependent on the metacentric height MG. When the body tilts, the center of buoyancy shifts to a new position (Bo’). This shifting also causes the wedge v’ to shift to a new position v. The moment due to the shifting of the buovant force BF(z) is must equal to moment due to wedge shift F(s) volume, v’ Volume, v Pitching " PLUID MECHANICS CHAPTER THREE Total Hydrostatic Force on Surfaces 93 & HYDRAULICS Moment due to shifting of BF = moment due to shifting of wedge BF (z) =F (s) BF=yVp F=yv z= MBo sin® y Vo MBo sin8 =yvs Us . = ——— Eq. 3 - 22 Mat Vp sin8 4 INITIAL VALUE OF MB, for small values of 8, (0 ~ 0 or 6 = 0): Metacenter Volume of wedge, v B/2 (B/2) tan 6 3 eee Wedge, volume = v Figure 3 - 11: Rectangular body Consider a body in the shape of a rectangular parallelepiped length L as shown in Figure 3 - 11; Volume of wedge, v = ¥2(B/2){(B/2) tan O)L Volume of wedge, v = LB? tan 0 For small values of 8, s ~ 4 B
- 54. PLUID MECHANICS ; CHAPTER THREE & HYDRAULICS Total Hydrostatic Force on Surfaces — “FOR RECTANGULAR SECTION 94 ‘CHAPTER THREE FLUID MECHANICS Total Hydrostatic Force on Surfaces & HYDRAULICS 0S MBo Vp sind + LB? tan@x2B MB, = ————_—— But for small values of 6, sin 0 = tan 0 Vp sin® 3 ro LB Vp MB, = (B/2) sec 6 But + LB? is the moment of inertia of the waterline section, | (B/2) tan 6 B/2 ays MB, = — Eq. 3-23 | (B/2) cos 8. ———+ Note: This formula can be applied to any section. B Centroid of wedge Since the metacentric height MG is dependent with MB,, the stability of a | from Eq. 3 - 22, floating body therefore depends on the moment of inertia of the waterline section. It can also be seen that the body is more stable in pitching than in MB, rolling because the moment of inertia in pitching is greater than that in rolling. tgs ~ Vpsin® Vp = BDL where L is the length perpendicular to the figure v = ¥2(B/2)[(B/2) tan OL l = + LB? #an 0 MOMENT i 4 Centroid of triangle, x The righting or overturning moment on a floating body is: Hy eG eS 3 0+(B/2)sec0 +(B/2)cos0 3 2 1+ cos* 0 = et 1 + cos0)= = : 6 6 cos8 R From geometry, le RM or OM = Wx = W(MG sin 6) Eq.3-24| y= ll R
- 55. FLUID MECHANICS CHAPTER THREE 97 CHAPTER THREE 96 FLUID MECHANICS Total Hydrostatic Force on Surfaces Total Hydrostatic Force on Surfaces & HYDRAULICS & HYDRAULICS Consider a pipe of diameter D and thickness f be subjected to a net pressure p. To determine the tangential stress in the pipe wall, let us cut a section of length s along the diameter. The forces acting on Projection of this section are the total pressure F due to ivetiatee | _ the internal pressure and this is to be . 52 4 resisted by T which is the total stress of MB. =-2 bt 60s" 0 the pipe wall. 24D cos? 6 (LLB? tano B 1+cos* 0 3 cos@ (BDL) sin ® LB* sin® MB, = 24 cos cos BDLsin®@ Applying equilibrium condition; mB, - 2 [se 24D cos? 6 ot Be epee MB, = ——~ (sec? @ + 1) but sec? 0 = 1 + tan26 a a pDs = oT fiwall B? T =Sr(sx ft) MB, ——[ (1+ tan?6) + 1] ean pDs = 2x t(sxt 24D Be 2 2 Rap 2* am) = 2, tan@ 7 D2! °3 Tangential stress, Sr = = Eq. 3-26 MB, >» Be tan? 6 | MB =| 1 + ——— Eq. 3-25 | To determine the longitudinal stress, let us cut the cylinder across its length as shown. [2Fy = 0] F=T F=pA STRESS ON THIN-WALLED PRESSURE VESSELS F=p 7D THIN-WALLED CYLINDRICAL TANK ae es . : , wall = 7 ie. . ae a fluid or gas under a pressure is subjected to tensile T=S, nDt ces, which resist bursting, developed across longitudinal and transvers tions. S sections » £D*=$,nDt D aaa Longitudinal stress, 51 = 7 Eq. 3 - 27 p = internal pressure - external pressure Eq. 3 - 28
- 56. CHAP 938 TER THREE FL Total Hydrostatic Force on Surfaces IP MECHANIG & HYDRAULICS SPHERICAL SHELL 7 Ifa spherical tank of diameter D and thickness t contains gas under a pressure of p, the stress at the wall can be expressed as: yf [ Z| | Wall stress, S = Ep 4t SPACING OF HOOPS OF A WOOD STAVE PIPE Eq. 3 - 30 where: S; = allowable tensile stress of the hoop Ai, = cross-sectional area of the hoop p = internal pressure in the pipe D = diameter of the pipe PLUID MECHANICS CHAPTER THREE 99 & HYDRAULICS Total Hydrostatic Force on Surfaces Solved Problems Problem 3 - 1 A vertical rectangular plane of height d and base b is submerged in a liquid with its top edge at the liquid surface. Determine the total force F acting on one side and its location from the liquid surface. Solution F=yhA h=d/2 A=bd F = y(d/2)(bd) F=Yayb d °* (bay(a/2) e=d/6 *# Pressure diagram Yp= h te (triangular prism) mayer ale Yp= 2d/3 Using the pressure diagram: F = Volume of pressure diagram F =%A(yd)(d)(b) = hy b The location of F is at the centroid of the pressure diagram. Note: For rectangular surface (inclined or vertical) submerged in a fluid with top- edge flushed on the liquid surface, the center of pressure from the bottom is 1/3 of its height.
- 57. FLUID MECHANICS CHAPTER THRE 100 : & HYDRAULICS Total Hydrostatic Force on Surfaces Problem 3 - 2 A vertical triangular surface of height d and horizontal base width b is submerged in a liquid with its vertex at the liquid surface. Determine the total force F acting on one side and its location from the liquid surface. i Solution Foy ha h = 2d A = Yabd Feyx2dx Vabd F= 1 ybd Y=h = 24/3 3 ibd (4 bd) (2d /3) e= e=d/12 Pressure diagram ‘ (pyramid) Yp = h tre Yp = d+d/12 = 3a/4 u Using the pressure diagram: F = Volume of pressure diagram F = 4 Abase x height F= = (bx yd)(d) = 4 yba? F is located at the centroid of the diagram, which is “% of the altitude from the base - Solution Problem 3 - 3 A vertical circular gate or radius r is submerged ina liquid with its top edged flushed on the liquid surface. Determine the magnitude and location of the total force acting on one side of the gate CHAPTER THREE PLUID MECHANICS Total Hydrostatic Force on Surfaces & HYDRAULICS 101 F=yhA F=y(r(nP) : F=nyr cg cp® 1 —OK— pL a Ay doy e= = =r/4 (nr*)(r) Wprrs@ Yp=rtr/4 : c Pressure diagram Yp = 51/4 (cylindrical wedge) Using the pressure diagram for this case is quiet complicated. With the shape shown, its volume can be computed by integration. Hence, pressure diagram is easy to use only if the area is rectangular, with one side horizontal. Problem 3 - 4 A vertical rectangular gate 1.5 m wide and 3 m high is submerged in water With its top edg@2 m below the water surface. Find the total pressure acting on one side of the gate and its location from the bottom. Solution F Z F=HyhA h =15+2=35m F = 9,81(3.5)[(1.5)(3)] F = 154.51 kN £ Le ~ HI jc i 3m cge (i AT ee ui 3 ae pe 0.214 m (1.5x 3)(3.5) y=1.5-e y=15-0.214 y = 1.286 m > e ——- y y (ee ee 1.5m
- 58. 1 02 CHAPTER THREE Total Hydrostatic Force on Surfaces | Using the pressure diagram: F = Volume of pressure diagram Fs — % 3| (1.5) 2 Fa 15.754 F = 15,75(9.81) F = 154.51 kN Pressure diagram (trapezoidal prism) Location of F: Ay = 2y(3) = 6y A2 = ¥a(3y)(3) = 4.5y A =A, + A2=10.5y [Ay = Zay] 10.5y y = 6y(1.5) + 4.5y(1) y = 1.286 m (much complicated to get than using the formula) FLUID MECHANICS & HYDRAULICS Problem 3 - 5 A vertical triangular gate with top base horizontal and 1.5 wide is 3m high. It is submerged in oil having sp. gr. of 0.82 with its top base submerged to a depth of 2m. Determine the magnitude and location of the total hydrostatic pressure acting on one side of the gate. FLUID MECHANICS CHAPTER THREE 103 4, HYDRAULICS Total Hydrostatic Force on Surfaces Bolution F=yhA h =2+ 4 (3) h =3m=¥V F = [9.81(0.82)](3)[2(1.5)(3) ] F=54.3 kN Ty, Seer Ay [4(1.5)@)]@) 0.167 m yah +e ¥p = 3.167 m from the oil surface é Problem 3'- 6 (CE Board May 1994) _A vertical rectangular plate is submerged half in oil (sp. gr. = 0.8) and half in: water such that its top edge is flushed with the oil surface. What is the ratio of the force exerted by water acting on the lower half to that by oil acting on the upper half? Solution 9 Force on upper half: Fo= Yoh A Fo = (Yw x 0.8)(d/4)[b(4/2)] ee Fo = 0.17 b d2 , Oil s = 0.80 Force on lower half: Fw = Peg? xA fo ; Pcg2 = Yo No + Yu Mw is: ie Para = (tw * 0.8)(a/2) + yl /) Oe ewes Pega = 0.65 Yo d ‘ b Fw= (0.65 tw d)(b(@/2)] a Fw = 0,325 yw b d? 0.325y,,bd7 Ratio = ——"# 0.1y ,,bd
- 59. FLUID MECHANICS CHAPTER THREE l 04 & HYDRAULICS Total Hydrostatic Force on Surfaces Problem 3 - 7 (CE Board May 1994) A vertical circular gate in a tunnel 8 m in diameter has oil (sp. gr. 0.8) on one side and air on the other side. If oil is 12 m above the invert and the air pressure is 40 kPa, where will a single support be located (above the invert of the tunnel) to hold the gate in position? Solution Oil; s = 0.8 Air; p = 40 kPa { invert Fou = You h A — Fou = (9.81 x 0.80)(8) x z (8)? Foi = 3,156 kN Ty Ay & (8)° #(8)°(8) z=4-e=35m =0.5m Paik = Pair Ay = 40 x + (8)? Fair = 2,011 KN The support must be located at point O where the moment due to Fair and Foi is zero. Since Foi > Fair, O must be below Foi. [ZMo = 0} Fou(z ~ y) = Fair(4 - y) (3,156)(3.5 - y) = 2,011(4 - y) 1.569(3.5 -y)=4-y 5.493 - 1.569y =4-y y= 262m HAPTER THREE : 105 FLUID MECHANICS : Total Hydrostatic Force on Surfaces & HYDRAULICS Problem 3 - 8 (CE Board May 1992) A closed cylindrical tank 2 m in diameter and 8 m deep with axis vertical contains 6 m deep of oil (sp. gr. = 0.8). The air above the liquid surface has a pressure of 0.8 kg/cm?. Determine the total normal force in kg acting on. the wall at its location from the bottom of the tank. Solution 2D Pac = 0.8 kg/cm? Rn Oil; s = 0.8 Fy = Pair A Pair = 0.8 kg/cm? = 8,000 kg/m? F, = 8,000(2mgx 2) = 32,0007 kg y=6+1=7m Fy = Peg A Peg = (1000 x 0.8)(3) + 8,000 Pex = 10,400 kg/m? Fy = 10,400(2n x 6) = 124,800n kg Solve for e: Fo=yYoh A 124,800n = (1000 x 0.8) ht (2n x 6) h=7y =13m 1, _ 3(2n)(6)° Ay (2nx6)(13) e = 0.23077 m yo=3-e=2.77m F=F,+ F,=156,800x kg | > Total normal force
- 60. / 1 06 CHAPTER THREE FLUID/MECHANICS HYDRAULICS Total Hydrostatic Force on Surfaces Fy = Fi yi + Foyp : (156,800) y = (32,000n)(7) + (124,800n)(2.77) -~ y = 3.63 m > Location of F from the bottom Using the pressure diagram: 8000 n(2) = 2xm | 800(6) = 4800 8000 Pressure Diagram P; = 8000(8)(2n) = 128,000n kg P2 = ¥(4,800)(6) (2) = 28,8007 kg P=P,+ P:=156,800x kg > Total normal force [Py = Pity + Po yo] (156,8007) y = (128,000n)(4) + (28,8007)(2) y=3.63m > Location of P from the bottom 6m Problem 3 - 9 In the figure shown, stop B will W.S. break if the force on it reaches 40 kN. Find the critical water depth. The length of the gate perpendicular to the sketch is 1.5m FLUID MECHANICS CHAPTER THREE 1 07 & HYDRAULICS Total Hydrostatic Force on Surfaces Solution [z Mhinge =0 ] Fz = 40(1) L= 1.5m cea hinge F=yh A=981h (1)(1.5) F=14.715h I. e= — + where AY (150)? 1 eS a = (1.5x1)h 12h I= I I> im 1 40 kN 0.5+e=0.5+ — 12h I N 14.715 h (os+—=| = 40 12h 0.5h + 0.08333 = 2.718 h =527m=h +05=5.77m > critical water depth Problem 3-10 » A vertical circular gate is submerged in a liquid so that its top edge is flushed with the liquid surface. Find the ratio of the total force acting on the lower half to that acting on the upper half. Solution co E I Ratio = 0.5756r E; Fi cg, Me k<—. 4b@b'T Yh Ag N yhy A; F, Ai = Ao Ratio = hy Ty, 1.424r 0.5756r u Ratio Ratio = 2.475