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# Fluid Mech. ALL Lecture Notes(2-16) for Exam 1 CIVILEN 3130

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CIVIL EN 3130 SPRING 2015 LECTURE 2 Contents 1 Continuum 2 2 De nition of a uid 5 3 Dimensions and units 7 Reading Sections 11 12 and 13 of the text book Homework NA CIVIL EN 3130 LECTURE 2 CONTINUUM 2 1 Continuum o Matter in both SOIid and Fluid form is made up of molecules atoms electrons protons neu trons quarks Higgs boson o In engineering applications we usually deal with pieces of matter that are very Large compared to these particles 0 Speci cally we are generally interested in studying the mechanical be havior of matter on a MlCrOSCODlC scale o This is what we call Engineering Mechanics o The problems of engineering mechanics are most precisely formulated using a mathematical or analytical basis 0 In order to do this we need to be able to de ne quantities such as density velocity etc in a mathematically tractable way 0 Speci cally quantities need to be de ned in a Pointwise manner ie by functions that are well de ned at each point within the matter 0 Considering the molecular structure of matter quantities such as den sity and velocity Are NOT well de ned pointwise To clarify this last point let s consider an example The Higgs boson often called the God Particle is an elementary particle that was theoretically proposed back in 1964 by Peter Higgs Its existence was only recently veri ed experimently in 2012 at the CERN laboratory in Geneva Switzerland resulting in a Nobel Prize for Professor Higgs in 2013 As you ll discover throughout these notes I like footnotes CIVIL EN 3130 LECTURE 2 CONTINUUM Example I Density POl l39ufsQ do f 5 ggwmc 1 W 7006 i m V o JIIT D oleLLlkr39 bel mb w Example II Velocity CIVIL EN 3130 LECTURE 2 CONTINUUM 4 0 With the so called COHtlnuum actual molecular structure of matter with a hypothetical continuous concept we replace the medium called a continuum o The continuum Completely lls space no holes or voids 0 And the continuum therefore has properties that can be described Point wise QUESTION Is the continuum concept valid ANSWER In general the answer is Yes o The spacing between molecules is generally very small compared to the scale of the problems we are considering 0 At normal pressures and temperatures molecular spacing is on the order of 1E396 and 1E397 for liquids and gases 0 The continuum concept will be valid for all the circum stances covered in this course 0 An area of uid mechanics where the continuum concept Breaks down is for rari ed gases encountered at very high altitudes o In that case the spacing between air molecules can become very Large and the continuum concept is no longer valid Will not be considered in this course o This type of uid CIVIL EN 3130 LECTURE 2 DEFINITION OF A FLUID 2 De nition of a uid 0 We generally recognize 3 states of matter SOlld Liquids and Gases It 0 The last two are both considered FlUldS 5 QUESTION What is the fundamental difference ANSWER Lack the ability to resist defor o In contrast to solids uids mation While remaining at rest 0 Because of this uids move continuously under the action of a Stress o More precisely Fluid is a substance that deforms continuously that is Any A When subjected to shear stress 739 gt 0 no matter how small that shear stress may be or phrased conversely A Fluid stress While remaining at is a substance that cannot support a non zero shear Rest CIVIL EN 3130 LECTURE 2 DEFINITION OF A FLUID 6 To Clarify and expand on this de nition let s consider the following experiment No 5 r Com13Mx Ali r I gt F 7 7 5 A EA B Substance t I lt A 2 mn new w TM M GBj HaAIL MOI CIVIL EN 3130 LECTURE 2 DIMENSIONS AND UNITS 7 3 Dimensions and units 0 In this course we will be dealing with a variety of uid characteristics 0 Therefore we need a system for describing these characteristics both Qualitativer and Quantitativer Qualitative Description Dimensions 0 The qualitative description identi es the Nature or type of the characteristic eg length time stress etc o It is most conveniently given in terms of certain Primary quantities or basic dimensions 0 For a number of problems in uids mechanics only 3 basic dimensions are required 1 Length L 2 Time T 3 Mass m 0 These primary quantities or basic dimensions can then be used to de scribe Secondary or derived quantities 0 Two examples of secondary quantities include Velocity LT Force MLTA2 0 Where the symbol is used to indicate the dimensions of the derived quantities in terms of the basic dimensions 0 Note that instead of using L T and M as the primary quantities we could L TF also use 0 Mass M would then be a derived quantity ie Fma 0 And other secondary quantities would then be expressed in terms of L T and F eg StressFL 2 CIVIL EN 3130 LECTURE 2 DIMENSIONS AND UNITS 8 Quantitative Description Units 0 The quantitative description provides a Numerical Measure of the characteristic o It requires both a number and a Standard by which various quantities can be compared 0 The most Widely used system of units in the world is the International system si o In the United States the Vs customary system similar to the British Imperial System is still Widely used SI 0 SI is an LTM system the units of the 3 basic dimensions are 1 L Meter 2 T Second 3 M Newton o The unit of the derived quantity force F is the ma o The N is de ned as the the amount of force required to give a mass of 1 kg an acceleration of 1 ms2 o Algebraically quotquot 1 m 0 CIVIL EN 3130 LECTURE 2 DIMENSIONS AND UNITS Quantitative Description Units cont d USC 0 USC is an LTF system the units of the 3 basic dimen SlOIlS are 1 L Foot 2 T Second 3 F Pound force o The unit of the derived quantity mass M is the Slug A slug is de ned as a mass that accelerates by 1 ft s2 When a force of 1 lbf is exerted on it Algebraically i 1 A basU339 IIIsa1 CIVIL EN 3130 SPRING 2015 LECTURE 3 Contents Reading Sections 15 and 16 of the text book Homework Problems 12 13 14 16 111 112 115 116 118 119 and 120 Odo Alum1K 3 agtNo F39WQL bN0439 LL n3m CuIonim MJ0L 20 W26 v lIS oooq Ms M1 112 N 5 C 377697 L10 coImEcTrso U 05 quot3 CIVIL EN 3130 LECTURE 3 VISCOSITY 2 1 Viscosity 0 From our earlier experiment we noted the following relationshi between shear stress and rate of angular deformation 7S U 7 f 1 M b 17 A 1 0 Recall that the constant of proportionality u is called the Viscosity of the uid o Viscosity is a measure of the uid s resistance to the Rate at which the uid ows 0 That is uids with a High viscosity eg molasses tar ow at a slower rate than those with a LOW viscosity eg water air 0 While Eq is a useful relationship it is of greater practical value to have this equation written in terms of a more measurable quantity of the ow such as VGIOCity rather than the angular rate of deformation 77 which is very dif cult to measure 0 To that end let s return to our previous experiment CIVIL EN 3130 LECTURE 3 VISCOSITY 3 YTq gtF FLUID U thichauif M y VOIOC39PH A dis l ribUH n Y9 7 Find Wadc 39 U 0 quot1 Aywmof In v aw 5mJ hag du lt 1 lyUU 37quot 4U Y W 11 f IO39 U o H fax 1339 dx veloci l39y39leme Pammen tqn 0 DL V d41th ooampK amp7 ARM9 op AASU M Dent if Q OH iv CIVIL EN 3130 LECTURE 3 VISCOSITY 4 0 To summarize we have the following relationship which is known as NeWtOn39S law of Viscosit ft El 3 2 Newtonian or o Fluids ar classi ed as NonNewtonian based on the relationship between applied shear stress and the resulting rate of deformation 0 When there is a Linear constant in 7 the uid is classi ed as Newtonian relationship between the two u is E 2 E a 0 When the relationship between the two is Nonlinear the quot 4 dY uid is classi ed as nonNewtonian o Gases and most common liquids tend to be NeWtOnian Some additional comments on viscosity 0 The dimensions of Viscosity in terms of F L and T can be determined directly from Newton s law of Viscosity ie F T L T FL 2 7 La A ILL 1 dudy LT L 5 o The SI unit of Viscosity is me n s o The USC unit of Viscosity is 94 o The Viscosity u is sometimes referred to as the absolute or dynamic Viscos ity to avoid confusing it with another quantity referred to as the kinematic viscosity 0 The kinematic Viscosity V is the ratio of Viscosity to mass density ie V E E 0 o Viscosity is a function of Temperature The Viscosity of a gas Increases with increasing tem perature but the Viscosity of a liquid Decreases with increasing temperature why See Figs Cl and C2 in Appendix C of the text for Viscosity values as a function of temperature for some common uids H xl CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 5 2 Mass weight and concentration variables In the previous lecture we de ned the Density of a uid as mass per unit volume PointWise this is de ned as Am p 2 11m AV gt0 A V 3 Where Am is the amount of mass contained Within AV Using density several related quantities are de ned The Volume per unit is the ie the reciprocal of the density 1 i p39 Vs m 4 Specific Volume I V 5 Mass 05 Units in SI are m3kg USC units are ft3 slug of a uid is de ned as its Specific Weight I Y Weight to density by the equation The per unit volume Thus speci c weight is related 1quot 3 Waiy v 5 lb 3 Units in SI are Nm3 USC units are The SDGClIlC Gravity I S of a uid is de ned as the ratio of the density p of the uid to the density of water speci ed temperature usually 4 C 2 392 In equation form at some s L 6 pH204 C Note this is a Dimensionless units quantity and thus has no CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 6 EXAMPLE PROBLEM The speci c weight of water at standard condi tions 4 C and atmospheric pressure is 981 kNm3 The speci c gravity of mercury is 1355 Compute a th thespeci c weight of mercury and c th ensity of merw SOLUTION IVL 5 39 qm 3 KW 17l MAIquot1l 3 5m 13 039 W twk hs39 399 I 7 Wt Mk 00 k m Iw Ooo kip13 13 SM I M DLMn33Jnl L3 I 3 77 a 11 SMYquot gt Y yw 7 Ym HS i7IKNm3gt Ym 3 7 quotNm e7 9 S the 7 139 SS000 kams Wm 11 0 quotJN i CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 7 Multicomponent mixtures For cases Where Several different forms of mass may be present in a uid we can de ne the following additional quantities The MiXture DenSity is de ned as Z Ami Z ApiAVi p 30 AV ea AV lt7 Where Ami AV and p are the mass volume and density respectively of the i th component of a mixture consisting of n total components The M833 FraCti0n of the i th component is de ned as PiAVz Ami Z 2 8 w pAV Am Note that w s 1 and is a dimensionlessquantity The Mass concentration of the i th component is de ned as Ami WAVE Ci AV AV 9 Note this quantity has the same dimensions as density The Volume Concentration of the i th component is de ned as AV AV Ci 10 Note that like mass fraction this quantity is dimensionless and c g 1 CIVIL EN 3130 LECTURE 3 TEMPERATURE AND THERMODYNAMIC VARIABLES 8 3 Temperature and thermodynamic variables SI Temperature o The Absolute temperature scale in the SI system has units of Kelvin K 0 Of more common use is the Relative temperature scale which has units of degrees Celsius 0 These absolute and relative temperature scales are related by the equa tion K 00 273 11 USC Temperature 0 The Absolute temperature scale in the USC system has units of degrees Rankine R 0 Of more common use is the RGIathe temperature scale which has units of degrees Fahrenheit F o In this case the absolute and relative temperature scales are related by the equation OR 2 OF 460 12 Conversion between degrees Fahrenheit and Celsius are given by 9 F 2 5 C 32 13 0 lt F 32 gt 14 CIVIL EN 3130 LECTURE 3 TEMPERATURE AND THERMODYNAMIC VARIABLES 9 Thermodynamic variables Your text book mentions several thermodynamic variables These are listed below QH 2 heat content cp 2 speci c heat at constant pressure CD 2 speci c heat at constant volume u intrinsic energy h enthalpy See page 16 of your text book for more details You are not expected to know the material related to these thermodynamic variables in great depth Prololm l Ll quotAir offers no resistancequot means 9 ASSUMEtun 0 SolidFL U QIZ A a 35 l L iyro all oLylyro J ELM LiNEHL CIVIL EN 3130 LECTURE 3 SOME USEFUL INFORMATION FOR THE HOMEWORK 10 4 Some useful information for the homework 41 More on the classi cation of substances 0 In the previous lecture we noted that substances can be classi ed as either Solid or Fluid based on how they respond to an applied shear stress o A Fluid was identi ed as a substance that deforms con tinuously under the action of a NONZERO shear stress no matter how small that shear stress may be 0 Again within the uid classi cation we identify two main different types of uids FLUID NEWTONIAN NONNEWTONIAN lt A linear r nship ionship between applied shear between applied shear stress and rate of deformation stress and rate of deformation CIVIL EN 3130 LECTURE 3 41 More on the classi cation of substances 11 To illustrate these relationships your text book provides the following Rheo logical diagram Rheology is the science dealing with the deformation and ow of matter Rain mlquotviieliirli39irilinIl II rill lilCElil ll iiiil Yield Elli SENSE T EillTrlz Ei k Note that in addition to Newtonian and non Newtonian uids 3 additional substances are plotted ThiXOtl OplC Substance TS Ideal Plastic IP and Ideal Fluid IF A TS is a particular type of rNewtorian uid hat shows a time dependent change in Viscosity the longer the uid undergoes shear stress the lower its Viscosity An IP is a SOIld that exhibits a linearrelation ship between applied shear stress and rate of deformation but only when the applied shear stress is greater than some de nite Bid StreSS W An IF is a Nonviscous and Incompressible uid It can be thought of as a limit quot0 ing case of a Newtonian uid with u gt 0 F r39l V O M CIVIL EN 3130 LECTURE 3 41 More on the classi cation of substances 12 HOMEWORK PROBLEM 12 Classify the substance that has the fol lowing rates of deformation and corresponding shear stresses d d rads 0 1 3 5 TkPa 15 20 30 40 SOLUTION 7 SIzPQSUJDMLJL mi Ai 099 NOT aquot mot Q V5 JUOly 0 O Sd193 l39mcL IS M n O IdemI PI I ty gtltpq J CIVIL EN 3130 LECTURE 3 42 The plate experiment revisited 13 42 The plate experiment revisited o In several of the homework problems you will encounter physical scenar ios similar to the plate experiment we discussed in the previous lecture 0 These problems do not in general explicitly state that 1 The uids are Newtonian and 2 The ow has a Linear velocituiistribution I However unless stated otherwise you are to assumw 0 Let s look at that particular case when both 1 and 2 are true yT Newtonian Fluid y0 l NLU L04lkA Fldiamp 39 t L 3 W as OOASM J Linw DN39I39r JoJHon 3quotquot A441 539 393 Slnpe aP HAL 39 30 U390 y Ay To 139 anuul 7K Assume Fi b fml lu CIVIL EN 3130 SPRING 2015 LECTURE 4 Contents Reading Sections 17 and 18 of the text book Homework Problems 144 156 and 158 AMT33 TIiQINK L XLquot5L N0 Tgtco SIT39Y IINCOMPaES 18LE CIVIL EN 3130 LECTURE 4 1 Pressure and a perfect gas Arman 0 509quot ab 11 Pressure FA Norn Ft 139 I39I PRESSURE AND A PERFECT GAS Force Fort L Consider a force acting on a triangular wedge of uid as shown above We de ne the Stress the force F per unit area A that is v F 5T ESL A Stress at a point is de ned in the sense of a limit that is AF AA Fm 574539 33 AAo acting on the sloped face of this element as 2 CIVIL EN 3130 LECTURE 4 11 Pressure 3 1 2 From equations and we can see that the DimenSionS of stress are F F d 9 quot 0 Stress 39 L A i terms of LT and F 39 m a 0 Stress 39 L T L in terms of LT and M The units of stress are 0 Pascal Pa in the SI system which is a N m2 TH 1 1 Cn v s ION 0 Common units used in the USG system are pounds per square foot pSF or pounds per square inch pSi Generally we express Stress in terms of two component stresses G The shear stress Which acts Para e or tangent to the local surface o The Greek letter Cquot is generally used to denote the shear stress component 0 The shear stress is computed by the equation Ft C A 2 The normal stress Which acts Perpendicular or normal to the local surface 0 The Greek letter 0 is generally used to denote the normal stress component 0 The normal stress is given by F n O T In uid mechanics the normal stress at a point in a uid relates to the Pressure p CIVIL EN 3130 LECTURE 4 11 Pressure Pressures can be speci ed in terms of PRESSURE 9 Absolute Pressure Fab which is the pressure rela tive t absolute zero pressure a pressure that would only occur in a perfect 0N L p 5 39aq lt2 Gage Pressure P GAME is the pressure measured relative to the local atmospheric pressure patm These two pressure measures are related by the equation fab fume farm j 3 Pressure will be discussed at greater length in Chapter 2 however before concluding the discussion here we note the following 0 Pressure p obviously has the same dimensions and units as Stress o Liquids can often sustain a considerable press1 with little or NO change in volume and thus density p N 0 That e nearly MRESR though we will comment on this at greater length in t e section on the bulk modulus of elasticity o The density p of a on the other hand will change under an applied pressure a F hp 0 This leads us to our next topic raid24ml an CIVIL EN 3130 LECTURE 4 12 A perfect gas 12 A perfect gas A PerfeCt Gas as de ned in your text book is a uid that satis es the following relationship betweenlaand density p gt at some constan absolute temperature T LfRT a 4 where R is the speci c gas constant or just gas constant Equation is referred to as the perfect or Ideal gas law R The gas constant 4 The units of R can be determined directly from equation Solving this equation for R we have R 2 P 79 l Thus N m3 1 mN I I 39 h Z o n S unltS We ave R m2 kg g 39 K i if l quot4lol oInUSCunltsR aslvqgt7 S Uq 9R I Tables C3 and Q4 in Appendix C of your text list values of R for several COIIlIIlOIl gases We note the following important distinction between Ideal Fluids and Ideal Gases 0 An ideal or perfect Gas is compressible according to equation and als o This is in sharp contrast to an ideal quUld which is incom pressible and has ZERO ViSCOSitY M CIVIL EN 3130 LECTURE 4 12 A perfect gas 6 EXERCISE Your text book lists several variations of the perfect gas law equation equation numbers 171 173 174 and 175 Derive these equations from the form of the perfect gas law given on the previous page SOLUTION 107RTL Ii39 KT 7EIL 121010 Densii39y g EV E 7 7MJ074 9 n OF ak 9 Mpmalu MOLLL C Lquot IX OF UL5Q QL CLS MHM kmsz a mz ma 1737 EVAMKT1075gt DelP z V5 Vf Z IZJUNIZiIZSSA oNS739ANT 179 mA K W 7 CIVIL EN 3130 LECTURE 4 12 A perfect gas 7 HOMEWORK PROBLEM 144 A gas at 20 C and 02 MPa abs has a volume of 40 L and a gas constant B 210 mNkgK Determine the density and mass of the gas SOLUTION 119 7 KT Ideal Gas Law D E 0 Solve for Density De a f Io RT 7 09XIOquot Nm L110 m mwaoK V DC K quE STWI a f 350394 k 5vf l Solve for Mass 61 m c V 7 m 3250hquot3gt1 aka 206 k3 T CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 8 2 Bulk modulus of elasticity As we saw in the previous section it is important to consider the following question with uids QUESTION HOW EASILY DOES THE VOLUME OF A GIVEN MASS AND THUS DENSITY CHANGE WITH A CHANGE IN PRESSURE That is HOW COMPRESSIBLE IS THE FLUID ANSWER o In the previous section we noted that liquids are nearly INCOMPRESSIBLE while gases Change in volume under compression 0 Beyond this general observation we would like to have a more Quantifiable measure of the compressibility of a uid 0 To this end a promrty that is commonly used to characterize compress ibility is th modulus of elastici or simply bulk modulus which we de ne below Bulk modulus of elasticity The bulk modulus is de ned as f JFv l 5 whers the differential change in Pressure needed to w create a differential change in VOIUme dV of some volume CV CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 9 We note the following about the BUIk MOdUIUS 0 Since the quantity dV V is DimenSionleSSQ the bulk modulus K has the same dimensions of pressure i FL 2 0 Thus SI units are the Pascal Pa 39Z 0 Common USC units are psi L 0 Large values for the bulk modulus indicate that a uid is relatively Incompressible 0 As expected values of K for commons liquids are Large ie very large changes in pressure are required to cause a signi cant change in VOIUme 0 Thus for most en ineering applications liquids can be considered as compressible 0 Tables C1 and C2 in Appendixg of your text list values of K for several common liquids 0 To gain some insight into the incompressibility of Water 7 let s consider the following example CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 10 EXAMPLE PROBLEM At standard atmospheric pressure and a temper ature of 60 F how muould be required to compress a unit volume M of water 1 V D SOLUTIOsz 1 lV 00 7 ED F TAIL e 2 K o39F 39Iooo F5 DefinitiorclLofB Ik Modulus I I lt K QLVV 7 J a if 3Hooof5 Doi j O 1954 ch H635 Fo Oo Comfr wSion D g 0 WW x w CIVIL EN 3130 SPRING 2015 LECTURE 5 Contents 1 Vapor pressure 2 Surface tension Reading Sections 19 and 110 of the text book Homework Problems 163 165 and 170 from the text book Answers to ODD Problems 163 a P m 165 lb 3 PAS 15535 m CIVIL EN 3130 LECTURE 5 VAPOR PRESSURE 2 1 Vapor pressure Vapor Pressure In order to understand the concept of we rst need to take a look at the processes of Evaporation and Condensation Evaporation E VHPO L A T 10 I o If a tainer open to the atmosphere under suitable conditions the amount of Decrease such as water or gasoline is placed in a con Liquid liquid in that container Will over time o This is a result of a portion of the Liquid converting to Vapor or a Gaseous State m the pro cess of Evaporation O Evaporation occurs because some of the liq uid molecules at the free surface of the liquid Will have enough Momentum to overcome intermolecular COheSive forces and escape into the atmosphere CIVIL EN 3130 LECTURE 5 VAPOR PRESSURE 3 CLOSEO CoNTAINErL Condensation D 9 M 3 0 mac 0 The process of Condensation is the reverse of the evap oration process that is it is the conversion of the Vapor back into the Liquid state 0 Suppose instead of an open container we now have a closed container with a small amount of open space between the Free surface of the liquid and the top of the container which we ll assume is initially a Vacuum 0 As evaporation occurs over time a greater number of vapor molecules ll the open space some of which condense back together over time to form liquid drops Getting back to vapor pressure 0 Eventually the rate at which molecules leave the liquid phase and enter the gas phase the EVaporation Rate and the rate at which the gas molecules return to liquid phase the Condensation Rate W l be equal 0 When these rates are equal we say an Equilibrium exists and both the Level of the liquid and the Amount of water vapor in the space above it remain Constant 0 When this equilibrium is reached the Vapor is said to be saturation and the pressure this vapor eX erts on the free surface of the Liquid is called the Vapor Pressure of the given liquid CIVIL EN 3130 LECTURE 5 VAPOR PRESSURE Some notes on vapor pressure and related phenomena o The vapor pressure of a liquid is a unique and characteristic property of the liquid and depends only on Temperature o This is due to the fact that vapor pressure is tied to MOIeCUIar Activity which is a function of temperature 0 Speci cally vapor pressure Increases with Increasing temperature 0 For example the vapor pressure of water is approxi mately 24 kappa at room temperature 20 C and 1013 kPa equal to standard atmospheric pressure at 100 C Boiling 0 Boiling of a liquid occurs when the Vapor Pressure of that liquid equals the sur rounding Atmospheric Pressure acting on the uid 0 Thus boiling can be induced at a xed PreSSUl39e Raising acting on the uid by the temperature and thus the vapor pressure of the uid 0 Or it can be induced at a xed uid Temperature by Lowering uid to the vapor pressure of the uid at that temperature the surrounding pressure acting on the boil at if the surrounding pres example water will Room Temperature 0 For sure is reduced to the vaDOF Pressure of water at that temperature 24 kPa 0 These relations also explain why water boils at LOwel39 temperatures at Higher elevations where atmo spheric pressure is lower Pressure Cooker Ma aiT 7T VAPo r2 PKESSUK E CIVIL EN 3130 LECTURE 5 VAPOR PRESSURE 5 Some notes on vapor pressure and related phenomena cont d Cavitation o In many situations it is possible to develop very LOW pressure in owing uid due to the Fluid39s Motion o If the Pressure is lowered to the vapor pressure of 7 the liquid Bomng will occur o This phenomenon may occur for example in ow through narrow q passages of a valVe or Pump 0 t gt 0 When this occurs vapor BUbbleS that are formed are moved by the ow into regions of Higher pressure o This will cause the sudden COIIapse of the va por bubble sometimes with such intensity as to actually cause Structural Damage to the System o The formation and subsequent collapse of vapor BUbbleS in owing uid is called CAVITATION CIVIL EN 3130 LECTURE 5 SURFACE TENSION 6 2 Surface tension Top surface Surface Tension 39 A L39 39d I I lt IqUI mo ecu es A i Z cohcsivv cquotvL Consider the volume of liquid shown in the gure above 0 For a liquid molecule in the interior of this volume COheS39Ve forces between molecules are Balanced in all directions 0 However for the layer of liquid molecules at the Free or top surface the cohesive forces from the layer of molecules below are NOT Balanced by cohesive forces from a sim ilar layer above o This results in the MOIGCUIGS of the top surface layer being pulled tightly to each other and the layer of molecules below 0 In effect this causes the surface to bW that is capable of supporting a TenSion hence the term SURFACE TENSION F 0 Speci cally surface TGHSiOH is th force per unit le that exists along a line in the surface That is L l39orLep SurfaceTension UST LenOHx 0 These forces act Normal to the line and l Tangent to the surface l o The units of surface tension are m in SI and U in USC CIVIL EN 3130 LECTURE 5 SURFACE TENSION 7 0 Surface tension is a property of the liquid and depends on Temperature and the other uid it is in contact With at the interface 0 Speci cally surface tension values Decrease as tempera Increase a m m i p T i 0 See Tables Cl and C2 of the text book for surface tension values as a function of temperature in contact With Air 0 Although surface tension forces are Negligible in many engineering problems they may be dominate in some M 0 Some examples Where surface tension plays a predominate role in uid mechanics phenomena include CAPll l ARY RISE of liquids in narrow tubes see Fig ure 16 of your text book The mechanics of Bubble Formation Movement of liquid through SO and other porous me dia Flow of Thin Films The formation of Liquid Drops 0 Let s take a look at an example With respect to this last item CIVIL EN 3130 LECTURE 5 SURFACE TENSION 8 EXAMPLE PROBLEM Calculate the pressure p relative to the local atmospheric pressure in a spherical liquid drop of radius R in terms of the surface tension ast 5 of the liquid ANALYSIS Pressure Force R Surface Tension lt J CIVIL EN 3130 SPRING 2015 LECTURE 6 Contents 1 Chapter summary 2 Reading Chapter 1 of the text book Homework NA CIVIL EN 3130 LECTURE 6 CHAPTER SUMMARY 2 1 Chapter summary We conclude this chapter by highlighting the major points we covered 9 Continuum J Considering the MOIGCUIar StrUCture of matter physical uid characteristics such as density are not well de ned at every point J Thus in order to formulate a mathematical basis for uid mechanics it is necessary to replace the actual molecular structure of uid with a hypothetical continuous medium called a J The continuum has NO h0es or VOidS and therefore has properties that can be described by continuous functions 2 De nition of a uid J A uid is de ned as a substance that deforms continuously Or FIOWS when subjected to a non zero shear stress T no matter how small that shear stress may be J The Rate of Deformation of the uid an be ob Dl 5 served to be directly proportional to the applied shear stress 739 ie 739 oc i J When the relationship between applied shear stress and rate of defor mation is Linear we call the uid aqgewtonian uid In non Newtonian uid there is aelationship between applied shear stress an rate of deformation Assume Newtonian Linear vel dudy CIVIL EN 3130 LECTURE 6 CHAPTER SUMMARY 3 Dimensions and units J In dealing with uid characteristics it is necessary to develop a System by which characteristics can be described both qualitatively and guantitatively S l Qualitativer in terms of primary quantities or basic dimensions LTM or LTF1 In turn these primary quantities are used to describe secondary quantities eg area stress velocity etc J These characteristics are described DSC J The Quantitative description consists of both a num ber and a unit by which various quantities can be compared Two com mon systems of units were discussed the International System SI and M the US Customary System USC Q1 Viscosity J From a simple plate experiment NGWtOH39S Law Of Viscosity t7 ududy lcan be observed where the the constant of proportionality u is termed the dynamic or absolute viscosity of the uid Kinematic J The viscosity of a uid is a measure of the uids resistance to the rate at which it ows Fluids with High viscosity ow at slower rates than those Wlth lOWGI39 VISCOSItIGS When subjected to the same shear stress J To arrive at Newton s law of viscosity it was shown that the an gular rate of deformation 7 of the uid is equal to the vertical Div Mass weight and concentration variables J For Simple Fluids Floats in of one molecular structure threejv te quantities can be de ned that are related to the density of the uid Sinks in water the W the speci c weight 7 2 pg and the speci c S 7 t 39t graVIyS ppggg a IVAHa J For Multicomponents mixtures we can de ne the quantities of mixture density mass fraction mass concentration and vol ume concentration m 2m l quoti CIVIL EN 3130 LECTURE 6 CHAPTER SUMMARY 4 Temperature and thermodynamic variable J Units of AbSOIUte temperature scales in the SI and USC systems are Kelvin gKg and degrees Rankine OR respectively J More common are the Relative temperature scales of de grees Celsius PC and degrees Fahrenheit OF Which are related through m the equationlon 9g5OC l 32l Pressure and a perfect gas uid is component the Two different mea J The pressure p in a related to Normal Stress sures of pressure are generally used absolute pressure pabs and gage pressure pgage J A perfect or Ideal Gas is a gas that satis es the perfect gas law relatin ssure p and density pzl p pRT iNher the speci c gas constant a d is the absolute temperature Bulk modulus of elasticity J The bulk modulus of elasticity K or simply bulk modulus is used as a measure of the compreSSibility of uids A lar e value of bulk modulus indicates that a uid is relatively J The bulk modulus is de ned as IE dp dV V kwhere dp is the differ Pressure ential change in needed to create a differential b change in volume dV of a volume V J Most common liquids have very Large bulk moduli and ca n lerefore be considered to bampincompressible gt Fab PATquot P6A61 VA c UUM CIVIL EN 3130 LECTURE 6 CHAPTER SUMMARY 5 Vapor pressure J Vapor pressure is de ned as the pressure exerted by a Vapor on its liquid counterpart when the two are in equilibrium and is a function of temperature only J Bomng of a liquid occurs when the vapor pressure of that liquid equals the surrounding atmospheric pressure acting on the uid J CaVitation is the formation and subsequent collapse of vapor bubbles in owing uid In severe cases it can cause structural damage to hydraulic systems Pump System F Surface Tension J Surface tension is a property of a liquid that causes the surface to act as a membrane that can support a ten sile force Speci cally it is a force per unit lengih 0 l J Phenomena in which surface tension is an important factor include such things as liquid drop formation capillary Rise and the ow of thin lms Bubble formation Homework Problem 163 CAVITATION 3 TABLE all 97 025 gQOL WOG CIVIL EN 3130 SPRING 2015 LECTURE 7 Contents Reading Section 21 of the text book Homework NA CIVIL EN 3130 LECTURE 7 INTRODUCTION TO FLUID STATICS 2 1 Introduction to uid statics 0 Just as in your study of solid mechanics we will begin our study of uid mechanics by considering uid Stat iCS 0 That is considering Newton s second law ZFMQ9 lt1 a we will speci cally consider the case where acceleration advdt0 in this chapter 0 This occurs when the uid is at Res39t or the entire uid body is moving with a constant velocity 0 In both cases there will be no Shea Iquot stresses in the uid and only forces due to normal stresses or pressure will be acting on the uid Maes 0 Thus our primary objective in this chapter is to formulatzrelationships for Pressure and its variation throughout a uid in the absence of shear stresses o This will allow us to calculate for example pressure forces exerted on Surfaces submerged such as dams and tank walls o In the nal section of this chapter we will consider the special case of a uid accelerating as a Rigid bOdy or in relative equilib rium where there are no shear stresses present 0 Although Are not strictly speaking rigid body motion problems uid statics problems they are covered in this chapter because the pressure relationships and the analysis of such problems are similar to those for uids At FESt CIVIL EN 3130 LECTURE 7 FORCE STRESS AND PRESSURE AT A POINT 2 Force stress and pressure at a point Illd Consider the pressure p at some arbitrary point located within a uid mass at rest The static pressure at a point is de ned as SEA AA A question that immediately arises is the following 3 How does the pressure p at a point vary with the orientation CT the pLane paSSihg through that A J pUiIIL QUESTION ANSWER To answer this question we consider the F Fee BOdY Diag ram on the following page 7 CIVIL EN 3130 LECTURE 7 IMQ 331 Fquot Fssin50 gtFA WW 52 womanma o 2F 0 F2 F5lt B W O PEWCCX Sy Cosbt Y39KX SQL Zgt 3EI GW J W S gs ms l PYgXJZFSSXSZ O FORCE STRESS AND PRESSURE AT A POINT 4 W 52 g 5w ngicyypgz gyz Gnul Terra ya 19yquot P 0 I 50W Slag 1 2quot 105 X 1 0 P2 9 RIP P5Py J CIVIL EN 3130 LECTURE 7 FORCE STRESS AND PRESSURE AT A POINT 5 CIVIL EN 3130 LECTURE 7 FORCE STRESS AND PRESSURE AT A POINT 6 CIVIL EN 3130 LECTURE 7 FORCE STRESS AND PRESSURE AT A POINT 7 This important result is known as PascaL I 5 LaW named in honor of Blaise Pasca1 1 We record this result below Pascal s law The pressure p at a point in a uid at rest is I ndependent of direction or equivalently we have The pressure p at a point in a uid at rest is the Same all directions Notes 0 This result also holds for uids in MOtion if there are no v shear stresses consider the exercise on the following page o If there are shear stresses present in the uid the normal stresses at a point are not necessarily The same in all directions x o In such cases we de ne the pressure as the Average of any three mutually perpendicular Normal stresses at that point see equation 213 of the text book p I Fz fPYTPle Shuxf Sireg 1Blaise Pascal June 19 1623 August 19 1662 was a French mathematician physicist and religious philosopher who made several important contributions to the eld of uid mechanics The SI unit of pressure is named in his honor Pascal s most in uential theological work the Pens es Thoughts which was not completed before his death at the young age of 39 is widely considered to be a masterpiece and includes the famous argument known as Pascal s Wager CIVIL EN 3130 LECTURE 7 FORCE STRESS AND PRESSURE AT A POINT 8 EXERCISE Extend the results of the previous analysis which demonstrated that the pressure at a point is independent of direction for a uid at rest to the case of a uid in motion that hasQEShear stresses ANALYSIS IF THERE ARE No SHEAR STRESSES THEN THE ONLY CHANGE To THE ANALYSIS IS THE ADDITION OF A NON ZERO RIGHT HAND SIDE IE 57 ZF may IT ZFZT maxLL WHERE 1y ANDQ Z ARE THE ACCELERATIONS IN THE y AND 2 DIRECTIONS RESPECTIVELY NEXT THE MASS CAN BE WRITTEN IN TERMS OF THE DENSITY 0 AND THE VOLUME V OF THE TRIANGULAR WEDGE OF FLUID LE 1 n CIVIL EN 3130 LECTURE 7 FORCE STRESS AND PRESSURE AT A POINT 9 FOLLOWING THE SAME LINES AS BEFORE AFTER COMMON TERMS ARE CANCELED WE HAVE RPI graced Pquot AZTSEEQQY p lt Fz Ps Eg zy z ii w AGAIN AS WE SHRINK TO A POINT 62 gt 0 WE HAVE Py GJTD P2 OLQ A 09 pAAf in Q QIUAd In MO Hoq j inCcho lbLm4quot CIVIL EN 3130 SPRING 2015 LECTURE 8 Contents Reading Section 22 of the text book Homework Problems 28 215 and 217 QUV I O SHEAtzVIsw31T7 C 2 3 KM SurCCLLL TLA SIOq lt7e CIVIL EN 3130 LECTURE 8 BASIC EQUATIONS OF FLUID STATICS 2 1 Basic equations of uid statics o In the previous lecture we addressed the question of how pressure at a point varies with D i FECt ion 0 Speci cally we showed that the P res S U re at a point is in dependent of direction for a uid with no Shea Iquot St res S o This result is known as Pascals LaW 0 However we now want to ask an important follow up question QUESTION How does the pressure vary from point to point ANSWER To answer this question we will again consider a small element of uid taken from some arbitrary position within the uid mass The F0 rce 5 acting on the element consist of two types 0 Body Forces A force that acts over the V01ume of a body or a force acting throughout the a mass of a body The most common body force and the only one we will consider here is the G ravity Force or weight of the ele ment W 0 SU FfaCe FOFCES A force that acts across an internal or external Su rface element in a body The only surface force considered here is due to Pressure re member we are looking at the speci c case where there if no shear stress CIVIL EN 3130 LECTURE 8 BASIC EQUATIONS OF FLUID STATICS 3 F053 5W V L 81962 EEW5 i I 62 F A 81963 ii A iygV ay366 8y2 7032 I 1 613 Vzriy39aa 2 m o The pressures on each face are expressed as Taylo Iquot Se ries expansions about the center point of the differential element 0 Recall that the Taylor series expansion of a function f about a point ais given by Oquot PNO P03 l l W M f ltagtltc a f 0 o The expressions for pressure in the gure above retain the rst TWO terms of the Taylor series expansion 0 We proceed in two steps First we Will look at the Taylor series expansions for the pressure Second making use of these expressions we Will write down the sum 01 forces in the 3 y and 2 directions The concept of a Taylor series was discovered by the Scottish mathematician James Gregory Go gure CIVIL EN 3130 LECTURE 8 BASIC EQUATIONS OF FLUID STATICS Q 1 7 k 1L L v imam TE 0L MP W 91 lt17 A J P F ffy39a V w i 2 EUM Mora3 S IZSFYSMQY lt LiL d7 2 l7 ydd l7 x 5may W W Lw q 52 9H r v j gt 3ng 1 gm Q7 D3 V 5v 3 D amp 9 731 XSYJE qux X CIVIL EN 3130 LECTURE 8 BASIC EQUATIONS OF FLUID STATICS 5 239 if 5 J2 quot nggyg t figMO lit 7 W J DEFINET SF JJFx HIFyg rIFz 32 QQx ay3QzQ Equot SMR V A1 dP A1 0 A W ozx a J T1 k 539 C X 1 CXJyJE DYMOL CIVIL EN 3130 LECTURE 8 BASIC EQUATIONS OF FLUID STATICS Cancel like terms PlEQEf 1 1amp7 A 0 F7 d Gradient of p Jx 139 3 T 3 f 0 an EVfX zf CIVIL EN 3130 LECTURE 8 BASIC EQUATIONS OF FLUID STATICS 7 CIVIL EN 3130 LECTURE 8 11 Pressure variation in a static uid 8 In summary the general equation of motion for a uid in which there are NO shear stresses is Vp Ykl9 1 A In the next subsection we look at how this equation simpli es for uids at rest 11 Pressure variation in a static uid 0 For a uid at RESt we have a 0 Thus Eq simpli es to Vp 7k0 quotVF YR J l quot Or in component form 8190 8190 3p 8 3 5 These equations show us that the P FESSU re dWm and y and therefore the pressure does not change from point to point as we move in the Ho r izontal It only changes as we move in the z direc tion in the ve rt lcal plane Sincel p depends on 2 only we can write the component equation for 2 above as an ordinary differential equation No acceleration d9 E 39Y 7 2 Equation T is the fundamental equation for uids at RESt CIVIL EN 3130 LECTURE 8 11 Pressure variation in a static uid 9 We note the following with respect to equation 111 o It can be used to determine how pressure changes with Elevation 0 The sign in front of 7 which is always positive indicates the Pressure gradient yl ggg galgl g bg 0 That is pressure decreases as we move Up in a uid at rest 0 The equation is valid for both con 5t ant values of speci c weight 7 subsection 111 of these notes as well as those that may have a z dependence subsection 112 of these notes r Pressure variation in a static incompressible uid 0 For an Incompressible uid there will be no change in volume and thus density when the uid mass is subjected to a compres sion force 0 Thus neglecting the small variations of g with elevation the speci c weight 7 pg of the uid is constant 0 That means equation T from the previous subsection can be di rectly integrated to obtain an expression for p as a function of Depth 0 We will consider this along with the case of a compressible uid CIVIL EN 3130 LECTURE 8 11 Pressure variation in a static uid 10 112 Pressure variation in a static compressible uid 0 When the compressible uid is a Pe rfect gas recall that we have the following relation the ideal gas law between pressure p and density p PIfRTl quotgt KT 0 Solving this equation for p and substituting the result into equation met 9i d1 RT 0 We consider the integration of this expression along with the case of an incompressible uid on the next two pages CIVIL EN 3130 LECTURE 8 11 11 Pressure variation in a static uid CASE 1 Incompressible Fluid w 2 Not a function CIVIL EN 3130 LECTURE 8 11 Pressure variation in a static uid 12 For pressure p at some depth h below the free surface pp ATM k Incompressible CASE 2 Compressible 3 SP d1 RT X f separate variables integrate assume gR T are constant PL 1339 5 1 amp l P 1739 79I P EL 7 39ZTZ31 2 Solve for p2 Compressible fluids 31 1 n deal gas Pa r ex LT CIVIL EN 3130 SPRING 2015 LECTURES 9 Contents 1 Units and scales of pressure measurement 2 2 Manometers 5 21 Standard manometers 6 22 Micromanometers 8 Reading Sections 23 and 24 of the text book Homework Problems 227 232 233 and 234 from the text book 3view from L8 i quot 7 d3 a Specific weight Incompressible Fluid Pl Yki 17 I P K fa 91 LKffg CIVIL EN 3130 LECTURE 9 UNITS AND SCALES OF PRESSURE MEASUREMENT 2 1 Units and scales of pressure measurement As we noted earlier pressure measurements are designated as either Absomte pressure or Gage pressure 0 AbSOIUte pressure is measured relative to a perfect vac uum absolute zero pressure b amp Panes 1 java l o Gage pressure is measured relative to the local atmospheric pressure 0 Absolute pressures are always Pos it ive o Gage pressures can be either positive l or negative depend ing on whether the pressure being measured is AbOVe or BEIOW the local atmospheric pressure 0 A negative gage pressure is referred to as a SUCtion or vacuum pressure 0 In your text book pressures given are gage unless speci cally marked absolute p5 ia o The concepts of absolute and gage pressure are illustrated in the gure below 4 Positive gage p Gage pressure 1 Local atmospheric pressure reference Pressure l JG age pressure 2 Nega39lt ive Gage P M ll Absolute pressure 1 Absolute pressure 2 Absolute zero pressure perfect vacuum CIVIL EN 3130 LECTURE 9 UNITS AND SCALES OF PRESSURE MEASUREMENT 3 Pvapor Patm lllll llll Mercury Mercury barometers o Atmospheric pressure is frequently measured using a Mercury Barometer 9 o In its simplest form a mercury barometer consists of a glass tube closed at one end with the other open end immersed in a container of mercury as shown in the gure above 0 The tube is initially lled with MerCU ry open end up and then turned upside down into the container o Eventually an Equilib rium will be reached where the weight of the mercury plus the vapor pressure force are balanced by the force due to atmospheric pressure 0 At this stage the pressure equation for the barometer can be written as fkun ygj39 g ff where 7H9 is the speci c weight of mercury o The vapor pressure for mercury is very small so we can write the approx imate relationship F K 7004704 39 3 i r The mercury barometer was invented in 1644 by the Italian physicist and mathematician Evangelista Torricelli October 15 1608 October 25 1647 CIVIL EN 3130 LECTURE 9 UNITS AND SCALES OF PRESSURE MEASUREMENT 4 0 With the use of a barometer pressure is frequently given in terms of the Height of the column of mercury 0 That is solving the pressure equation on the previous page for h we have h patm 7H9 0 Using the appropriate values we have P 3135 5 f LtM Ol3973 Iquot ang VH3 quot39 r ll tam mm HJ o This is a frequently used measure of atmospheric pressure EXERCISE What is the minimum length of tube that would be required for a water barometer to measure atmospheric pressure SOLUTION lo 291er YHa96Qx L lH39g Pedal rh 739 H Hlo 750mm tgqqr 7i 713 CIVIL EN 3130 LECTURE 9 MANOMETERS 5 2 Manometers 0 As discussed in the previous section pressure is often measured in terms of the Height of a column of liquid in a tube 0 Pressure measuring devices that employ this technique are called Manometers o The Mercury Ba rome39ter discussed in the previous lecture is one type of manometer however there are many other types of manometers used 0 Three common types of standard ManomEte r are the a Piezometer b V tube manometer and C Incline tube manometer o The rst two types of these manometers are described in detail in the following pages CIVIL EN 3130 LECTURE 9 21 Standard manometers 6 21 Standard manometers a Piezometer tube 0 I VE IV Y 7 13 A lit 139 0 The piezometer tube is the simplest type of Manomete r o It consists of a VG rtical tUbe open at the top and connected to a container in which the pressure is desired see Figure above and Figure 210a of your text book 0 Liquid rises in the tube until an Equilibrium is reached at which point in time the pressure in the container is given by PA I39M Ca6 E J where y is the speci c weight of the liquid in the container 0 Although the piezometer tube is a simple and accurate device it has several disadvantages if the in the container is than the atmospheric pressure other 1 It is only suitable Greater pressure wise air would be sucked into the system 2 The pressure to be measured must be Re LatiVe Ly small able The uid in t iner in which the pressure is to be measured must be a Liquid rather than a Gas V so the required height of the column is reason CIVIL EN 3130 LECTURE 9 21 Standard manometers b U tube manometer o The uid in the manometer is called the start at lqu 0 To overcome the dif culties noted with the piezometer tube another type of manometer that is widely used consists of a tube formed into the shape of a U as shown above also see Figures 210 b and 210 c of your text book Gage Fluid 0 To nd the pressure p A in the terms of the various column heights you Point A and work your around to the open end To begin we note the pressures at points A and 1 are the same Moving from point 1 to point 2 the pressure will increase by an amount I Next we note the pressure at Point 2 is equal to the pressure at P0 int 3 Finally as we move up from Point 3 to the free sur face or meniscus of the gage uid the pressure decreases by an amount In equation form these various steps can be expressed as nix 0 CIVIL EN 3130 LECTURE 9 22 Micromanometers 8 b U tube manometer cont d o A major advantage of the U tube manometer lies in the fact that the Gage Flu 1d can be different than the uid in the container in which the pressure is to be determined o For example the uid in A can be a Liquid or a Gas o If A does contain a gas the contribution of the gas column to the pressure is almost always negligible so that VA 2 R9 o If the pressure p A is large than a Dense gage uid such as mercury can be used so that a reasonable column height can be main tained o The U tube manometer is also widely used to measure the D if fe ren ce in pressure between two containers in a given system 0 This type of manometer is called a U tube manometer see Figures 211a and b of your text Your text book outlines a general solution procedure for manometer problems see page 50 We consider an example on the following page 22 Micromanometers We will not be too concerned with micromanometers however please read pages 52 and 53 of your text book for general information CIVIL EN 3130 LECTURE 9 22 Micromanometers 9 HOMEWORK PROBLEM 227 In Fig 244 A contains water and the manometer uid has a speci c gravity of 294 When the left meniscus is at zero on the scale p A 100 mm H20 Find the reading of the right meniscus for p A 8 kPa With no adjustment of the U tube or scale SOLUTION CIVIL EN 3130 LECTURE 9 22 Micromanometers 10 CIVIL EN 3130 LECTURE 9 22 Micromanometers 11 CIVIL EN 3130 SPRING 2015 LECTURE 12 Contents Reading Section 26 of the text book Homework Problems 284 289 and 299 from the text book CIVIL EN 3130 LECTURE 12 FORCE COMPONENTS ON CURVED SURFACES 1 Force components on curved surfaces 0 When the submerged surface is curved rather than planar as shown in the gure above the differential pressure forces dF on the surface vary in Direction 0 At any particular point on the curved surface the associated differ ential force dF may have both Horizonal de and Vertical dFy force components 0 Thus to nd the Magnitude of the resultant force we must rst sum integrate the horizontal and vertical differential force components separately 0 We denote the total sum of these horizontal and vertical component forces on the curved surface by FX and FX respectively 0 These component forces can then be used to nd the Magnitude of the total resultant force via the stan dard equation F 1 W W 0 Let s look at how we nd these component forces beginning with the horizontal force Fm CIVIL EN 3130 LECTURE 12 11 Horizontal force component 11 Horizontal force component 3 L i dAx dA cos6 V 9 yravrzc Hgy Referring to the gure above the component of the differential force in the c direction is given by ampFx9 JFLCMB 3 use Summing integrating these components over the entire surface gives Fx A remade The product dA cos6 E dAm is the horizontal projection of the the dif ferential area dA and the pressure p equation above as l so we can write the Fx 4 4 73le I qx A The integral expression in the above equation is equal to the product ngx Am Where gm is the y coordinate of the centroid of the horizontally projected area CIVIL EN 3130 LECTURE 12 11 Horizontal force component Thus we can calculate the horizontal component of force on a curved area by calculating the force on the horizontal PPOJeCtlon of the area that is Fquot t KX Ax 1 Where 561 2 321 is the depth of the uid at the centroid of the horizontal pro d jection Am f J39 In a similar way it can be shown that the line of action of the horizontal force component passes through the Pressure center of the projected area Ax CIVIL EN 3130 LECTURE 12 12 Vertical force component 12 Vertical force component i dAy dA cos L Referring to the gure above the component of the differential force in the y direction is given by JFcos L05 Again summing integrating these components over the entire surface gives h P as The product dA cos E dA is the vertical projection of the the differential area dA and the pressur so the equation above can be written as I F r IdAV JV l 39 In 39 39 I v Where we have de ned 395 l cl A of a prism of height y With a base39 of area dAy Which is equal to the volume CIVIL EN 3130 LECTURE 12 12 Vertical force component Z T S 8 Liquid B 73 hi 2 12 O I Vl 7A Liquid A 7A Thus the vertical force component is given by W Fy YV39 2 where V is the T0131 VOIDme between the curved sur face and the free surface Note that this expression is equivalent to the Weight of the liquid in the Volume V39 A special case to consider is when the liquid is BGIOW the curved surface and the pressure is known at some point say 0 see gure above In that case we use the following procedure 1 An imaginary or equivalent free surface s s is constructed some height h 1 Fp Y above 0 where p0 is the pressure at point 0 and y is the speci c weight of the liquid in contact with the curved surface 2 The vertical component of the pressure force on the curved surface is then the Weight of the imaginary volume vertically above the curved surface see page 68 of your text book Finally it can be easily shown see page 69 of your text book that the line of action of the vertical force component passes through the CentrOid of the volume V real or imaginary CIVIL EN 3130 SPRING 2015 LECTURE 11 Contents Reading Section 25 of the text book Homework Problems 238 247 265 266 and 272 from the text book Review B j Atmospheric Pressure Pressure measurements Gage L Absolute gt Vacuum 1 Piezometer I p 2 U tube manometer LJEEE 3 Incline tube Manometers CIVIL EN 3130 LECTURE 11 FORCES ON PLANE AREAS 2 1 Forces on plane areas 0 The calculation of forces on submerged Surfaces due to uid pressure is an essential part of the design process for dams storage tanks ships etc 0 From our previous lectures we can already deduce a cou ple things regarding these forces and pressures for static uids 1 The forces acting on the submerged surfaces Will be Perpendicular to them since there are no shear stresses 2 For incompressible uids the pressure on the submerged surfaces will vary Linearly With depth p 2 7h 0 Building on this information the goal of this section is to develop techniques for determining the DiI GCtiOI l Location and Magnitude of resultant 7 forces acting on submerged surfaces 0 In some cases these quantities may be the end result that we seek in others they may simply be a means to an end 0 Let s begin by looking at the simple case of the resultant force on a horizontal submerged surface CIVIL EN 3130 LECTURE 11 11 Horizontal surfaces 3 11 Horizontal surfaces 0994 ATM L l if 3 Speci c weight y 1 F h HHHHHHJ L Consider a cross section of a tank of liquid as shown above We know that the uniform pressure along the bottom of the tank is simply RYH L39 J Over some small differential area dA of the base of the tank the corresponding olA EF AQ EEEdF and the total resultant force acting on the base of the tank is the sum integral of differential force dF is simply these differential forces that is F 1 P 1 A where A is the total area of the base of the tank For this case note that the pressure p 2 7h is constant and can therefore be pulled out of the integral so the resultant force computed by equation 7 is simply Resultant E The next question we want to ask is where this resultant force acts ie where it s line of action is located We will denote the coordinates of this point often called the center of pressure or pressure center by CIVIL EN 3130 LECTURE 11 11 Horizontal surfaces 4 I a Rectangular tank base b Consider the case when the base of the tank is a rectangle as shown in Figure a above For this case intuitively we know that the resultant force will act in the center of the rectangle Formally we can show this by equating the moments of t e resultant force and the moments of the distributed pressure force about the y and M iii336 s respectively Pressilire center T A A 2 8 Cl F A PX A I F A 9 y 2 l X As we noted above the uniform pressure p in this case can be removed from the integrals and dividing the equations by F we have the following expressions for 33p and yp yPAXdAI lSAIdA J where we have used the fact thatF 1A You may recognize the integral eX pressions in the equations above as the de nitions of the coordinates of the centroid of the area A that is I Xr PTCAXOM V Till ydA Thus for horizontal surfaces the line of action of the resultant or pressure center is located at the centroid of the area A ie 33p 2 5 yp g This is true not only for the simple case of a rectangular horizontal surface as considered above but also for some arbitrary shape as shown above in Figure Horizontal Surfaces 7 PA I X97 X 70 739 CIVIL EN 3130 LECTURE 11 12 Inclined surfaces 5 12 Inclined surfaces v ATM o Flu l As in the case of the horizontal surface let s begin by looking at the pressure along the inclined surface Referring to the gure above at some point 33 y the pressure PY TYsm zgt dfgt ltgt Inserting this expression into our Equation 1 for the resultant force we have F Ygiq 9 SA YdA39 is given by Note that in this case the full expression for the pressure cannot be pulled out of the integral because of its y dependence CM 2 dasdy However 7 and sin 9 both constant have been removed You may note that the remaining integral expression is equal to the product QA Thus the resultant force in this case is given by F YhA vi 7R 3 whers the depth of the uid at the centroid 5 1 of area A CIVIL EN 3130 LECTURE 11 13 Center of pressure 6 13 Center of pressure To determine the line of action of the resultant force or the pressure center for this more general case we again equate the moments of the resultant lc e and the mo mWWrce about the 31 and az axes see Equation 2 Solving for the pressure center coordinates 33 and yp in this case we have l L XP JA Pxom 1 7 FJAFYJA Again note that in this case unlike the case for the horizontal surface the pressure cannot removed from the integral because of its y dependence Substituting our ex pressions for p and F from the previous page into the equation for yp we have 1 1 39 9 dA 39 9 2dA mp ygA sin 6 Liz11 my 7 yp ygA sin 6 A 7311 y Canceling out 7 and sin6 we have M Wadi th 9A Yf VA39 In this case the integral expression in the equation for mp is called the product of inertia with respect to the a and y axes that is IXY SA Ya The integral expression in the equation for yp is called the second moment of the area or moment of inertia with respect to the m axis that is Thus for inclined surfaces the coordinates of the pressure center are given by Ixy IX CIVIL EN 3130 LECTURE 11 13 Center of pressure 7 These expressions can be written in terms of the product of inertia and the second moment of area about the centroidal axes denoted 363 and 36 respectively by using the respective parallel axis theorems IX7 jiny A279 IX le ry Q See appendix A of your text book for more information on the centroid second mo ment of area product of inertia and parallel axis theorem Using the above expres sions the coordinates of the pressure center for an inclined surface can be written as A I 2 2 lt4gt Xe 9A IVr W V q Note that for the case of an inclined surface the pressure center is not located at the centroid of the submerged area A Howeveris interesting to note that if f 7 the submerged area is symmetric With respect to either of the centroidal axes then f ce my 9that case CIVIL EN 3130 LECTURE 11 14 The pressure prism 8 Lecture 11 Continued 2915 14 The pressure prism M Col a b For submerged surfaces of relatively simple shape such as a rectangle a useful graphical approach can be used in lieu of the equations derived in the previous sec tions to determine the magnitude and location center of pressure of the resultant force acting on the submerged surface To illustrate this approach consider the situation depicted in the gure above where we wish to determine the resultant force acting on the side of the tank Using Equation 3 of section 12 the resultant force can be computed as where we have used the fact that ft y sin 90 2 g h 2 for this case The pressure distribution shown in gure a runs across the entire width w of the tank as shown in gure b and the magnitude of the resultant force is equal to the volume of this shape the so called pressure prism that is where wh is the area A of the side of the tank CIVIL EN 3130 LECTURE 11 14 The pressure prism 9 l a b The resultant force acts through the centroid of the pressure prism not the cen troid of area A For the volume of gure b of the previous page this occurs at a distance h 3 from the bottom of the tank since the centroid of a triangle is located a distance h 3 above its base It can easily be shown that this result is consistent with that obtained from Equations 4 derived in the previous section The pressure prism approach can also be used for cases where the submerged surface does not extend all the way to the free surface and or the submerged surface is inclined at some arbitrary angle 6 as illustrated in gures a and b above In both cases shown above the pressure prisms are trapezoidal instead of triangular but the magnitude of the resultant force is still equal to the volume of the pressure prism and the pressure center can be computed as the centroid of the pressure prism I39 i IIIlquot 39 m QfHFJ J gg TC i IF quotl 1 PF 3939 r39 Iqu mW F H39h quot 39 quot E mum Dnnuunm 0 a Emma H Eumau u nn luu n u h nun lunm mu an I I 3 EH CE IL 3 H E y 2 it cu I m L IIJ Ir La n E IIII nummczxuu Ian 11P E39 n Eda 39 I f I IJ39I39 f FIT I 39 539 LI II r quotIlII I n m m h F W k H m W m m d u I m m n H u MU m H m M u d II m n m u w m I m a a a I u I I I I u m m d I n m H w m u m F W E I I h T 4 y n I J I I I1 I 39IIII a 39 I I i IJI HH39F139i rrllI I 39ll l I I u39 I I l lul 5 39 u 39 I LI h l 1II39I IfI I quotnhTi H I I I39llI39I39l I39IIII II 1 I I ml q ll l I I II I I w FILL HAl39 l 3939 I 39 I39Iu39 I I II fluh and1 llquot39i 39 I39Iquot quot 39l 39 IIquot T U 2 G m I h I J I 4 i I 4 4quot 3 hm1 quotFEW Il m llIF lIrll llII IIJ II I F I I II E I Iii I I I I1IlhIIII 39ll39 i I 3 mama Dzde Em J39 l l n a I W a m m w b h m M u a E a Mo n n n m 1 I W m m I I IIlI I I Fli I I l l I I Will Irl II allI1 I I I I u 1 I I uLI IIII II a m m m u n in w E m m I I II III IIIIHJIJH wwwdiIIIIIIFW HIII a I III a I L I I I I I I I L l IIIIIJ III hIl E I 1 I I W2 IIIIlI III ll IIITII II I MIIUI n W F h m m b m E w m E W m m m P a m u I u u m I I I I 11I1I I III IIIFI III lTILIIII IIIII IIIrllllrlhi IIIIIIIIII iii LII IIIIIIIII IriIIIII TinIll u I m n m b u m m m an I ma EH I n a I II I I I II IIIInl I II ILTII II I IIII I IlllI1IIILqIII IIII IIIllllLILIlIIII I1I III IIIIIIi M m n r n m m u III I WWI IIIIEHUI III IIII uIIII11FI IIlII1IaIrLTII1III1IULTIIIIIIII1 IIIII m n M w m IJIII 111 I I I III I II ITIIII m 1M I mm I EFIIHJEEMquot J an 3 L w 4 a I l7 39I I II I I WEI FE 3353 39quotc39 quot TIIIIII II IIIII I I I I I I I I I 1I HannaMEIER DEInFmIHl ul uniU n nB HEM WIN m M A M M m I Eul u DD IUEnIHHHnI Iu a Ii 1 magma Swim 39 III quotH r m I n 4 a r r m n j r I awnalImrI I 1 Qnl WHWJIH 1 I In fray at mnII1lnu wwmu Hmarnyd II III I1 l I I 39 39IIIIII WEEWE Emmydh m i u n u l u I I I I I I II I I I I I 39IJr I I I IIII I i I 15 I II1II II I IIImHnI u IIIInIIIIIIIIunaIIHitli IIIIIgI I I l a u n I l m IIFFa BEEF lm I Irv uh BF main I 5 la Iquot E Eta Erm I a J I I 3 a I I39llEIIMIEE in V l III I a E I I I m u I I u a a I I I I m I m I u u I L E I u Fl I l I llI IIIIJ VIIIIi I I I I EEIIE u INFIFIIIIII Il II I I I r I a n um H1 I I I I I l H H rI H u m I I I I I I i I I m m u n I I III I I I I II I I I m I m l I II I ll I39ll II II I I E III I Ii I I I I I II III Ii I 1 I IIIIIJIIIII n u I u 2 1 m I J I I I w Ira I I u I I II I IIIIIII r IIEITIIIIIIIFI l I I I HI ILII I I u nI E F IanIIIIEIIIIIIIIII IIFIIILF E II I nIIlij W I u n I E III I39llI I z 3 I I m 1 I I I II malaria I DIJ IFEIHIEIIIEIFHIFHHIGFHJ I m m w m M I am HIEIEBIFIEIEMIIE HEELTTi cuu DIIJI n rr 2 II EEDIHILMIFHHIHIWIunHn ma I e III m I m I I I I m n u H n a In I kilnIE 1n III aing n m n I w III ldann I ll Inquot E Ijulrl if Ihl m m ur I q I I III Jr an III II Iu nl I I E II E unhnu li a I in Dunn n m n m I m m I I m I I E I u I I I EEEF I I g I n I m I I I I I T II I II I II I l I I I I II iir Jill IIIlILIIInII IIIII II39IL in I1IJII II IIII d11IIIJIII Ir quotPI I Ill In 1 n I I H m W W I II I I1 I Edrlluiil I m l I I l h Inlunulmmnunmn nl annual quotha I I gnu FIE I I I m I Film Till EHIEH Gdhr rnu Ian u I I m nnuII If I IthFELE UIJHNIHnHma Inna I I I n l m l l 1I I IIIIIIII Innaf l 39uflni IIIIII II IiiIII I I I I HI I I II I I I I I I I n I III IIIII IJr II I 11 IIII I IIIIL ll1II I I i TI1IIil IIIIEEIWII IIIILIHIIll I m n I I quotHJrIE I r III11139 III11 1II IITIIIIIII II III I n u I I n I I I IlIri til IIIFIleIIIII m I FI1I1 IIIJII lllwiFiIIIIIILIIEIIIIII lull I I W I u I I III F a Ih analI III n W TEELIIIII39IIIEI JEEII Inlll llj EIHIIEIJIIIHMIIIn MIIHdI 1 In nr in I Hun nn IniIFE ma UnIE JIIIJFIIII III I E III CIVIL EN 3130 SPRING 2015 LECTURE 14 Contents Reading Sections 27 and 28 of the text book Homework Problems 2104 and 2107 from your text book CIVIL EN 3130 LECTURE 14 BUOYANT FORCE 2 1 Buoyant force 0 The resultant force exerted on a oating or submerged body by a static uid is called the Buoyant Force which we denote by F B o The buoyant force Always acts vertically Upward o It is equal to the Specific Weight of the uid times the Volume of the uid displaced by the body or equivalently the Weight of the uid displaced by the body O In equation form we have Fbth 1 0 To understand how we arrive at this equation let s consider the following NET HORIZONTAL FORCE IS ZERO CIVIL EN 3130 LECTURE 14 BUOYANT FORCE 3 Elt AK Cv Tn v I I H YVU FL I YV39L NET WEI LT t LL JfL Fd FluioL 7 YWL Via JVJ Submerged D Buoyant quot Force FB 7 CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES 4 o It can easily be shown that the Buoyant Force acts through the CentPOid of the ispla edmelmeof uid see pages 73 and 74 of your text book o This holds for both Smeng ed and Floating obj ects o The centroid of the displaced volume of uid is called the center of Buoyancy 2 Stability of oating and submerged bodies 0 Another important thing to consider with submerged or oating bodies is their Stability o A submerged or oating body is said to be in Stable equilibrium if when displaced it Returns to its equilibrium position 0 Conversely a body is in UnStable equilibrium if when displaced even slightly it Moves to a new equilibrium position o The stability of a body is determined by the relative positions of its centers of Buoyancy and Gravity 0 Consider the example below of a completely submerged body Center of Buoyancy Center of Gravity Restoring Couple Stable CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERCED BODIES 5 V co 0 q CB Overturning Couple Unstable 0 Thus a completely submerged body is in Stable equilibrium if its center of gravity falls BGIOW the center of buoyancy 0 Conversely a completely submerged body is in Unstable equi librium if its center of gravity is Above the center of buoyancy o For Floating bodies the stability problem is more complicated since as the body rotates the location of the center of buoyancy Which is the centroid of the displaced volume May change o For example consider a oating body such as a barge that rides low in the wa ter as shown below Which can be Stable even though the center of gravity lies Above the center of buoyancy I It I If I CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES EXAMPLE PROBLEM A spherical buoy has a diameter of 15 m weighs 850 kN and is anchored to the sea oor with a cable as shown in the gure below Although the buoy normally oats on the surface at certain times the water depth increases so that the buoy is completely immersed as illustrated For this condition what is the tension in the cable SOLUTION 2 T Seawater y 101 kNm3 Cable FEDgt 39 3 11 Ifmg J 4 150 W N ISOE 2 a t T a T w i r W l W a Va jfhsf twj p 1 MM r r llii fa J E rz gt F l397vgwo i IV a l quoti 3 W1 W oarsT Fl irt 7 U N 4 ha M 9 G if itquot N 39 m human WWW 39 9330 A 1 m CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES 7 CIVIL EN 3130 SPRING 2015 LECTURE 15 Contents 1 Relative equilibrium 2 11 Uniform linear acceleration OJ Reading Section 29 of the text book Homework Problems 2121 2123 2126 and 2128 from your text book CIVIL EN 3130 LECTURE 15 RELATIVE EQUILIBRIUM 2 1 Relative equilibrium 0 Up to this point we have been primarily concerned With uids at REST 0 However recall that in a previous lecture we developed the general EQUATION OF MOTION for a uid un dergoing acceleration under the assumption that there are no SHEAR SRESSES o In vector form this equation is 7 F l 1 EL 1 J Where a axi ayj azk is the acceleration vector recall i j and k are the unit vectors in the 3 y and 2 directions respectively and Vp is the PRESSURE GRADIENT de ned by o In scalar form the equations are Where we have used the fact that p 79 CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration 3 11 o A general class of problems involving uid motion when equations 1 2 and 3 are applicable that is when there is acceleration but no shearing stresses occurs when a mass of uid undergoes RIGIDBODY MOTION o For example if a container of uid ACCELERATION along a after some initial sloshing has died out with each particle having the same straight path the uid will move as a RIGIDBODY acceleration rigid body the uid is said to be in RELATIVE EQUILIBRIUM 0 When moving as a 0 Two cases are considered in your text book 1 2 UNIFORM LINEAR ACCELERATION UNIFORM ROTATION ABOUT A VERTICAL AXIS 0 We will only be concerned with the rst of these two Uniform linear acceleration 0 Suppose an open container of liquid is translating along a straight path with a constant acceleration a as illustrated below 0 The equations of motion 1 2 and 3 in this case are d 0quot YIaigt 911 oLy 3 lto d2 v d f DU 3 2 I MLFIXA UNLIKE STATIC CASE 39 em ATM CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration 4 0 Note that in this case the PRESSURE p is a function of BOTH the a and y directions 0 Thus the total differential dp of the pressure is given by d d AFraEAxdny J39 0 Substituting the expressions for 81383 and 819 g from the previous page CAL 9L lt13 Y 0 XmlH T 8 h J 39 V l o This expression can be integrated to obtain 07 PXIVj 5 x YKI P L Whoi0 f o where c is a constant of integration 0 The constant of integration can be evaluated provided we know the pres sure at ONE POINT say p0 and assign this point the coor dinates a y O 0 Thus our pressure equation is CLx O Pr SX VIT7FO 4 0 Let s verify that this equation would simplify to our usual p yh in the case of a static uid with a free surface CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration 5 EXERICISE Verify that in the case of a static uid with a free surface the general pressure equation 4 from the previous page simpli es to the equation pw SOLUTION CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration p0 L1nes of constant pressure p1 P2 y L 0 Another important piece of information we can obtain from the pressure relationship is the SLOPE of the free surface of an accelerat ing liquid 0 Setting p 0 in equation 4 and solving fo e obtain a a gtltXF Y 739 f173 V Y339 I o This is the equation of a line y 2 ma I refer to gure above With Qv q I l SLOPEm39 am 9 pM l39xbfl 3900 work v wyg 0 Furthermore this slope is the slope of all lines of constant pressure throughout the liquid 0 To see this rst note that along lines of constant pressure our previously derived expression for the total pressure differential is zero that is 1 AF 0 0 Solving this expression for dyda slope of lines of CONSTANT PRESURE 7 we have 1 ax Galr3 0 Thus lines of constant pressure are PARALLEL to the free surface quot L M 4 4 Tz fIYLGviE rm 6 Wank Luannw m 53m nests M W xx 05 m1 cm Titusquot139 Farm 1 AQELEIA r K M A Farmd 9F X a Y 3939 TK T I L m 65 or coH M r P smi C PAnALLEL 11 FREE SoILFALE U lFotw 3 Ac eLELATuN quotquot VMI Raw 939 MOTH i P15IE 9 AW Pun39r P 339 3 ufL 3 5 I P P 39a 9 FREE SnpAce 7 SOLVE m V Fuut w 5P gaming J andquot Y i x a i my 2quot Y39 W K 9 EE EMEW 0F LEE Artfulimu39 u xpt w I NE Am i mnw N yulmscnm CIVIL EN 3130 SPRING 2015 LECTURE 11 Contents Reading Section 25 of the text book Homework Problems 238 247 265 266 and 272 from the text book Review B j Atmospheric Pressure Pressure measurements Gage L Absolute gt Vacuum 1 Piezometer I p 2 U tube manometer LJEEE 3 Incline tube Manometers CIVIL EN 3130 LECTURE 11 FORCES ON PLANE AREAS 2 1 Forces on plane areas 0 The calculation of forces on submerged Surfaces due to uid pressure is an essential part of the design process for dams storage tanks ships etc 0 From our previous lectures we can already deduce a cou ple things regarding these forces and pressures for static uids 1 The forces acting on the submerged surfaces Will be Perpendicular to them since there are no shear stresses 2 For incompressible uids the pressure on the submerged surfaces will vary Linearly With depth p 2 7h 0 Building on this information the goal of this section is to develop techniques for determining the DiI GCtiOI l Location and Magnitude of resultant 7 forces acting on submerged surfaces 0 In some cases these quantities may be the end result that we seek in others they may simply be a means to an end 0 Let s begin by looking at the simple case of the resultant force on a horizontal submerged surface CIVIL EN 3130 LECTURE 11 11 Horizontal surfaces 3 11 Horizontal surfaces 0994 ATM L l if 3 Speci c weight y 1 F h HHHHHHJ L Consider a cross section of a tank of liquid as shown above We know that the uniform pressure along the bottom of the tank is simply RYH L39 J Over some small differential area dA of the base of the tank the corresponding olA EF AQ EEEdF and the total resultant force acting on the base of the tank is the sum integral of differential force dF is simply these differential forces that is F 1 P 1 A where A is the total area of the base of the tank For this case note that the pressure p 2 7h is constant and can therefore be pulled out of the integral so the resultant force computed by equation 7 is simply Resultant E The next question we want to ask is where this resultant force acts ie where it s line of action is located We will denote the coordinates of this point often called the center of pressure or pressure center by CIVIL EN 3130 LECTURE 11 11 Horizontal surfaces 4 I a Rectangular tank base b Consider the case when the base of the tank is a rectangle as shown in Figure a above For this case intuitively we know that the resultant force will act in the center of the rectangle Formally we can show this by equating the moments of t e resultant force and the moments of the distributed pressure force about the y and M iii336 s respectively Pressilire center T A A 2 8 Cl F A PX A I F A 9 y 2 l X As we noted above the uniform pressure p in this case can be removed from the integrals and dividing the equations by F we have the following expressions for 33p and yp yPAXdAI lSAIdA J where we have used the fact thatF 1A You may recognize the integral eX pressions in the equations above as the de nitions of the coordinates of the centroid of the area A that is I Xr PTCAXOM V Till ydA Thus for horizontal surfaces the line of action of the resultant or pressure center is located at the centroid of the area A ie 33p 2 5 yp g This is true not only for the simple case of a rectangular horizontal surface as considered above but also for some arbitrary shape as shown above in Figure Horizontal Surfaces 7 PA I X97 X 70 739 CIVIL EN 3130 LECTURE 11 12 Inclined surfaces 5 12 Inclined surfaces v ATM o Flu l As in the case of the horizontal surface let s begin by looking at the pressure along the inclined surface Referring to the gure above at some point 33 y the pressure PY TYsm zgt dfgt ltgt Inserting this expression into our Equation 1 for the resultant force we have F Ygiq 9 SA YdA39 is given by Note that in this case the full expression for the pressure cannot be pulled out of the integral because of its y dependence CM 2 dasdy However 7 and sin 9 both constant have been removed You may note that the remaining integral expression is equal to the product QA Thus the resultant force in this case is given by F YhA vi 7R 3 whers the depth of the uid at the centroid 5 1 of area A CIVIL EN 3130 LECTURE 11 13 Center of pressure 6 13 Center of pressure To determine the line of action of the resultant force or the pressure center for this more general case we again equate the moments of the resultant lc e and the mo mWWrce about the 31 and az axes see Equation 2 Solving for the pressure center coordinates 33 and yp in this case we have l L XP JA Pxom 1 7 FJAFYJA Again note that in this case unlike the case for the horizontal surface the pressure cannot removed from the integral because of its y dependence Substituting our ex pressions for p and F from the previous page into the equation for yp we have 1 1 39 9 dA 39 9 2dA mp ygA sin 6 Liz11 my 7 yp ygA sin 6 A 7311 y Canceling out 7 and sin6 we have M Wadi th 9A Yf VA39 In this case the integral expression in the equation for mp is called the product of inertia with respect to the a and y axes that is IXY SA Ya The integral expression in the equation for yp is called the second moment of the area or moment of inertia with respect to the m axis that is Thus for inclined surfaces the coordinates of the pressure center are given by Ixy IX CIVIL EN 3130 LECTURE 11 13 Center of pressure 7 These expressions can be written in terms of the product of inertia and the second moment of area about the centroidal axes denoted 363 and 36 respectively by using the respective parallel axis theorems IX7 jiny A279 IX le ry Q See appendix A of your text book for more information on the centroid second mo ment of area product of inertia and parallel axis theorem Using the above expres sions the coordinates of the pressure center for an inclined surface can be written as A I 2 2 lt4gt Xe 9A IVr W V q Note that for the case of an inclined surface the pressure center is not located at the centroid of the submerged area A Howeveris interesting to note that if f 7 the submerged area is symmetric With respect to either of the centroidal axes then f ce my 9that case CIVIL EN 3130 LECTURE 11 14 The pressure prism 8 Lecture 11 Continued 2915 14 The pressure prism M Col a b For submerged surfaces of relatively simple shape such as a rectangle a useful graphical approach can be used in lieu of the equations derived in the previous sec tions to determine the magnitude and location center of pressure of the resultant force acting on the submerged surface To illustrate this approach consider the situation depicted in the gure above where we wish to determine the resultant force acting on the side of the tank Using Equation 3 of section 12 the resultant force can be computed as where we have used the fact that ft y sin 90 2 g h 2 for this case The pressure distribution shown in gure a runs across the entire width w of the tank as shown in gure b and the magnitude of the resultant force is equal to the volume of this shape the so called pressure prism that is where wh is the area A of the side of the tank CIVIL EN 3130 LECTURE 11 14 The pressure prism 9 l a b The resultant force acts through the centroid of the pressure prism not the cen troid of area A For the volume of gure b of the previous page this occurs at a distance h 3 from the bottom of the tank since the centroid of a triangle is located a distance h 3 above its base It can easily be shown that this result is consistent with that obtained from Equations 4 derived in the previous section The pressure prism approach can also be used for cases where the submerged surface does not extend all the way to the free surface and or the submerged surface is inclined at some arbitrary angle 6 as illustrated in gures a and b above In both cases shown above the pressure prisms are trapezoidal instead of triangular but the magnitude of the resultant force is still equal to the volume of the pressure prism and the pressure center can be computed as the centroid of the pressure prism CIVIL EN 3130 SPRING 2015 LECTURE 12 Contents Reading Section 26 of the text book Homework Problems 284 289 and 299 from the text book CIVIL EN 3130 LECTURE 12 FORCE COMPONENTS ON CURVED SURFACES 1 Force components on curved surfaces 0 When the submerged surface is curved rather than planar as shown in the gure above the differential pressure forces dF on the surface vary in Direction 0 At any particular point on the curved surface the associated differ ential force dF may have both Horizonal de and Vertical dFy force components 0 Thus to nd the Magnitude of the resultant force we must rst sum integrate the horizontal and vertical differential force components separately 0 We denote the total sum of these horizontal and vertical component forces on the curved surface by FX and FX respectively 0 These component forces can then be used to nd the Magnitude of the total resultant force via the stan dard equation F 1 W W 0 Let s look at how we nd these component forces beginning with the horizontal force Fm CIVIL EN 3130 LECTURE 12 11 Horizontal force component 11 Horizontal force component 3 L i dAx dA cos6 V 9 yravrzc Hgy Referring to the gure above the component of the differential force in the c direction is given by ampFx9 JFLCMB 3 use Summing integrating these components over the entire surface gives Fx A remade The product dA cos6 E dAm is the horizontal projection of the the dif ferential area dA and the pressure p equation above as l so we can write the Fx 4 4 73le I qx A The integral expression in the above equation is equal to the product ngx Am Where gm is the y coordinate of the centroid of the horizontally projected area CIVIL EN 3130 LECTURE 12 11 Horizontal force component Thus we can calculate the horizontal component of force on a curved area by calculating the force on the horizontal PPOJeCtlon of the area that is Fquot t KX Ax 1 Where 561 2 321 is the depth of the uid at the centroid of the horizontal pro d jection Am f J39 In a similar way it can be shown that the line of action of the horizontal force component passes through the Pressure center of the projected area Ax CIVIL EN 3130 LECTURE 12 12 Vertical force component 12 Vertical force component i dAy dA cos L Referring to the gure above the component of the differential force in the y direction is given by JFcos L05 Again summing integrating these components over the entire surface gives h P as The product dA cos E dA is the vertical projection of the the differential area dA and the pressur so the equation above can be written as I F r IdAV JV l 39 In 39 39 I v Where we have de ned 395 l cl A of a prism of height y With a base39 of area dAy Which is equal to the volume CIVIL EN 3130 LECTURE 12 12 Vertical force component Z T S 8 Liquid B 73 hi 2 12 O I Vl 7A Liquid A 7A Thus the vertical force component is given by W Fy YV39 2 where V is the T0131 VOIDme between the curved sur face and the free surface Note that this expression is equivalent to the Weight of the liquid in the Volume V39 A special case to consider is when the liquid is BGIOW the curved surface and the pressure is known at some point say 0 see gure above In that case we use the following procedure 1 An imaginary or equivalent free surface s s is constructed some height h 1 Fp Y above 0 where p0 is the pressure at point 0 and y is the speci c weight of the liquid in contact with the curved surface 2 The vertical component of the pressure force on the curved surface is then the Weight of the imaginary volume vertically above the curved surface see page 68 of your text book Finally it can be easily shown see page 69 of your text book that the line of action of the vertical force component passes through the CentrOid of the volume V real or imaginary I39 i IIIlquot 39 m QfHFJ J gg TC i IF quotl 1 PF 3939 r39 Iqu mW F H39h quot 39 quot E mum Dnnuunm 0 a Emma H Eumau u nn luu n u h nun lunm mu an I I 3 EH CE IL 3 H E y 2 it cu I m L IIJ Ir La n E IIII nummczxuu Ian 11P E39 n Eda 39 I f I IJ39I39 f FIT I 39 539 LI II r quotIlII I n m m h F W k H m W m m d u I m m n H u MU m H m M u d II m n m u w m I m a a a I u I I I I u m m d I n m H w m u m F W E I I h 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IIIIIJIIIII n u I u 2 1 m I J I I I w Ira I I u I I II I IIIIIII r IIEITIIIIIIIFI l I I I HI ILII I I u nI E F IanIIIIEIIIIIIIIII IIFIIILF E II I nIIlij W I u n I E III I39llI I z 3 I I m 1 I I I II malaria I DIJ IFEIHIEIIIEIFHIFHHIGFHJ I m m w m M I am HIEIEBIFIEIEMIIE HEELTTi cuu DIIJI n rr 2 II EEDIHILMIFHHIHIWIunHn ma I e III m I m I I I I m n u H n a In I kilnIE 1n III aing n m n I w III ldann I ll Inquot E Ijulrl if Ihl m m ur I q I I III Jr an III II Iu nl I I E II E unhnu li a I in Dunn n m n m I m m I I m I I E I u I I I EEEF I I g I n I m I I I I I T II I II I II I l I I I I II iir Jill IIIlILIIInII IIIII II39IL in I1IJII II IIII d11IIIJIII Ir quotPI I Ill In 1 n I I H m W W I II I I1 I Edrlluiil I m l I I l h Inlunulmmnunmn nl annual quotha I I gnu FIE I I I m I Film Till EHIEH Gdhr rnu Ian u I I m nnuII If I IthFELE UIJHNIHnHma Inna I I I n l m l l 1I I IIIIIIII Innaf l 39uflni IIIIII II IiiIII I I I I HI I I II I I I I I I I n I III IIIII IJr II I 11 IIII I IIIIL ll1II I I i TI1IIil IIIIEEIWII IIIILIHIIll I m n I I quotHJrIE I r III11139 III11 1II IITIIIIIII II III I n u I I n I I I IlIri til IIIFIleIIIII m I FI1I1 IIIJII lllwiFiIIIIIILIIEIIIIII lull I I W I u I I III F a Ih analI III n W TEELIIIII39IIIEI JEEII Inlll llj EIHIIEIJIIIHMIIIn MIIHdI 1 In nr in I Hun nn IniIFE ma UnIE JIIIJFIIII III I E III CIVIL EN 3130 SPRING 2015 LECTURE 14 Contents Reading Sections 27 and 28 of the text book Homework Problems 2104 and 2107 from your text book CIVIL EN 3130 LECTURE 14 BUOYANT FORCE 2 1 Buoyant force 0 The resultant force exerted on a oating or submerged body by a static uid is called the Buoyant Force which we denote by F B o The buoyant force Always acts vertically Upward o It is equal to the Specific Weight of the uid times the Volume of the uid displaced by the body or equivalently the Weight of the uid displaced by the body O In equation form we have Fbth 1 0 To understand how we arrive at this equation let s consider the following NET HORIZONTAL FORCE IS ZERO CIVIL EN 3130 LECTURE 14 BUOYANT FORCE 3 Elt AK Cv Tn v I I H YVU FL I YV39L NET WEI LT t LL JfL Fd FluioL 7 YWL Via JVJ Submerged D Buoyant quot Force FB 7 CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES 4 o It can easily be shown that the Buoyant Force acts through the CentPOid of the ispla edmelmeof uid see pages 73 and 74 of your text book o This holds for both Smeng ed and Floating obj ects o The centroid of the displaced volume of uid is called the center of Buoyancy 2 Stability of oating and submerged bodies 0 Another important thing to consider with submerged or oating bodies is their Stability o A submerged or oating body is said to be in Stable equilibrium if when displaced it Returns to its equilibrium position 0 Conversely a body is in UnStable equilibrium if when displaced even slightly it Moves to a new equilibrium position o The stability of a body is determined by the relative positions of its centers of Buoyancy and Gravity 0 Consider the example below of a completely submerged body Center of Buoyancy Center of Gravity Restoring Couple Stable CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERCED BODIES 5 V co 0 q CB Overturning Couple Unstable 0 Thus a completely submerged body is in Stable equilibrium if its center of gravity falls BGIOW the center of buoyancy 0 Conversely a completely submerged body is in Unstable equi librium if its center of gravity is Above the center of buoyancy o For Floating bodies the stability problem is more complicated since as the body rotates the location of the center of buoyancy Which is the centroid of the displaced volume May change o For example consider a oating body such as a barge that rides low in the wa ter as shown below Which can be Stable even though the center of gravity lies Above the center of buoyancy I It I If I CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES EXAMPLE PROBLEM A spherical buoy has a diameter of 15 m weighs 850 kN and is anchored to the sea oor with a cable as shown in the gure below Although the buoy normally oats on the surface at certain times the water depth increases so that the buoy is completely immersed as illustrated For this condition what is the tension in the cable SOLUTION 2 T Seawater y 101 kNm3 Cable FEDgt 39 3 11 Ifmg J 4 150 W N ISOE 2 a t T a T w i r W l W a Va jfhsf twj p 1 MM r r llii fa J E rz gt F l397vgwo i IV a l quoti 3 W1 W oarsT Fl irt 7 U N 4 ha M 9 G if itquot N 39 m human WWW 39 9330 A 1 m CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES 7 CIVIL EN 3130 SPRING 2015 LECTURE 15 Contents 1 Relative equilibrium 2 11 Uniform linear acceleration OJ Reading Section 29 of the text book Homework Problems 2121 2123 2126 and 2128 from your text book CIVIL EN 3130 LECTURE 15 RELATIVE EQUILIBRIUM 2 1 Relative equilibrium 0 Up to this point we have been primarily concerned With uids at REST 0 However recall that in a previous lecture we developed the general EQUATION OF MOTION for a uid un dergoing acceleration under the assumption that there are no SHEAR SRESSES o In vector form this equation is 7 F l 1 EL 1 J Where a axi ayj azk is the acceleration vector recall i j and k are the unit vectors in the 3 y and 2 directions respectively and Vp is the PRESSURE GRADIENT de ned by o In scalar form the equations are Where we have used the fact that p 79 CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration 3 11 o A general class of problems involving uid motion when equations 1 2 and 3 are applicable that is when there is acceleration but no shearing stresses occurs when a mass of uid undergoes RIGIDBODY MOTION o For example if a container of uid ACCELERATION along a after some initial sloshing has died out with each particle having the same straight path the uid will move as a RIGIDBODY acceleration rigid body the uid is said to be in RELATIVE EQUILIBRIUM 0 When moving as a 0 Two cases are considered in your text book 1 2 UNIFORM LINEAR ACCELERATION UNIFORM ROTATION ABOUT A VERTICAL AXIS 0 We will only be concerned with the rst of these two Uniform linear acceleration 0 Suppose an open container of liquid is translating along a straight path with a constant acceleration a as illustrated below 0 The equations of motion 1 2 and 3 in this case are d 0quot YIaigt 911 oLy 3 lto d2 v d f DU 3 2 I MLFIXA UNLIKE STATIC CASE 39 em ATM CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration 4 0 Note that in this case the PRESSURE p is a function of BOTH the a and y directions 0 Thus the total differential dp of the pressure is given by d d AFraEAxdny J39 0 Substituting the expressions for 81383 and 819 g from the previous page CAL 9L lt13 Y 0 XmlH T 8 h J 39 V l o This expression can be integrated to obtain 07 PXIVj 5 x YKI P L Whoi0 f o where c is a constant of integration 0 The constant of integration can be evaluated provided we know the pres sure at ONE POINT say p0 and assign this point the coor dinates a y O 0 Thus our pressure equation is CLx O Pr SX VIT7FO 4 0 Let s verify that this equation would simplify to our usual p yh in the case of a static uid with a free surface CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration 5 EXERICISE Verify that in the case of a static uid with a free surface the general pressure equation 4 from the previous page simpli es to the equation pw SOLUTION CIVIL EN 3130 LECTURE 15 11 Uniform linear acceleration p0 L1nes of constant pressure p1 P2 y L 0 Another important piece of information we can obtain from the pressure relationship is the SLOPE of the free surface of an accelerat ing liquid 0 Setting p 0 in equation 4 and solving fo e obtain a a gtltXF Y 739 f173 V Y339 I o This is the equation of a line y 2 ma I refer to gure above With Qv q I l SLOPEm39 am 9 pM l39xbfl 3900 work v wyg 0 Furthermore this slope is the slope of all lines of constant pressure throughout the liquid 0 To see this rst note that along lines of constant pressure our previously derived expression for the total pressure differential is zero that is 1 AF 0 0 Solving this expression for dyda slope of lines of CONSTANT PRESURE 7 we have 1 ax Galr3 0 Thus lines of constant pressure are PARALLEL to the free surface quot L M 4 4 Tz fIYLGviE rm 6 Wank Luannw m 53m nests M W xx 05 m1 cm Titusquot139 Farm 1 AQELEIA r K M A Farmd 9F X a Y 3939 TK T I L m 65 or coH M r P smi C PAnALLEL 11 FREE SoILFALE U lFotw 3 Ac eLELATuN quotquot VMI Raw 939 MOTH i P15IE 9 AW Pun39r P 339 3 ufL 3 5 I P P 39a 9 FREE SnpAce 7 SOLVE m V Fuut w 5P gaming J andquot Y i x a i my 2quot Y39 W K 9 EE EMEW 0F LEE Artfulimu39 u xpt w I NE Am i mnw N yulmscnm CIVIL EN 3130 SPRING 2015 LECTURE 16 Contents 1 Introduction to uid dynamics 2 2 Flow concepts and kinematics 21 Analysis approaches 22 Flow classi cation 23 Pathlines streaklines and streamlines IOOCJO Reading Section 31 of the text book Homework None CIVIL EN 3130 LECTURE 16 INTRODUCTION TO FLUID DYNAMICS 2 1 Introduction to uid dynamics 0 In the rst two chapters we studied the basic properties of uids and considered various situations involving uids that are either 1 AT REST STATIC or are 2 moving in a rather simple way such as rigid body motion 0 Although we saw that there are a number of situations where one of these two conditions is met most engineering uid mechanics problems are concerned with FLUID FLOW o This fact should not come as a surprise given that the application of the slightest SHEAR STRESS to a uid will cause it to ow 0 Thus in this chapter we will begin our study of uid dynamics 0 In order to analyze the problems of uid dynamics we will need at our disposal a basic set of equations that can be used to predict the FLUID MOTION that results from the application of spe ci c forces 0 These equations can be derived from the following fundamental princi ples or laws of physics CD CONSERVATION OF MASS lt2 NEWTONS SECOND LAw FMA 3 FIRST LAW OF THERMODYNAMICS CONSERVATION OF ENERGY C4 SECOND LAW OF THERMODYNAMICS INCREASING ENTROPY INEQUALITY o The derivation and subsequent application of the basic equations ob tained from the consideration of these laws will be the primary focus of this chapter CIVIL EN 3130 LECTURE 16 FLOW CONCEPTS AND KINEMATICS 3 21 Flow concepts and kinematics Prior to considering the equations of uid dynamics we will discuss vari ous aspects of uid motion without being concerned with the actual forces necessary to produce such motion The study of the motion of particles or bodies collections of particles again without consideration of what causes the motion is known as KINEMATICS KINEMATIC GREEK The variables considered in kinematics are the POSITION velocity and acceleration of particles As we shall see below in the next subsection there are two general ap proaches that can be used to describe these and other variables the LAGRANGIAN and EULERIAN descriptions J Using these kinematic variables and descriptions uid ow can be clas si ed in a number of ways This will be covered in subsection 22 Additionally various ow concepts can be developed to help visualize and analyze ow elds Speci cally in subsection 2 3 we will look at the concepts Of PATHLINE STREAKLINE and STREAM LIN E Analysis approaches 0 As we mentioned above there are two general approaches used in analyz ing uid mechanics problems or problems in any branch of engineering mechanics or physical sciences for that matter 0 We consider these on the next two pages beginning with the LAGRANGIAN DESCRIPTION XYZ CIVIL EN 3130 LECTURE 16 21 Analysis approaches 4 Lagrangian Description 1 3 j E O O O PARTICLE A K I 5 F10 Consider a mass of particles as shown above which may represent a SOLID or a FLUID This collection of particles is an example of a SYSTEM By de nition a system is a xed amount of matter that may may move distort in shape and interact with its surroundings but that will always THE SAME contain particles It may consist of a relatively LARGE amount of mass or it may be in nitesimally small ie a single particle In any case we follow the particles of the system as they move about PROPERTIES and determine how the associated with those particles velocity acceleration temperature etc change with time eg velocity of particle A 21A FREE BODY DIAGRAMS When we draw in statics or dy namics the body that we consider is our system an identi ed portion of matter that we follow during its interactions with its surroundings TRACK the movement of the particles and identify how the properties associ Such a description of a body in which we ated with that particle change as a function of time is called a LAG RANG IAN description The Lagrangian description is named for Joseph Louis Lagrange January 25 1736 April 10 1813 an Italian mathematician and astronomer who made a number of outstanding contributions to mathematics and classical mechanics CIVIL EN 3130 LECTURE 16 21 Analysis approaches Eulerian Descript ionb x39 g f l 9 AvT l FLOW X l CONTROL l VOLUME I n 39 l o In uid dynamics it is often quite dif cult to identify and keep track of a speci c quantity of matter or system as is done in a LAGRANGIAN DESCRIPTION o This is due to the fact that the FLUID PARTICLES move about quite freely unlike a SOLID that may deform but usu ally remains relatively easy to identify 0 For example it is much easier to follow a branch oating in a river than it is to identify and follow a speci c portion of water in the river consisting of the SAME PARTICLES 0 Thus in uid dynamics it is often more convenient to adapt an alterna tive approach where we focus on a FIXED REGION in space as opposed to a xed set of particles 0 Such a xed region in space as shown above is an example of a CONTROL VOLUME the size and shape of which are arbitrary and may even be in nitesimally small ie a single point 0 Now instead of tracking individual particles as in the Lagrangian approach we obtain information in terms of what happens at FIXED POINTS in space within our control volume as par ticles ow past those points 0 A description is thus given of the properties of the uid at each SPATIAL POINT 3 y and 2 within our control volume as a function of time eg velocity v va y z t o This type of description is known as a EULERIAN descrip tion bThe Eulerian description is named for Leonhard Euler April 15 1707 September 18 1783 a Swiss mathematician and physicist renowned for his work in mechanics optics and astronomy FOR THIS COURSE CIVIL EN 3130 LECTURE 16 21 Analysis approaches 6 In summary 0 With the LAGRANGIAN with a as it MOVES ABOUT description which is associated SYSTEM we follow the uid and observe its behavior 0 With the EULERIAN a CONTROL VOLUME the uid s behavior at a description which is associated with we remain stationary and observe FIXED LOCATION 0 These two descriptionsc are illustrated below Mam CAccording to one of the classical text books on uid mechanics Lamb Hydrodynamics 6th ed Cam bridge University Press 1937 both the Lagrangian and Eulerian descriptions are in reality due to Euler it is interesting to note that Lagrange was a doctoral student of Euler s Any attempt to correct this inaccuracy however would most likely cause more confusion than bene t CIVIL EN 3130 LECTURE 16 22 Flow classi cation 7 22 Flow classi cation Flow can be classi ed depending on how the ow eld changes in both SPACE and TIME quotTIMEquot Temporal classi cations M L til STEADY FLOW G 0111 STEADY FLOW the velocity eld and all other uid properties at all points in space do not vary in time EG AVIATO o For example if the velocity at a particular point in space is 3 m s in the 3 direction it Will remain that amount and in that direction for all time if the ow is steady UNSTEADY FLOW o In contrast to this UNSTEADY FLOW is when the velocity or any other uid property at any point in space varies With time EG AVIATO o In reality almost all ows are unsteady in some sense 0 As might be expected unsteady ows are usually more dif cult to analyze than STEADY FLOWS AV BE LO ERAGE FLUID HAVIOR OVER A NG TIME 0 However in the analysis of uid ow one can often make the ASSUMPTION of steady ow Without compromising the usefulness of the results CIVIL EN 3130 LECTURE 16 22 Flow classi cation 8 Spatial classi cations quotSPACEquot Uniform ow 0 In uid variables in time as UNIFORM FLOW the velocity eld and all other are the same at all points in space at a given instant AVAX AVAYAVAZ 0 0 Note that NO CHANGE that in any of the uid variables in space at this de nition simply states there is a particular I o It makes no comment about the properties in NSTANT IN TIME CHANGE in the uid TIME Nonuniform o oIIl NONUNIFORM FLOW W the velocity eld or other uid variables time vary from point to point in space at any instant in E G AVAX ilAvAvylAvAZ 710 Your text book also describes What are called TURBULENT LAMINAR and ows in this chapter however we Will postpone this discussion to a more appropriate time CIVIL EN 3130 LECTURE 16 23 Pathlines streaklines and streamlines 9 TO VISUALIZE THE FLOW 23 Pathlines streaklines and streamlines Although uid motion can be quite COMPLICATED there are several geometric concepts that have been developed to help visualize and an alyze ow elds To this end we brie y discuss PATHLINES STREAKLINES and STREAMLINES in this sub section PATHLINE o A PATHLINE is a line or the trajectory TRACED OUT by a given particle as it ows 0 A PATHLINE can be produced in the laboratory by marking a uid particle dying a small uid element and taking a time exposure photograph of its motion EXPERIMENTAL Q STREAKLINE O A STREAKLINE is a line JOINING all particles in a ow that have previously passed through a common xed point 0 A STREAKLINE can be obtained in the laboratory by taking instantaneous photographs of marked particles that all passed through a given location in the ow eld at some time STREAMLINE o A S T R EA M LIN E is a line that is everywhere TANGENT to the velocity eld 0 The s TR EAM U N E is used more in mLYTm work than experimental V CIVIL EN 3130 LECTURE 16 23 Pathlines streaklines and streamlines 10 Some notes on these concepts o For STEADY FLOWS pathlines streamlines and streak lines are all the SAME o For UNSTEADY FLOWS none of these three types of lines need be the SAM E 0 See page 117 of your book for more information PROBLEM SOLUTIONS Q39q Problem 299 j 570 r39wmww E KJ FLgryvl H Wig173 239 1 FL 9311 Hi FL 2 r 39 WOW 4 Ar A 1 a I Lampt 5 W M W mm PROBLEM SOLUTIONS m w SUM Ma aN72 Q d f w W 39 0 Jb vr i Flef iwayltggtto g a Fwy t Uc ryg quot3 kiwqiii u 4 I quot quot quot quot quot w quotcquot 3 SCamp A37 gj f H 7l JHzQ C ifquot W QQS 4 I Q U a gig D Em LIESmquot It 1 W 7 V I 15 quot lt1 U A quot r 9 1 3 ma Ma PROBLEM SOLUTIONS Problem 2121 0 Q I I a i A tuwHWWMWWWMM I 50pr 1 3 r1 xiW 3 4 0w 2 01 I 5 739 039 3 k l ngw W m M N mw fi em MMMMM W 5 WW mm F55 2 We Paarff quot397 Ont01A H 39 39 quot W gt Hm f H 1 VJ y1 x P w u x v V 3 a 0gaqb 1L POW 3 3 43 LiaPH Pox7 I3 2 9 z 63 lieH U gt 3 7 3 39P Fquot Mm C 3Y39L 3 19 gs f39 Di z 2 334133 21 quotw fig Rag pm m on H 3 2 3 aFgtgt 79 gt 03 n to y 2 H PROBLEM SOLUTIONS Problem 2123 9 L13 1 7quot Slept mg Frw 3vt fm M a w i A quoti 300V 1 a 7 L WA 6 quot 3Zerigq g A Hquot WP M MMquot quot 57 L n 1 gm 6 I 7 5 q ax 0 Q05 5 I V V r i h M m mm a O 39 344 I I 0739 P A F quot 5 quot 39 quotM1 EVA 2 LPN WANT x 39o3mltm5i gt a 39m 39 I VGJ 1 2 x 21gt 224 U rm 1 L J B c fgtf 0333331 1quot w ECLJT quotQ 1 Oilzgznlj X 63 5 36 Om2 m I 9 gt9 I 2 2 Qt ui xy 4 1 P y g Fly 3 n yy VA I 39 V f FOzVT A P qJonz qQ V2 09232 1 909 k 2quot 8 2 pfxg vowza s gem 13 my mmmz rquot l 93 kfy l 63 00 3 wawfs z w i O 10WZZ 1 403quot quot 7 a M PROBLEM SOLUTIONS Problem 1128 31 132 39 4353 1 i N39t NJTOTAL gi ox 9 EiHmO g 5130 4 1 Sma i 2 306 m3 39 I 1 g 57 V 35 X tquot m3 x 019M 99636 MA2gtIl l giggj a J m V Cg PJ 039679JN ii m H I 21 tm j wsww R 3 Wk ffq h 39 Mw 344gt n gm w633 95 Mn Q 62353 Ms392 quot39 quot Whvw you was IN93 AL I 39quot W 129339g 5 m 3 3917quotquot mm Q fwfi 9 i K 4 5 i133 6ng 331quot 30 3 i Watquot 4 I a a 39 Okay wmsrm 0 a 7quot 3 r 2041 THE SbofE or T39Hg Ffl SvQFAca 2 11 a f II K 93903 7 3 m muwmww yuwnwmww LI 3 1 WW v 39939 M 023e SOLVE CIVIL EN 3130 SPRING 2015 LECTURE 4 Contents Reading Sections 17 and 18 of the text book Homework Problems 144 156 and 158 AMT33 TIiQINK L XLquot5L N0 Tgtco SIT39Y IINCOMPaES 18LE CIVIL EN 3130 LECTURE 4 1 Pressure and a perfect gas Arman 0 509quot ab 11 Pressure FA Norn Ft 139 I39I PRESSURE AND A PERFECT GAS Force Fort L Consider a force acting on a triangular wedge of uid as shown above We de ne the Stress the force F per unit area A that is v F 5T ESL A Stress at a point is de ned in the sense of a limit that is AF AA Fm 574539 33 AAo acting on the sloped face of this element as 2 CIVIL EN 3130 LECTURE 4 11 Pressure 3 1 2 From equations and we can see that the DimenSionS of stress are F F d 9 quot 0 Stress 39 L A i terms of LT and F 39 m a 0 Stress 39 L T L in terms of LT and M The units of stress are 0 Pascal Pa in the SI system which is a N m2 TH 1 1 Cn v s ION 0 Common units used in the USG system are pounds per square foot pSF or pounds per square inch pSi Generally we express Stress in terms of two component stresses G The shear stress Which acts Para e or tangent to the local surface o The Greek letter Cquot is generally used to denote the shear stress component 0 The shear stress is computed by the equation Ft C A 2 The normal stress Which acts Perpendicular or normal to the local surface 0 The Greek letter 0 is generally used to denote the normal stress component 0 The normal stress is given by F n O T In uid mechanics the normal stress at a point in a uid relates to the Pressure p CIVIL EN 3130 LECTURE 4 11 Pressure Pressures can be speci ed in terms of PRESSURE 9 Absolute Pressure Fab which is the pressure rela tive t absolute zero pressure a pressure that would only occur in a perfect 0N L p 5 39aq lt2 Gage Pressure P GAME is the pressure measured relative to the local atmospheric pressure patm These two pressure measures are related by the equation fab fume farm j 3 Pressure will be discussed at greater length in Chapter 2 however before concluding the discussion here we note the following 0 Pressure p obviously has the same dimensions and units as Stress o Liquids can often sustain a considerable press1 with little or NO change in volume and thus density p N 0 That e nearly MRESR though we will comment on this at greater length in t e section on the bulk modulus of elasticity o The density p of a on the other hand will change under an applied pressure a F hp 0 This leads us to our next topic raid24ml an CIVIL EN 3130 LECTURE 4 12 A perfect gas 12 A perfect gas A PerfeCt Gas as de ned in your text book is a uid that satis es the following relationship betweenlaand density p gt at some constan absolute temperature T LfRT a 4 where R is the speci c gas constant or just gas constant Equation is referred to as the perfect or Ideal gas law R The gas constant 4 The units of R can be determined directly from equation Solving this equation for R we have R 2 P 79 l Thus N m3 1 mN I I 39 h Z o n S unltS We ave R m2 kg g 39 K i if l quot4lol oInUSCunltsR aslvqgt7 S Uq 9R I Tables C3 and Q4 in Appendix C of your text list values of R for several COIIlIIlOIl gases We note the following important distinction between Ideal Fluids and Ideal Gases 0 An ideal or perfect Gas is compressible according to equation and als o This is in sharp contrast to an ideal quUld which is incom pressible and has ZERO ViSCOSitY M CIVIL EN 3130 LECTURE 4 12 A perfect gas 6 EXERCISE Your text book lists several variations of the perfect gas law equation equation numbers 171 173 174 and 175 Derive these equations from the form of the perfect gas law given on the previous page SOLUTION 107RTL Ii39 KT 7EIL 121010 Densii39y g EV E 7 7MJ074 9 n OF ak 9 Mpmalu MOLLL C Lquot IX OF UL5Q QL CLS MHM kmsz a mz ma 1737 EVAMKT1075gt DelP z V5 Vf Z IZJUNIZiIZSSA oNS739ANT 179 mA K W 7 CIVIL EN 3130 LECTURE 4 12 A perfect gas 7 HOMEWORK PROBLEM 144 A gas at 20 C and 02 MPa abs has a volume of 40 L and a gas constant B 210 mNkgK Determine the density and mass of the gas SOLUTION 119 7 KT Ideal Gas Law D E 0 Solve for Density De a f Io RT 7 09XIOquot Nm L110 m mwaoK V DC K quE STWI a f 350394 k 5vf l Solve for Mass 61 m c V 7 m 3250hquot3gt1 aka 206 k3 T CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 8 2 Bulk modulus of elasticity As we saw in the previous section it is important to consider the following question with uids QUESTION HOW EASILY DOES THE VOLUME OF A GIVEN MASS AND THUS DENSITY CHANGE WITH A CHANGE IN PRESSURE That is HOW COMPRESSIBLE IS THE FLUID ANSWER o In the previous section we noted that liquids are nearly INCOMPRESSIBLE while gases Change in volume under compression 0 Beyond this general observation we would like to have a more Quantifiable measure of the compressibility of a uid 0 To this end a promrty that is commonly used to characterize compress ibility is th modulus of elastici or simply bulk modulus which we de ne below Bulk modulus of elasticity The bulk modulus is de ned as f JFv l 5 whers the differential change in Pressure needed to w create a differential change in VOIUme dV of some volume CV CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 9 We note the following about the BUIk MOdUIUS 0 Since the quantity dV V is DimenSionleSSQ the bulk modulus K has the same dimensions of pressure i FL 2 0 Thus SI units are the Pascal Pa 39Z 0 Common USC units are psi L 0 Large values for the bulk modulus indicate that a uid is relatively Incompressible 0 As expected values of K for commons liquids are Large ie very large changes in pressure are required to cause a signi cant change in VOIUme 0 Thus for most en ineering applications liquids can be considered as compressible 0 Tables C1 and C2 in Appendixg of your text list values of K for several common liquids 0 To gain some insight into the incompressibility of Water 7 let s consider the following example CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 10 EXAMPLE PROBLEM At standard atmospheric pressure and a temper ature of 60 F how muould be required to compress a unit volume M of water 1 V D SOLUTIOsz 1 lV 00 7 ED F TAIL e 2 K o39F 39Iooo F5 DefinitiorclLofB Ik Modulus I I lt K QLVV 7 J a if 3Hooof5 Doi j O 1954 ch H635 Fo Oo Comfr wSion D g 0 WW x w CIVIL EN 3130 SPRING 2015 LECTURE 2 Contents 1 Continuum 2 2 De nition of a uid 5 3 Dimensions and units 7 Reading Sections 11 12 and 13 of the text book Homework NA CIVIL EN 3130 LECTURE 2 CONTINUUM 2 1 Continuum o Matter in both SOIid and Fluid form is made up of molecules atoms electrons protons neu trons quarks Higgs boson o In engineering applications we usually deal with pieces of matter that are very Large compared to these particles 0 Speci cally we are generally interested in studying the mechanical be havior of matter on a MlCrOSCODlC scale o This is what we call Engineering Mechanics o The problems of engineering mechanics are most precisely formulated using a mathematical or analytical basis 0 In order to do this we need to be able to de ne quantities such as density velocity etc in a mathematically tractable way 0 Speci cally quantities need to be de ned in a Pointwise manner ie by functions that are well de ned at each point within the matter 0 Considering the molecular structure of matter quantities such as den sity and velocity Are NOT well de ned pointwise To clarify this last point let s consider an example The Higgs boson often called the God Particle is an elementary particle that was theoretically proposed back in 1964 by Peter Higgs Its existence was only recently veri ed experimently in 2012 at the CERN laboratory in Geneva Switzerland resulting in a Nobel Prize for Professor Higgs in 2013 As you ll discover throughout these notes I like footnotes CIVIL EN 3130 LECTURE 2 CONTINUUM Example I Density POl l39ufsQ do f 5 ggwmc 1 W 7006 i m V o JIIT D oleLLlkr39 bel mb w Example II Velocity CIVIL EN 3130 LECTURE 2 CONTINUUM 4 0 With the so called COHtlnuum actual molecular structure of matter with a hypothetical continuous concept we replace the medium called a continuum o The continuum Completely lls space no holes or voids 0 And the continuum therefore has properties that can be described Point wise QUESTION Is the continuum concept valid ANSWER In general the answer is Yes o The spacing between molecules is generally very small compared to the scale of the problems we are considering 0 At normal pressures and temperatures molecular spacing is on the order of 1E396 and 1E397 for liquids and gases 0 The continuum concept will be valid for all the circum stances covered in this course 0 An area of uid mechanics where the continuum concept Breaks down is for rari ed gases encountered at very high altitudes o In that case the spacing between air molecules can become very Large and the continuum concept is no longer valid Will not be considered in this course o This type of uid CIVIL EN 3130 LECTURE 2 DEFINITION OF A FLUID 2 De nition of a uid 0 We generally recognize 3 states of matter SOlld Liquids and Gases It 0 The last two are both considered FlUldS 5 QUESTION What is the fundamental difference ANSWER Lack the ability to resist defor o In contrast to solids uids mation While remaining at rest 0 Because of this uids move continuously under the action of a Stress o More precisely Fluid is a substance that deforms continuously that is Any A When subjected to shear stress 739 gt 0 no matter how small that shear stress may be or phrased conversely A Fluid stress While remaining at is a substance that cannot support a non zero shear Rest CIVIL EN 3130 LECTURE 2 DEFINITION OF A FLUID 6 To Clarify and expand on this de nition let s consider the following experiment No 5 r Com13Mx Ali r I gt F 7 7 5 A EA B Substance t I lt A 2 mn new w TM M GBj HaAIL MOI CIVIL EN 3130 LECTURE 2 DIMENSIONS AND UNITS 7 3 Dimensions and units 0 In this course we will be dealing with a variety of uid characteristics 0 Therefore we need a system for describing these characteristics both Qualitativer and Quantitativer Qualitative Description Dimensions 0 The qualitative description identi es the Nature or type of the characteristic eg length time stress etc o It is most conveniently given in terms of certain Primary quantities or basic dimensions 0 For a number of problems in uids mechanics only 3 basic dimensions are required 1 Length L 2 Time T 3 Mass m 0 These primary quantities or basic dimensions can then be used to de scribe Secondary or derived quantities 0 Two examples of secondary quantities include Velocity LT Force MLTA2 0 Where the symbol is used to indicate the dimensions of the derived quantities in terms of the basic dimensions 0 Note that instead of using L T and M as the primary quantities we could L TF also use 0 Mass M would then be a derived quantity ie Fma 0 And other secondary quantities would then be expressed in terms of L T and F eg StressFL 2 CIVIL EN 3130 LECTURE 2 DIMENSIONS AND UNITS 8 Quantitative Description Units 0 The quantitative description provides a Numerical Measure of the characteristic o It requires both a number and a Standard by which various quantities can be compared 0 The most Widely used system of units in the world is the International system si o In the United States the Vs customary system similar to the British Imperial System is still Widely used SI 0 SI is an LTM system the units of the 3 basic dimensions are 1 L Meter 2 T Second 3 M Newton o The unit of the derived quantity force F is the ma o The N is de ned as the the amount of force required to give a mass of 1 kg an acceleration of 1 ms2 o Algebraically quotquot 1 m 0 CIVIL EN 3130 LECTURE 2 DIMENSIONS AND UNITS Quantitative Description Units cont d USC 0 USC is an LTF system the units of the 3 basic dimen SlOIlS are 1 L Foot 2 T Second 3 F Pound force o The unit of the derived quantity mass M is the Slug A slug is de ned as a mass that accelerates by 1 ft s2 When a force of 1 lbf is exerted on it Algebraically i 1 A basU339 IIIsa1 CIVIL EN 3130 SPRING 2015 LECTURE 4 Contents Reading Sections 17 and 18 of the text book Homework Problems 144 156 and 158 AMT33 TIiQINK L XLquot5L N0 Tgtco SIT39Y IINCOMPaES 18LE CIVIL EN 3130 LECTURE 4 1 Pressure and a perfect gas Arman 0 509quot ab 11 Pressure FA Norn Ft 139 I39I PRESSURE AND A PERFECT GAS Force Fort L Consider a force acting on a triangular wedge of uid as shown above We de ne the Stress the force F per unit area A that is v F 5T ESL A Stress at a point is de ned in the sense of a limit that is AF AA Fm 574539 33 AAo acting on the sloped face of this element as 2 CIVIL EN 3130 LECTURE 4 11 Pressure 3 1 2 From equations and we can see that the DimenSionS of stress are F F d 9 quot 0 Stress 39 L A i terms of LT and F 39 m a 0 Stress 39 L T L in terms of LT and M The units of stress are 0 Pascal Pa in the SI system which is a N m2 TH 1 1 Cn v s ION 0 Common units used in the USG system are pounds per square foot pSF or pounds per square inch pSi Generally we express Stress in terms of two component stresses G The shear stress Which acts Para e or tangent to the local surface o The Greek letter Cquot is generally used to denote the shear stress component 0 The shear stress is computed by the equation Ft C A 2 The normal stress Which acts Perpendicular or normal to the local surface 0 The Greek letter 0 is generally used to denote the normal stress component 0 The normal stress is given by F n O T In uid mechanics the normal stress at a point in a uid relates to the Pressure p CIVIL EN 3130 LECTURE 4 11 Pressure Pressures can be speci ed in terms of PRESSURE 9 Absolute Pressure Fab which is the pressure rela tive t absolute zero pressure a pressure that would only occur in a perfect 0N L p 5 39aq lt2 Gage Pressure P GAME is the pressure measured relative to the local atmospheric pressure patm These two pressure measures are related by the equation fab fume farm j 3 Pressure will be discussed at greater length in Chapter 2 however before concluding the discussion here we note the following 0 Pressure p obviously has the same dimensions and units as Stress o Liquids can often sustain a considerable press1 with little or NO change in volume and thus density p N 0 That e nearly MRESR though we will comment on this at greater length in t e section on the bulk modulus of elasticity o The density p of a on the other hand will change under an applied pressure a F hp 0 This leads us to our next topic raid24ml an CIVIL EN 3130 LECTURE 4 12 A perfect gas 12 A perfect gas A PerfeCt Gas as de ned in your text book is a uid that satis es the following relationship betweenlaand density p gt at some constan absolute temperature T LfRT a 4 where R is the speci c gas constant or just gas constant Equation is referred to as the perfect or Ideal gas law R The gas constant 4 The units of R can be determined directly from equation Solving this equation for R we have R 2 P 79 l Thus N m3 1 mN I I 39 h Z o n S unltS We ave R m2 kg g 39 K i if l quot4lol oInUSCunltsR aslvqgt7 S Uq 9R I Tables C3 and Q4 in Appendix C of your text list values of R for several COIIlIIlOIl gases We note the following important distinction between Ideal Fluids and Ideal Gases 0 An ideal or perfect Gas is compressible according to equation and als o This is in sharp contrast to an ideal quUld which is incom pressible and has ZERO ViSCOSitY M CIVIL EN 3130 LECTURE 4 12 A perfect gas 6 EXERCISE Your text book lists several variations of the perfect gas law equation equation numbers 171 173 174 and 175 Derive these equations from the form of the perfect gas law given on the previous page SOLUTION 107RTL Ii39 KT 7EIL 121010 Densii39y g EV E 7 7MJ074 9 n OF ak 9 Mpmalu MOLLL C Lquot IX OF UL5Q QL CLS MHM kmsz a mz ma 1737 EVAMKT1075gt DelP z V5 Vf Z IZJUNIZiIZSSA oNS739ANT 179 mA K W 7 CIVIL EN 3130 LECTURE 4 12 A perfect gas 7 HOMEWORK PROBLEM 144 A gas at 20 C and 02 MPa abs has a volume of 40 L and a gas constant B 210 mNkgK Determine the density and mass of the gas SOLUTION 119 7 KT Ideal Gas Law D E 0 Solve for Density De a f Io RT 7 09XIOquot Nm L110 m mwaoK V DC K quE STWI a f 350394 k 5vf l Solve for Mass 61 m c V 7 m 3250hquot3gt1 aka 206 k3 T CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 8 2 Bulk modulus of elasticity As we saw in the previous section it is important to consider the following question with uids QUESTION HOW EASILY DOES THE VOLUME OF A GIVEN MASS AND THUS DENSITY CHANGE WITH A CHANGE IN PRESSURE That is HOW COMPRESSIBLE IS THE FLUID ANSWER o In the previous section we noted that liquids are nearly INCOMPRESSIBLE while gases Change in volume under compression 0 Beyond this general observation we would like to have a more Quantifiable measure of the compressibility of a uid 0 To this end a promrty that is commonly used to characterize compress ibility is th modulus of elastici or simply bulk modulus which we de ne below Bulk modulus of elasticity The bulk modulus is de ned as f JFv l 5 whers the differential change in Pressure needed to w create a differential change in VOIUme dV of some volume CV CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 9 We note the following about the BUIk MOdUIUS 0 Since the quantity dV V is DimenSionleSSQ the bulk modulus K has the same dimensions of pressure i FL 2 0 Thus SI units are the Pascal Pa 39Z 0 Common USC units are psi L 0 Large values for the bulk modulus indicate that a uid is relatively Incompressible 0 As expected values of K for commons liquids are Large ie very large changes in pressure are required to cause a signi cant change in VOIUme 0 Thus for most en ineering applications liquids can be considered as compressible 0 Tables C1 and C2 in Appendixg of your text list values of K for several common liquids 0 To gain some insight into the incompressibility of Water 7 let s consider the following example CIVIL EN 3130 LECTURE 4 BULK MODULUS OF ELASTICITY 10 EXAMPLE PROBLEM At standard atmospheric pressure and a temper ature of 60 F how muould be required to compress a unit volume M of water 1 V D SOLUTIOsz 1 lV 00 7 ED F TAIL e 2 K o39F 39Iooo F5 DefinitiorclLofB Ik Modulus I I lt K QLVV 7 J a if 3Hooof5 Doi j O 1954 ch H635 Fo Oo Comfr wSion D g 0 WW x w CIVIL EN 3130 SPRING 2015 LECTURE 3 Contents Reading Sections 15 and 16 of the text book Homework Problems 12 13 14 16 111 112 115 116 118 119 and 120 Odo Alum1K 3 agtNo F39WQL bN0439 LL n3m CuIonim MJ0L 20 W26 v lIS oooq Ms M1 112 N 5 C 377697 L10 coImEcTrso U 05 quot3 CIVIL EN 3130 LECTURE 3 VISCOSITY 2 1 Viscosity 0 From our earlier experiment we noted the following relationshi between shear stress and rate of angular deformation 7S U 7 f 1 M b 17 A 1 0 Recall that the constant of proportionality u is called the Viscosity of the uid o Viscosity is a measure of the uid s resistance to the Rate at which the uid ows 0 That is uids with a High viscosity eg molasses tar ow at a slower rate than those with a LOW viscosity eg water air 0 While Eq is a useful relationship it is of greater practical value to have this equation written in terms of a more measurable quantity of the ow such as VGIOCity rather than the angular rate of deformation 77 which is very dif cult to measure 0 To that end let s return to our previous experiment CIVIL EN 3130 LECTURE 3 VISCOSITY 3 YTq gtF FLUID U thichauif M y VOIOC39PH A dis l ribUH n Y9 7 Find Wadc 39 U 0 quot1 Aywmof In v aw 5mJ hag du lt 1 lyUU 37quot 4U Y W 11 f IO39 U o H fax 1339 dx veloci l39y39leme Pammen tqn 0 DL V d41th ooampK amp7 ARM9 op AASU M Dent if Q OH iv CIVIL EN 3130 LECTURE 3 VISCOSITY 4 0 To summarize we have the following relationship which is known as NeWtOn39S law of Viscosit ft El 3 2 Newtonian or o Fluids ar classi ed as NonNewtonian based on the relationship between applied shear stress and the resulting rate of deformation 0 When there is a Linear constant in 7 the uid is classi ed as Newtonian relationship between the two u is E 2 E a 0 When the relationship between the two is Nonlinear the quot 4 dY uid is classi ed as nonNewtonian o Gases and most common liquids tend to be NeWtOnian Some additional comments on viscosity 0 The dimensions of Viscosity in terms of F L and T can be determined directly from Newton s law of Viscosity ie F T L T FL 2 7 La A ILL 1 dudy LT L 5 o The SI unit of Viscosity is me n s o The USC unit of Viscosity is 94 o The Viscosity u is sometimes referred to as the absolute or dynamic Viscos ity to avoid confusing it with another quantity referred to as the kinematic viscosity 0 The kinematic Viscosity V is the ratio of Viscosity to mass density ie V E E 0 o Viscosity is a function of Temperature The Viscosity of a gas Increases with increasing tem perature but the Viscosity of a liquid Decreases with increasing temperature why See Figs Cl and C2 in Appendix C of the text for Viscosity values as a function of temperature for some common uids H xl CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 5 2 Mass weight and concentration variables In the previous lecture we de ned the Density of a uid as mass per unit volume PointWise this is de ned as Am p 2 11m AV gt0 A V 3 Where Am is the amount of mass contained Within AV Using density several related quantities are de ned The Volume per unit is the ie the reciprocal of the density 1 i p39 Vs m 4 Specific Volume I V 5 Mass 05 Units in SI are m3kg USC units are ft3 slug of a uid is de ned as its Specific Weight I Y Weight to density by the equation The per unit volume Thus speci c weight is related 1quot 3 Waiy v 5 lb 3 Units in SI are Nm3 USC units are The SDGClIlC Gravity I S of a uid is de ned as the ratio of the density p of the uid to the density of water speci ed temperature usually 4 C 2 392 In equation form at some s L 6 pH204 C Note this is a Dimensionless units quantity and thus has no CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 6 EXAMPLE PROBLEM The speci c weight of water at standard condi tions 4 C and atmospheric pressure is 981 kNm3 The speci c gravity of mercury is 1355 Compute a th thespeci c weight of mercury and c th ensity of merw SOLUTION IVL 5 39 qm 3 KW 17l MAIquot1l 3 5m 13 039 W twk hs39 399 I 7 Wt Mk 00 k m Iw Ooo kip13 13 SM I M DLMn33Jnl L3 I 3 77 a 11 SMYquot gt Y yw 7 Ym HS i7IKNm3gt Ym 3 7 quotNm e7 9 S the 7 139 SS000 kams Wm 11 0 quotJN i CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 7 Multicomponent mixtures For cases Where Several different forms of mass may be present in a uid we can de ne the following additional quantities The MiXture DenSity is de ned as Z Ami Z ApiAVi p 30 AV ea AV lt7 Where Ami AV and p are the mass volume and density respectively of the i th component of a mixture consisting of n total components The M833 FraCti0n of the i th component is de ned as PiAVz Ami Z 2 8 w pAV Am Note that w s 1 and is a dimensionlessquantity The Mass concentration of the i th component is de ned as Ami WAVE Ci AV AV 9 Note this quantity has the same dimensions as density The Volume Concentration of the i th component is de ned as AV AV Ci 10 Note that like mass fraction this quantity is dimensionless and c g 1 CIVIL EN 3130 LECTURE 3 TEMPERATURE AND THERMODYNAMIC VARIABLES 8 3 Temperature and thermodynamic variables SI Temperature o The Absolute temperature scale in the SI system has units of Kelvin K 0 Of more common use is the Relative temperature scale which has units of degrees Celsius 0 These absolute and relative temperature scales are related by the equa tion K 00 273 11 USC Temperature 0 The Absolute temperature scale in the USC system has units of degrees Rankine R 0 Of more common use is the RGIathe temperature scale which has units of degrees Fahrenheit F o In this case the absolute and relative temperature scales are related by the equation OR 2 OF 460 12 Conversion between degrees Fahrenheit and Celsius are given by 9 F 2 5 C 32 13 0 lt F 32 gt 14 CIVIL EN 3130 LECTURE 3 TEMPERATURE AND THERMODYNAMIC VARIABLES 9 Thermodynamic variables Your text book mentions several thermodynamic variables These are listed below QH 2 heat content cp 2 speci c heat at constant pressure CD 2 speci c heat at constant volume u intrinsic energy h enthalpy See page 16 of your text book for more details You are not expected to know the material related to these thermodynamic variables in great depth Prololm l Ll quotAir offers no resistancequot means 9 ASSUMEtun 0 SolidFL U QIZ A a 35 l L iyro all oLylyro J ELM LiNEHL CIVIL EN 3130 LECTURE 3 SOME USEFUL INFORMATION FOR THE HOMEWORK 10 4 Some useful information for the homework 41 More on the classi cation of substances 0 In the previous lecture we noted that substances can be classi ed as either Solid or Fluid based on how they respond to an applied shear stress o A Fluid was identi ed as a substance that deforms con tinuously under the action of a NONZERO shear stress no matter how small that shear stress may be 0 Again within the uid classi cation we identify two main different types of uids FLUID NEWTONIAN NONNEWTONIAN lt A linear r nship ionship between applied shear between applied shear stress and rate of deformation stress and rate of deformation CIVIL EN 3130 LECTURE 3 41 More on the classi cation of substances 11 To illustrate these relationships your text book provides the following Rheo logical diagram Rheology is the science dealing with the deformation and ow of matter Rain mlquotviieliirli39irilinIl II rill lilCElil ll iiiil Yield Elli SENSE T EillTrlz Ei k Note that in addition to Newtonian and non Newtonian uids 3 additional substances are plotted ThiXOtl OplC Substance TS Ideal Plastic IP and Ideal Fluid IF A TS is a particular type of rNewtorian uid hat shows a time dependent change in Viscosity the longer the uid undergoes shear stress the lower its Viscosity An IP is a SOIld that exhibits a linearrelation ship between applied shear stress and rate of deformation but only when the applied shear stress is greater than some de nite Bid StreSS W An IF is a Nonviscous and Incompressible uid It can be thought of as a limit quot0 ing case of a Newtonian uid with u gt 0 F r39l V O M CIVIL EN 3130 LECTURE 3 41 More on the classi cation of substances 12 HOMEWORK PROBLEM 12 Classify the substance that has the fol lowing rates of deformation and corresponding shear stresses d d rads 0 1 3 5 TkPa 15 20 30 40 SOLUTION 7 SIzPQSUJDMLJL mi Ai 099 NOT aquot mot Q V5 JUOly 0 O Sd193 l39mcL IS M n O IdemI PI I ty gtltpq J CIVIL EN 3130 LECTURE 3 42 The plate experiment revisited 13 42 The plate experiment revisited o In several of the homework problems you will encounter physical scenar ios similar to the plate experiment we discussed in the previous lecture 0 These problems do not in general explicitly state that 1 The uids are Newtonian and 2 The ow has a Linear velocituiistribution I However unless stated otherwise you are to assumw 0 Let s look at that particular case when both 1 and 2 are true yT Newtonian Fluid y0 l NLU L04lkA Fldiamp 39 t L 3 W as OOASM J Linw DN39I39r JoJHon 3quotquot A441 539 393 Slnpe aP HAL 39 30 U390 y Ay To 139 anuul 7K Assume Fi b fml lu CIVIL EN 3130 SPRING 2015 LECTURE 3 Contents Reading Sections 15 and 16 of the text book Homework Problems 12 13 14 16 111 112 115 116 118 119 and 120 Odo Alum1K 3 agtNo F39WQL bN0439 LL n3m CuIonim MJ0L 20 W26 v lIS oooq Ms M1 112 N 5 C 377697 L10 coImEcTrso U 05 quot3 CIVIL EN 3130 LECTURE 3 VISCOSITY 2 1 Viscosity 0 From our earlier experiment we noted the following relationshi between shear stress and rate of angular deformation 7S U 7 f 1 M b 17 A 1 0 Recall that the constant of proportionality u is called the Viscosity of the uid o Viscosity is a measure of the uid s resistance to the Rate at which the uid ows 0 That is uids with a High viscosity eg molasses tar ow at a slower rate than those with a LOW viscosity eg water air 0 While Eq is a useful relationship it is of greater practical value to have this equation written in terms of a more measurable quantity of the ow such as VGIOCity rather than the angular rate of deformation 77 which is very dif cult to measure 0 To that end let s return to our previous experiment CIVIL EN 3130 LECTURE 3 VISCOSITY 3 YTq gtF FLUID U thichauif M y VOIOC39PH A dis l ribUH n Y9 7 Find Wadc 39 U 0 quot1 Aywmof In v aw 5mJ hag du lt 1 lyUU 37quot 4U Y W 11 f IO39 U o H fax 1339 dx veloci l39y39leme Pammen tqn 0 DL V d41th ooampK amp7 ARM9 op AASU M Dent if Q OH iv CIVIL EN 3130 LECTURE 3 VISCOSITY 4 0 To summarize we have the following relationship which is known as NeWtOn39S law of Viscosit ft El 3 2 Newtonian or o Fluids ar classi ed as NonNewtonian based on the relationship between applied shear stress and the resulting rate of deformation 0 When there is a Linear constant in 7 the uid is classi ed as Newtonian relationship between the two u is E 2 E a 0 When the relationship between the two is Nonlinear the quot 4 dY uid is classi ed as nonNewtonian o Gases and most common liquids tend to be NeWtOnian Some additional comments on viscosity 0 The dimensions of Viscosity in terms of F L and T can be determined directly from Newton s law of Viscosity ie F T L T FL 2 7 La A ILL 1 dudy LT L 5 o The SI unit of Viscosity is me n s o The USC unit of Viscosity is 94 o The Viscosity u is sometimes referred to as the absolute or dynamic Viscos ity to avoid confusing it with another quantity referred to as the kinematic viscosity 0 The kinematic Viscosity V is the ratio of Viscosity to mass density ie V E E 0 o Viscosity is a function of Temperature The Viscosity of a gas Increases with increasing tem perature but the Viscosity of a liquid Decreases with increasing temperature why See Figs Cl and C2 in Appendix C of the text for Viscosity values as a function of temperature for some common uids H xl CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 5 2 Mass weight and concentration variables In the previous lecture we de ned the Density of a uid as mass per unit volume PointWise this is de ned as Am p 2 11m AV gt0 A V 3 Where Am is the amount of mass contained Within AV Using density several related quantities are de ned The Volume per unit is the ie the reciprocal of the density 1 i p39 Vs m 4 Specific Volume I V 5 Mass 05 Units in SI are m3kg USC units are ft3 slug of a uid is de ned as its Specific Weight I Y Weight to density by the equation The per unit volume Thus speci c weight is related 1quot 3 Waiy v 5 lb 3 Units in SI are Nm3 USC units are The SDGClIlC Gravity I S of a uid is de ned as the ratio of the density p of the uid to the density of water speci ed temperature usually 4 C 2 392 In equation form at some s L 6 pH204 C Note this is a Dimensionless units quantity and thus has no CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 6 EXAMPLE PROBLEM The speci c weight of water at standard condi tions 4 C and atmospheric pressure is 981 kNm3 The speci c gravity of mercury is 1355 Compute a th thespeci c weight of mercury and c th ensity of merw SOLUTION IVL 5 39 qm 3 KW 17l MAIquot1l 3 5m 13 039 W twk hs39 399 I 7 Wt Mk 00 k m Iw Ooo kip13 13 SM I M DLMn33Jnl L3 I 3 77 a 11 SMYquot gt Y yw 7 Ym HS i7IKNm3gt Ym 3 7 quotNm e7 9 S the 7 139 SS000 kams Wm 11 0 quotJN i CIVIL EN 3130 LECTURE 3 MASS WEIGHT AND CONCENTRATION VARIABLES 7 Multicomponent mixtures For cases Where Several different forms of mass may be present in a uid we can de ne the following additional quantities The MiXture DenSity is de ned as Z Ami Z ApiAVi p 30 AV ea AV lt7 Where Ami AV and p are the mass volume and density respectively of the i th component of a mixture consisting of n total components The M833 FraCti0n of the i th component is de ned as PiAVz Ami Z 2 8 w pAV Am Note that w s 1 and is a dimensionlessquantity The Mass concentration of the i th component is de ned as Ami WAVE Ci AV AV 9 Note this quantity has the same dimensions as density The Volume Concentration of the i th component is de ned as AV AV Ci 10 Note that like mass fraction this quantity is dimensionless and c g 1 CIVIL EN 3130 LECTURE 3 TEMPERATURE AND THERMODYNAMIC VARIABLES 8 3 Temperature and thermodynamic variables SI Temperature o The Absolute temperature scale in the SI system has units of Kelvin K 0 Of more common use is the Relative temperature scale which has units of degrees Celsius 0 These absolute and relative temperature scales are related by the equa tion K 00 273 11 USC Temperature 0 The Absolute temperature scale in the USC system has units of degrees Rankine R 0 Of more common use is the RGIathe temperature scale which has units of degrees Fahrenheit F o In this case the absolute and relative temperature scales are related by the equation OR 2 OF 460 12 Conversion between degrees Fahrenheit and Celsius are given by 9 F 2 5 C 32 13 0 lt F 32 gt 14 CIVIL EN 3130 LECTURE 3 TEMPERATURE AND THERMODYNAMIC VARIABLES 9 Thermodynamic variables Your text book mentions several thermodynamic variables These are listed below QH 2 heat content cp 2 speci c heat at constant pressure CD 2 speci c heat at constant volume u intrinsic energy h enthalpy See page 16 of your text book for more details You are not expected to know the material related to these thermodynamic variables in great depth Prololm l Ll quotAir offers no resistancequot means 9 ASSUMEtun 0 SolidFL U QIZ A a 35 l L iyro all oLylyro J ELM LiNEHL CIVIL EN 3130 LECTURE 3 SOME USEFUL INFORMATION FOR THE HOMEWORK 10 4 Some useful information for the homework 41 More on the classi cation of substances 0 In the previous lecture we noted that substances can be classi ed as either Solid or Fluid based on how they respond to an applied shear stress o A Fluid was identi ed as a substance that deforms con tinuously under the action of a NONZERO shear stress no matter how small that shear stress may be 0 Again within the uid classi cation we identify two main different types of uids FLUID NEWTONIAN NONNEWTONIAN lt A linear r nship ionship between applied shear between applied shear stress and rate of deformation stress and rate of deformation CIVIL EN 3130 LECTURE 3 41 More on the classi cation of substances 11 To illustrate these relationships your text book provides the following Rheo logical diagram Rheology is the science dealing with the deformation and ow of matter Rain mlquotviieliirli39irilinIl II rill lilCElil ll iiiil Yield Elli SENSE T EillTrlz Ei k Note that in addition to Newtonian and non Newtonian uids 3 additional substances are plotted ThiXOtl OplC Substance TS Ideal Plastic IP and Ideal Fluid IF A TS is a particular type of rNewtorian uid hat shows a time dependent change in Viscosity the longer the uid undergoes shear stress the lower its Viscosity An IP is a SOIld that exhibits a linearrelation ship between applied shear stress and rate of deformation but only when the applied shear stress is greater than some de nite Bid StreSS W An IF is a Nonviscous and Incompressible uid It can be thought of as a limit quot0 ing case of a Newtonian uid with u gt 0 F r39l V O M CIVIL EN 3130 LECTURE 3 41 More on the classi cation of substances 12 HOMEWORK PROBLEM 12 Classify the substance that has the fol lowing rates of deformation and corresponding shear stresses d d rads 0 1 3 5 TkPa 15 20 30 40 SOLUTION 7 SIzPQSUJDMLJL mi Ai 099 NOT aquot mot Q V5 JUOly 0 O Sd193 l39mcL IS M n O IdemI PI I ty gtltpq J CIVIL EN 3130 LECTURE 3 42 The plate experiment revisited 13 42 The plate experiment revisited o In several of the homework problems you will encounter physical scenar ios similar to the plate experiment we discussed in the previous lecture 0 These problems do not in general explicitly state that 1 The uids are Newtonian and 2 The ow has a Linear velocituiistribution I However unless stated otherwise you are to assumw 0 Let s look at that particular case when both 1 and 2 are true yT Newtonian Fluid y0 l NLU L04lkA Fldiamp 39 t L 3 W as OOASM J Linw DN39I39r JoJHon 3quotquot A441 539 393 Slnpe aP HAL 39 30 U390 y Ay To 139 anuul 7K Assume Fi b fml lu

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#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.