Fluid Mechanics Week 1-2 Notes (Lecture 2-5)
Fluid Mechanics Week 1-2 Notes (Lecture 2-5) CIVILEN 3130
Popular in FLUID MECHANICS
verified elite notetaker
Popular in Department
This 40 page Class Notes was uploaded by Aaron Bowshier on Friday January 23, 2015. The Class Notes belongs to CIVILEN 3130 at Ohio State University taught by Colton Conroy in Spring2015. Since its upload, it has received 237 views.
Reviews for Fluid Mechanics Week 1-2 Notes (Lecture 2-5)
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 01/23/15
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
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'