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# CE 3130 Midterm 2 Study Guide & Unit 3 Notes CIVILEN 3130

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This 76 page Study Guide was uploaded by Aaron Bowshier on Saturday April 4, 2015. The Study Guide belongs to CIVILEN 3130 at Ohio State University taught by Colton Conroy in Spring2015. Since its upload, it has received 268 views.

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CE 3130 FLUID MECHANICS SP 2015 MIDTERM 2 MIDTERM 2 FLUID DYNAMICS 1 STUDY GUIDE CONTENTS INTRODUCTION 1 GENERAL ASSUMPTIONS 1 CONTINUITY CONSERVATION OF MASS 1 STEADY STATE LINEAR MOMENTUM 2 Equilibrium 2 CONSERVATION OF ENERGY 2 POWER 2 BERNOULLI EQUATION 2 PROBLEMS 3 LARGE RESERVOIRS amp STORAGE TANKS ENERGYCONTNUITY 3 PIPELINES 3 FREE JETS 3 DIMENSIONAL ANALYSIS 4 BUCKINGHAM I39I THEOREM 4 Method Of Repeating Variables 4 PREDICTION EQUATION 6 Dimensional Analysis Example 7 INTRODUCTION System a collection of unchanging contentsmass particles Control Volume fixed space Lagrangian Description Analyzing the properties associated with the change of motion in a system of particles as a function of time Unsteady Flow Eulerian Description Observing the properties of fluid change at fixed spatial points Steady Flow l ALLA 4 I liliLl 141 I LI 1 l o 0 I FLow l 0 I O W I xgx Xz l CoNTRO L VOLUVAE GENERAL ASSUMPTIONS ASSUME o CONTROL VOLUME 0 STEADY FLowthe flow variables do not change in time o INCOMPRESSIBLE FLUIDpin pout CONTINUITY CONSERVATION OF MAss 3 A 2 Control Surface Area L2 Q volumetric f low rate Discharge m1 2 m2 PQ Q W1 pVAhn pVAout Incompressible Fluid pm pout Qin Qout Z VAin VAout STEADY STATE LINEAR MOMENTUM szzzmv Emv out in Newton s 2quot l Law Equilibrium F ma p A 2px 20utmvx2inmvx amp ZFy 20utmvy 2inmvy CONSERVATION OF ENERGY 1St Law of Thermodynamics Reversibilitywhen a process can be made to take place in such a way that it can be REVERSED that is made to return to its original state without change in the system or its surroundings z elevation head hL quotheadlossquot 2 losses due to friction hp 2 Energy from pump hT 2 losses from turbine p v2 p v2 y 9 out y g in POWER PP YQhPI PP Yth BERNOULLI EQUATION 1 2 1 2 171 511771 P921 2 172 511772 P922 PROBLEMS LARGE RESERVOIRS amp STORAGE TANKS ENERGYCONTNUITY Solution Strategy 2 v 1 Draw Control Volume H E our vm t V ngL nOZZle 2 Analysis of free Surface PO 3 Energy Bout Em hL identify unknown r f quotfi variables 5 mob l gt HEF pl 3i 4411 hL hqz E i 29 out quot 2L9 in l u gt2gtd vout 39 L 2 2quot hL E l E A 39 1212 4 Continuity to eliminate velocity gt 17m vout Ao ut 39 in quotquotquotquotquotquotquotquotquotquotquot quot PIPELINES 1 ControlVolume 2 Consider Forces weight of fluid 2 W pressure forces 2 F p A 3 MOMENTUM EQ 2 PW Zoutmv Zinmv j irwr me 2 w gt 192142 COS92 p1A1 cos01 va Emu in out in PaAgcos 4539 390 A s I o I rtd va Cos quotlsj39MV39 cos 5 w w FREEJETS OUT ASSOCIATED PROBLEMS 315 339 362 0 Neglect 0 Elevation changes gt p2p1 pv12 v22 1 F gt 93 2 m p lz v32 Friar jet i quot A me 0 Surface Tension Elli LJ V gt Pjet PATM 4333 1m39r 0 Weight of Liquid o Losses Due to Impact 1 Draw Control Volumeinletoutlet 2 Draw Forces on CV momentum fluxes gt 17m vout 170 3 Momentum Eq gt 2 PW Zoutmv Zmi hv DIMENSIONAL ANALYSIS Dimensional Analysis is an experimental study of fluids problems to check that the dimensions of each side of the equation are the same Similitude is a tool that is frequently used to help determine the behavior of fullscale Prototypes from experimental results obtained on similar usually smaller Models of prototype Model representation of a physical system used to predict its behavior BUCKINGHAM I39I THEOREM Buckingham l39l TheoremEquation having k variables is dimensionally the same being reduced to a relationship among k r independent dimensionless products where r is the minimum number of reference dimensions required to describe the variables F L T 7T1 107T2I7T3I mink r Method of Repeating Variables Example Problem A thin rectangular plate having a width w and a height h is located so that it is normal to a moving stream of fluid Assume the drag D that the fluid exerts on the plate is a function of w and h the fluid viscosity and density p and p respectively and the velocity v of the fluid approaching the plate Determine a suitable set of pi terms to study this problem experimentally 1 List all variables k of variables v Dpa D 2 Express variable in basic dimensions 1quot of basic dimensions M FLT U5n3 MI LIT D F 3 Vic7quot 5L E lt l 3 3 Determine the of T terms Ti terms k r Tf lj erquot 1 i j 4 Select repeating variables Never choose dependent var repeating var 1quot Hal LHW 39E Gui or Ll 1 I og 5 Obtain pi terms by forming products of repeating var and nonrep to Tr D u v fc V e mr L L T MLquot L H L T J gtco Forw 0 qb393 oquot L 9 a bo To r C s 7 o 1 b a LgTFTDHJVA l D I 39 anJ 6 Repeat Step 5 for each of the remaining nonrepeating variables REPEAT For hwgy bc 1710 E quotfillAbel39ng 7 543 1T3 Inc V 39b 39l 73G AIL 7 Check all pi terms are Dimensionless 8 Expressed final form as a relationship among rtterms 23mm i393Il 5 PREDICTION EQUATION Model is operated under conditions HZm 172 H3m 173 Hnm n ASSUME pmodez Pprototype H1 Him Example on next page Dimensional Analysis Example to 1m a a hem v and 39 andthe 1 mm is im 39 that in the 39 SDLUTIDM DIIIIJENBIMLLEBS murals sums THAT THE RE Is mE or THE RELEVANT 1n TERMB REM RE I SIMILITUDE ff um x gums F39s airs in 7 0m ii WM Li I i III CIVIL EN 3130 SPRING 2015 LECTURE 19 Contents 1 The conservation of mass 3 2 Introduction to the energy equation 10 Reading Sections 33 and 34 of the text book Homework Problems 32 and 34 from your text book CIVIL EN 3130 LECTURE 19 2 o In the previous lecture we discussed two different analysis approaches that can be used in analyzing mechanics problems 0 These are the Lagrangian and the Eulerian approaches o The LAG RANGIAN approach makes use of a xed quantity CONTROL SURFACE of matter known as a SYSTEM that 1s tracked as 1t moves about and interacts with its surroundings V I o The EULERIAN approach on the other hand makes use of I a xed region in space the CONTROL VOLUME CONTROL J l g VOLUME 0 In the Eulerian approach we observe the uid particles that are within the control volume at some instant of time and those that are passing through the control surface the boundary of the control volume 0 We also noted in our previous lecture that for our uid dynamics prob lems we would be deriving a set of basic equations from some of the fundamental principles of physics 0 In this lecture we introduce both the system and control volume forms of the MASS and ENERGY equations that we will be using 0 These equations can be derived by considering the principles of the CONSERVATION of mass and the rst law of THERMODYNAMICS respectively for a uid CRHLEN3HMLECTURE19 THECONSERWMHONCHFMASS 3 1 The conservation of mass 0 By de nition a SYSTEM is a collection of unchanging con tents so the conservation of mass principle for a system is simply time rate of change 0 I of system mass L or in equation form Where msyS is the total mass of the system 0 The total system mass is the sum integral of the differential masses dm of all the differential elements dV making up the system that is W4 folV ST 39SY Miyg 0 Thus the system form of the conservation of mass can be written as d a W 01V 9 lt1 CIVIL EN 3130 LECTURE 19 THE CONSERVATION OF MASS 4 OUTFLOW SURFACE INFLOW SURFACE CONTROL SURFACE o The conservation of mass principle for a CONTROL VOLUME is given by time rate of change net rate of ow of mass net rate of ow of mass I of total mass in I into the control volume out of the control volume the control volume 0 The left hand side of this statement can be expressed mathematically simply as PARTIAL DE IVATIVE time rate of change J 19 dl 2 of total mass in V C t Where the integral is taken over the entire control volume cv o Expressing the right hand side of this statement mathematically requires a little more work CIVIL EN 3130 LECTURE 19 THE CONSERVATION OF MASS 5 ZtZtll dt Located on out ow control surface 7 lcosdJ d F TT1DT 0 Let s begin by looking at the AMOUNT OF MASS that passes through some differential area dA on an out ow portion of the control surface in a small amount of time dt 0 Consider the FLUID PARTICLES that are located within the patch dA at time t 751 as shown in the gure above 0 These particles are traveling with a speed v so at time 752 751 dt they have traveled a total distance DLv DT 0 Thus the amount of mass dmout carried OUT of the control volume across the area element dA in the time interval dt is v N dmout pdV 0 CM dln pdA dllos6 pdA V dt cos6 dV n dz 0 Making use of the dot product identity V f1 V cos 9 where fl is the unit thus 1 outward normal of the area dA we have OZMWTI 024 it 0 And the INSTANTANEOUS RATE at which the mass is be ing carried out of the control volume across the the area element CM is dm t t dm dt t t A dmom3 lim 0u1 0u1 pVndA cit gt0 dt CIVIL EN 3130 LECTURE 19 THE CONSERVATION OF MASS 6 0 Now by integrating over the entire OUTFLOW portion of the control surface csout we obtain the net rate of mass ow or mass uX out of the control volume mom I dmout Z 0 V 39 fl o In a similar way the NFLOW portion of the control surface csin can be analyzed so the net rate of mass ow into the control volume is min pV dA 4 CSin 0 Thus the statement of the conservation of mass for a control volume can be expressed in equation form using equations 2 3 and 4 MASS IN MASS OUT 3 apdV pV dA pV dA 5 CV CSin V j CSout V J Eq 2 Eq 3 Eq 4 0 Now let s look at how this equation simpli es for some special cases 0 For example in the case of STEADY ow the ow variables do not change in time thus the left hand side of equation 5 is zero and we simply have that the net mass in ow is equal to the net mass out ow that is No STORAGE NOLOSS pV dA pV dA CSin 0 Furthermore if the in ow and out ow control surfaces can be de ned in such a way that the outward normal vectors f1 are parallel to the ow velocities then V f1 v and we have THE MASS THAT MA 4 MA LSM39 C our FLOWS OUT CIVIL EN 3130 LECTURE 19 THE CONSERVATION OF MASS 7 f g va o In the general case the density 0 and the velocity v are spatially V VARYING functions over the control surface 0 However in many cases it is reasonable to use spatially AVERAGED values of the density and velocity over the in ow and out ow control surfaces which we can pull out of the integral that is 7 73 COAS lvml39 pvdA pinUm dA where Am and A011 are the total areas of the in ow and out ow control pin ruin Ain pout vout Aout 0 v R5 pout vout CSout surfaces respectively 0 Using these approximations the conservation of mass for steady ow can then be written simply as MASS IN MASS OUT fm Vin FOlT VolT AOVT I L3 T 7 and thus represents a volumetric ow rate often called the discharge and denoted by Q ie QVA equation can be written as o It can be noted that the product 72A has dimensions of Thus the conservation of mass in Q fatTQOU V 6 Finally for an incompressible uid pin pout constant Thus for O steady incornpressible ow of single stream of uid we have 70 Con an gt39in 7 0lT QM 397 QOUT 7 lgt 0 You will be using this equation frequently CIVIL EN 3130 LECTURE 19 THE CONSERVATION OF MASS 8 EXAMPLE 31 At section 1 of a pipe system carrying water Fig 38 the velocity is 30 ft s and the diameter is 20 ft At section 2 the diameter is 30 ft Find the discharge and the velocity at section 2 SOLUTION CS OUT I CS N g W WATER v2 m ASSUM E STEADY FLOW CONMASSEQ AQPUT QIQQ vi39 Vl Am Q1 VIAI 3390 CHIQ Mr G 1 L 33 74 9LHSgt YLT71H V T 137 9 J CIVIL EN 3130 LECTURE 19 THE CONSERVATION OF MASS 9 CS OUT CSIN39I r VA l lzjvj j 0 We conclude this secctsiothg noting two things 0 First for STEADY ow involving multiple inlets and or out lets we must SUM the in ow and out ow mass ow rates that is EM uow lt8gt 0 Second note that the general conservation of mass equation for the con trol volume equation 5 can be written as NET FLUX QA Q where the second integral is over the entire CONTRO SURFACE both in ow and out ow portions 0 Comparing this with the system form of the conservation of mass equa tion we can see that for a CONTROL VOLUME and a SYSTEM that are coincident at some instant of time the two forms of the equations are related by Oi 0 V Q 0H 74 d dV 8 dV AdA 9 p p pV n dt sys cv cs REYNOLDS TRANSPORT THEOREM 0 We will make use of this relationship in subsequent developments CIVIL EN 3130 LECTURE 19 INTRODUCTION TO THE ENERGY EQUATION 10 2 Introduction to the energy equation 0 The ENERGY EQUATION that we Will be using for our problems can be derived from the rst law of thermodynamics o The rst law of thermodynamics for a system is of 1 net time rate of energy 1 net time rate of energy I I ofthe I I added by added by I 0f the System l o In equation form this can be written as d 6 10 Qnet Wnet dt sys sys If LAGRANGIAN where e is the INTERNAL ENERGY per unit mass we will say more about this in the next lecture Qnet is the net rate of HEAT TRANSFER into the system and Wnet is the net rate of WORK TRANSFER into the system The rates of heat and work transfer are considered if the net rates are into the system and if they are coming out o The CONTROL VOLUME form of the equation can be ob tained by making use of equation 9 on the previous page that related v J the system and control volume forms but lookmg now at energy per unit mass 6 9 a pdV epdV epV dA 11 CIVIL EN 3130 LECTURE 19 INTRODUCTION TO THE ENERGY EQUATION II 0 Using equations 10 and 11 the control volume form of the rst law of thermodynamics is i QNV t eV39 cm My CV J4 NE CIVIL EN 3130 SPRING 2015 LECTURE 20 Contents 1 Summary of the continuity equation 2 2 The energy equation 21 Steady state energy equation 22 Reversibility irreversibility and losses 23 The Bernoulli equation CONCECJO Reading Sections 34 and 35 of the text book Homework Problems 315 313 331 334 CIVIL EN 3130 LECTURE 20 SUMMARY OF THE CONTINUITY EQUATION 1 Summary of the continuity equation 0 In the previous lecture we derived the basic conservation of mass equation or CONTINUITY EQUATION for a uid for both a system and a control volume 0 As previously noted the control volume or EULERIAN form of the equation is generally more useful in uid mechanics than the system or LAG RANGIAN form of the equation 0 We summarize What we found in the case of the control volume below For a CONTROL VOLUME the basic continuity equation has theform i VQA Jb if Q J v J k I v Where Q the time rate of change of the total mass in the control volume cv lt2 2 the net rate of ow of mass through the control surface cs For INCOMPRESSIBLE ow involving one stream of uid the basic continuity equation for the control volume reduced to the sim ple equation Q 711141 712142 Where 0 is taken as thef the component of the velocity nor mal to the control surface area A and Q is the volumetric ow rate or DISCHARGE which has dimensions LA3T CIVIL EN 3130 LECTURE 20 THE ENERGY EQUATION 3 2 The energy equation 0 In the previous lecture we also introduced the basic energy equation which was obtained from the FIRST LAW OF THERMODYNAMIOS in both system and control vol ume form 0 The basic energy equation for the control volume is summarized below For a CONTROL VOLUME the basic energy equation as the d 5 form ON 5 Fquot 39 l J A 39 5 fv39ndA WNET NET C 0i C LS V o o o C where the time rate of change of the total stored energy in CV f d the net rate of ow of total stored energy through cs a 39 I the net time rate of energy addition by work transfer into CV r the net time rate of energy addition by heat transfer into CV 0 Terms and Q in the energy equation above involve the TOTAL STORED ENERGY per unit mass 8 This consists of three parts Total stored energy per unit mass 6 Potential energy per unit mass Kinetic energy per unit mass Intrinsic energy per unit mass 0 Let s look at each of these energy terms CIVIL EN 3130 LECTURE 20 THE ENERGY EQUATION Potential energy per unit mass You may recall that GRAVITATIONAL potential energy is given by the equation Potential energy mg Z Where 2 is the vertical height above an arbitrarily assigned reference point or datum Thus the potential energy per unit mass is given by 3 2 Potential ener er unit mass 39 3 Z gy p l M a Dimensions of this term are energy or work per luanit mass or FL 01 units of this term are Newton meters per kilogram USC units of this term are lb ft slugs which reduces to dimensions of 73 SI Kinetic energy per unit mass You may recall that kinetic energy is given by the equation I 1 Kinetic energy Q Thus the kinetic energy per unit mass is given by 1 1 l 1 L Kinetic energy per unit mass T c I The dimensions and units of this term are the same as noted above for the potential energy term Intrinsic energy per unit mass gtltgtlt The intrinsic energy per unit mass denoted by u in your text book is due to molecular spacing and forces dependent on p 0 or T We will discuss this at greater length in the last subsection of today s notes CIVIL EN 3130 LECTURE 20 THE ENERGY EQUATION 0 Using the information from the previous page the total stored energy per unit mass 6 is given by a V 4r lt 6 q 1 1 51 l39 le 1 Pot Kln Int 0 We now move on to term the net time rate of energy addition by work transfer into the control volume Wnet o The net work transfer rate Wnet also called POWER can be separated into two parts Wnet W8 Wp where WS is the work transfer rate due to SH EAR forces such as the torque exerted on a rotating shaft and WP is the work transfer rate across the control surface by PRESSURE forces 0 Using the formula for power this latter term WP can be shown to be equal to WP pvndA 2 0 Using equations 1 and 2 we can now write the energy equation as a p 02 A 39 39 epdv gz u pVndAWsQnet at W CS p 2 0 As we did for the continuity equation we will now look at how this equation simpli es for steady ow CIVIL EN 3130 LECTURE 20 21 Steady state energy equation 6 21 Steady state energy equation STEADY FLOW o In the case of the rst integral of the energy equation involving the time derivative is zero and the energy equation simpli es to p v2 A gzz3u pVndAZWS l Qnet o If the properties within the parentheses 92 p p 1122 and u are all as UNIFORMLY sumed to be distributed over the control sur face areas then the integration simply becomes 2 p 3902 Zgz u m p 2 out 2 gz1 9v upv dA 2 CS p p 3902 E gz u m p 2 in where m is the mass ow rate which is equal to va 0 Thus the energy equation becomes 2 p v p v 39 39 Zltgzz3u m Zltgzz3u MZWS l Qnet out in ONE STEADY 0 Now if we are considering only stream of uid entering and leaving the control volume the energy equation can be written simply as where we have divided through by the mass ow rate m CIVIL EN 3130 LECTURE 20 22 Reversibility irreversibility and losses 7 22 Reversibility irreversibility and losses 0 A PROCESS can be de ned as the PATH of the succession of states through which a system passes such as the changes in velocity elevation pressure density or temperature 0 When a process can be made to take place in such a way that it can be REVERSED that is made to return to its origi nal state without change in the system or its surroundings it is said to be REVERSIBLE o In reality any actual process is IRREVERSIBLE due to ef fects such as friction o This irreversibility represents a loss in useful or available energy P V o In the energy equation the terms 7 139 and I t represent the useful or available energy 0 The remaining terms in the equation represent the loss of useful energy Speci cally considering the loss in a steady stream of one uid m m Ws Qnet loss 2 um uout 0 Using this de nition the steady state energy equation from the previous page can be written as 2 2 gz l Z Qv gz l Z Qv loss p 2 out p 2 in 0 Finally we note that this equation is often written in terms of energy per unit weight instead of energy per unit mass by dividing through by g P v P l L Td l f d l a 13 E Y 9 8i L OUT where we have de ned in 2 loss g 0 Note the dimension are simply L units m or ft This length is referred to as head for instance the 2 term is referred to as the elevation head CIVIL EN 3130 LECTURE 20 23 The Bernoulli equation 8 23 The Bernoulli equation 0 In the case of negligible losses the steady state energy equation for one STREAM of uid written between any two points 1 and v2 v2 gz 9 gz1 9 p 2 1 p 2 2 0 Now if the uid under consideration is also incompressible p1 2 p2 p 2 is simply then multiplying through by p we obtain P v r 32d v f 7 54 o This is the celebrated BERNOULLI EQUATION named in honor of Daniel Bernoulli a o More generally this equation can be written as P43400432 o This same equation can also be derived by applying Newton s sec ond law F 2 ma along a streamline of STEADY FRICTIONALESS INCOMPRESSIBLE ow 3 0 To use this equation correctly it is always important to keep in mind the basic ASSUMPTIONS used in its derivation Daniel Bernoulli February 8 1700 March 8 1782 was one of the many prominent mathematicians in the Bernoulli family who made many important contributions to the elds of mathematics and mechanics He has been described as by far the ablest of the younger Bernoullis CIVIL EN 3130 SPRING 2015 LECTURE 23 Contents 1 The system linearmomentum equation 3 2 The control volume linearmomentum equation 5 21 Basic equation 5 22 Steady state equation analysis 6 Reading Sections 36 and 37 of the text book Homework Problems 362 363 367 372 373 385 and 3115 CIVIL EN 3130 LECTURE 23 2 0 Up to this point in our study of FLUID DYNAMICS we have derived and subsequently applied two important equations for analyzing uid dynamics problems 0 These two equations the continuity and energy equations were ob tained from the following laws of physics Q THE LAW OF CONSERVATION OF MASS THE FIRST LAw OF THERMODYNAMICS ENERGY EQ E OUT E IN Hc HL HP o In this lecture we will derive the LINEAR MOMENTUM EQUATION for a uid 0 As was done for both the continuity and energy equations we will con sider the momentum equation for both a SYSTEM and a CONTROL VOLUME 0 These equations can be obtained from the consideration of NEWTONS SECOND LAw o The form of Newton s second law with which we are all familiar is ZFZMQ 0 You have used this equation in your study of RIGID BODY motion 0 Recall that a rigid body is a body or a system of particles in which the distance between any two given points of the body remains CONSTANT in time Le DEFORMATION is neglected 0 Here we will consider a slightly different form of Newton s second law which is more applicable for FLUID MOTION CIVIL EN 3130 LECTURE 23 THE SYSTEM LINEAR MOMENTUM EQUATION 3 1 The system linearmomentum equation 0 Newton s second law is basically a statement of the CONSERVATION OF LINEAR MOMENTUM o For a system this can be expressed in words as the time rate of change 1 the sum of the l of the linear momentum J l external forces J of a system acting on the system 0 Recall that MoMENTUM mentum of a small particle of mass dm with a velocity V is vdmPW is velocity times mass so the mo momentum SYSTEM 0 The total momentum of the entire can be ob tained by summing integrating the momentum of all the particles in the system total linear momentum of system W V W 0 Thus the conservation of linear momentum of the system can be ex pressed in equation form as aw Ky iit S vfdlZZ f lt1 o In the following exercise we demonstrate how the ABOVE expression with which we are all familiar can be obtained from this equation in the special case when the system being considered is a CIVIL EN 3130 LECTURE 23 THE SYSTEM LINEAR MOMENTUM EQUATION 4 EXERCISE Derive the familiar E F 2 ma form of Newton s second law from equation 1 of the previous page for the special case of rigid body motion SOLUTION START WITH EQ 1 ad F iy FOR A RIGID BODY EVERY PARTICLE HAS THE SAME VELOCITY V OR V SPATIALLY CONSTANT oI CIVIL EN 3130 LECTURE 23 THE CONTROL VOLUME LINEAR MOMENTUM EQUATION 5 2 The control volume linearmomentum equation 2 1 Basic equation 0 As was the case for the EN ERGY and CONTINUITY equations we will primarily be con cerned with the control volume form of the linear momentum equation 0 Again we will make use of the equation used in previous lectures relat ing the system and control volume forms of conservation equations that is QUANTITYMASS d l 8 A REYNOID39s diV diV NpV dA TRANSPORT dt sys cv cs THEOREM 0 Setting N1MAssMASS in the equation above we ob tain the equation relating the system and control volume forms of the CONTINUITY EQ 0 Setting NEENEReylMAss we obtain the equation relating the system and control volume forms of the ENERGY EQ 0 Finally by setting NVMOMENTUMMASS we ob tain the equation relating the system and control volume forms of the MOMENTUM EQ o The CONTROL VOLUME form of the momentum equation can then be obtained by comparing equation 1 of the previous page for a system to the equation above with N V 0 Speci cally the control volume form of the momentum equation takes the form am A VfCll l v avo 26 2 l Jt a CIVIL EN 3130 LECTURE 23 22 Steady state equation analysis 6 22 Steady state equation analysis so the term of the momentum equation will 0 Again we will primarily be analyzing steady ows TIME DERIVATIVE be ZERO 0 Thus the steady state form of the control volume momentum equation is M FLOWLACROSS VpV dA ZFCV 0 As was the case for the continuity and energy equations if we de CONTROL VOLUME ne our such that the ow velocity 1 LET OUTLET is perpendicular to the and control surfaces and make the approximation that ow properties are UNIFORM over it then the integral in the steady state form of the equation is simply vag v lwi L VpVf ndA gmmv where again m is the mass ow rate dimensions of MT 1 STEADY STATE 0 Thus our linear momentum equation for a control volume is mam2w o The right hand side MOM ENTUM EXCHANGES this the out of and into the con terms of equation are trol volume respectively 0 Let s look at a simple example from your text book in order to understand how we apply this equation CIVIL EN 3130 LECTURE 23 22 Steady state equation analysis EXAMPLE 39 Suppose in Figs 320 and 321 the following pipe conditions apply The in ow and out ow radii are 25 and 15 cm respectively the in ow and out ow angles with respect to the horizontal 61 and 62 are 45 and 30 respectively Q is 50 L s the area average inlet and outlet pressures are 85 kPa and 583 kPa and the total uid weight in the pipe is 20 N Find the horizontal and vertical force required to hold the pipe in place ie the resultant force vector F SOLUTION Dam A Q FED Q In ng Iz MOMEATIM CIVIL EN 3130 LECTURE 23 22 Steady state equation analysis 8 if f I chVxfm ra P A L05 qu Aagpsjog ax L D my c0570 m C03 LL W 1M Co TIAUiTY SJ FindFAME quotLILVI V1 M Mt 099 50 local J 5 OE 23 Q WAIgta 03905 y 41 TrOI5mL Q 17 07 mS MOMENTUM EQUATION gt 7 XIO7 nggtltVltOQ M 05 52 5quot VKOIffcOS 7039 FAX xzyomg rpmb CIVIL EN 3130 SPRING 2015 LECTURE 24 Contents 1 Applications of the linear momentum equation 2 11 Flow through pipelines 2 12 Flow of free jets 3 Reading Section 37 of the text book Homework Problems 362 363 367 372 3 385 and CIVIL EN 3130 LECTURE 24 APPLICATIONS OF THE LINEAR MOMENTUM EQUATION 2 1 Applications of the linear momentum equation 11 Flow through pipelines 0 So far we have looked at a MOMENTUM EQUATION ex ample involving uid ow in a pipe 0 In these types of problems we are typically asked to solve for the ANCHORING FORCES that are required to hold the pipe in place o In addition to the anchoring forces the other forces that are generally present in these problems include the WEIGHT of the uid in the pipe and the PRESSURE FORCES exerted on the INFLOW and OUTFLOW control surfaces 0 Additionally there are typically MOM ENTUM EXCHANGES into and out of the control volume through the in ow and out ow control surfaces respectively 0 Special care must be taken to insure that correct signs are used for the FORCE and MOMENTUM EXCHANGE V30 tors 2 lolAILoS ngo mlv Cos HSj M39 05 Q5 vVV GUT CIVIL EN 3130 LECTURE 24 12 Flow of free jets 12 Flow of free jets Free jet 1 Nozzle I 0 Another typical set of problems that is analyzed using the momentum equation involves the ow of what are called FREE JETS o A FREE JET is a owing stream or jet of liquid that is not contained within a pipe and is OPEN to the air or perhaps some other uid 0 Free jets are observed for example in ow issuing from a NOZZLE or DUCT as shown in the gure above o The study of free jets is of practical importance because the FORCE that arises fromafree jet IMPACTING a surface is often used to power machinery o For example the FORCE the jet of liquid exerts on the vanes of a turbine wheel as shown in the gure above will act to set the wheel in motion 0 A full analysis of the ow of FREE JETS is a very com plex uid dynamics problem however these problems can be made more tractable through a number of simpli cations o For example given that the dimensions of a free jet are relatively SMALL we can typically neglect the WEIGHT of the liquid comprising the free jet 0 We introduce the additional simpli cations made in our analysis of free jets by considering some examples A vane is a thin at or curved object that is rotated about an axis by a ow of uid or that rotates to cause a uid to ow or that redirects a ow of uid eg the vanes of a windmill CIVIL EN 3130 LECTURE 24 12 Flow of free jets HOMEWORK PROBLEM 362 What force F Fig 361 is required to hold the plate from owing oil Sp gr 083 for V0 2 20 ms SOLUTION l F 1 i OIL 7 LF CD our i i f 9 id F x I f I 4 l 4 a Z MVK ZENMVO OUT I S9 CIVIL EN 3130 LECTURE 24 12 Flow of free jets 5 0 With the previous problem we took the PRESSURE within the free jet to be EQUAL TO the local atmospheric pressure 0 In reality due to the effects of SURFACE TENSION the pressure within a jet will be slightly GREATER than the local atmospheric pressure 0 Related to this idea you may recall in our discussion on sur face tension we demonstrated that the pressure within a spherical LIQUID DROP relative to local atmospheric pressure is h 2 given by t e expression DROP a R where 051 is the surface tension of the liquid and R is the RADIUS of the drop o A similar expression can be derived for the pressure within a CYLINDRICAL free jet of radius R 0 Speci cally the PRESSURE within a free jet when consid ering surface tension is given by the expression P R Fmse IET Z 0 However for most problems involving FREE JETS it is rea sonable to NEGLECT this pressure o Neglecting this pressure along with losses due to impact and the small changes in elevation if any between different sections of a free jet leads to an important result regarding the VELOCITY of the free jet at different sections 0 We consider these two SIMPLIFICATIONS to our problems for free jets in the following exercise and example respectively CIVIL EN 3130 LECTURE 24 12 Flow of free jets EXERCISE Derive the expression given on the previous page for the pres sure Within a free jet when considering surface tension and demonstrate that it is reasonable to neglect this pressure for homework problem 362 SOLUTION CIVIL EN 3130 LECTURE 24 12 Flow of free jets FOR THE FREE JET OF HW PROBLEM 3622 amp R25MM Us r quot 7 9 x 0 Nr39l Gil a F L 31quot 2 qt N15 2v g T r fQoVo IAN 59th 15 X163 a quotF 3 QOVD 0K 9 NtaLuml an3104 in 4rd CIVIL EN 3130 LECTURE 24 12 Flow of free jets EXAMPLE 315 Fluid issues from a long slot and strikes against a smooth inclined at plate Fig 327 Determine the division of ow and the force exerted on the plate neglecting losses due to impact SOLUTION CIVIL EN 3130 LECTURE 24 12 Flow of free jets CIVIL EN 3130 LECTURE 24 12 Flow of free jets 10 In summary we employ the following simpli cations to problems involving free jets G Neglect weight of liquid in free jet gt p free jet 2 p local atm Neglect surface tension of free jet 0 free jet is gt gt the same at C3 Neglect elevation changes if any between different sections of free jet all sections Q1 Neglect losses due to impact General solution strategy for momentum problems Construct an appropriate CONTROL VOLUME and identify the INLET and OUTLET control sur faces Identify the FORCES acting on this control volume and the MOMENTUM EXCHANGES through the control sur faces Using the control volume diagram write down the STEADY STATE linear momentum equation ZFCV out Special care must be taken to insure that proper SIGNS are used for the force and momentum exchange vectors Q1 Identify the unknowns in the momentum equations and use the CONTINUITY and0139 ENERGY equations as appropriate to determine unknown Ivelocitiesflow rates and or at the control surfaces 5 Use these values in the MOMENTUM equations to solve for the unknown FORCES CIVIL EN 3130 SPRING 2015 LECTURE 26 Contents 1 Introduction 2 2 Dimensional Analysis 3 3 The Buckingham H theorem 7 4 Common dimensionless groups in uid mechanics 11 Reading Sections 51 52 and 53 of the text book Homework Posted on Carmen CIVIL EN 3130 LECTURE 26 INTRODUCTION 2 1 Introduction 0 Up to this point we have mainly solved our uid mechanics problems using EQUATIONS and ANALYTICAL proce dures 0 However a large number of problems encountered in uid mechanics can not be solved by ANALYSIS imental results alone and rely in part on exper 0 Thus in this chapter we will brie y cover some ideas and techniques related to the EXPERIMENTAL chanics problems investigation of uid me look at and main concepts S I M I LITUDE 0 Speci cally we will two DIMENSIONAL ANALYSIS 0 Dimensional analysis is a conceptual tool that all of us have used that DIMENSIONS involves the of physical quantities o For example we often use dimensional analysis to check that equations HOMOGENEOUS that is to check that SAME are dimensionally the dimensions of each side of the equation are the o In the next few pages we will demonstrate that the concepts of dimensional analysis are a powerful tool in deriving suitable DIMENSION LESS combinations of variables that are helpful in the experimental study of uids problems 0 SIMILITUDE is a tool that is frequently used to PROTOTYPES help determine the behavior of full scale from experimental results obtained on similar MODELS usually smaller of prototypes DIMENSIONLESS 0 Generally the parameters mentioned above that are related to a particular problem must be equal to achieve SIMILARITY or similitude between the prototype and the model 0 We will discuss this concept at greater length in the next lecture CIVIL EN 3130 LECTURE 26 DIMENSIONAL ANALYSIS 3 2 Dimensional Analysis 0 To motivate our interest in DIMENSIONAL ANALYSIS we consider the following ow problem that we wish to examine by experi mental methods Flow scenario Steady ow of an incompressible Newtonian uid through a long smooth walled horizontal circular i Experimental Objective 9 a quotI To characterize the pressure drop per unit length of pipe that occurs as a result of losses due to friction o This problem although it appears to be simple cannot in general be solved by ANALYSIS alone and requires the use of supporting experimental data 0 To set up an experiment for this problem one of the rst things we want to do is think about what VARIABLES or FACTORS that will affect the pressure drop per unit length Ap L 0 We might expect this list to include such things as the uid velocity v the PIPE DIAMETER D the uid density 0 and the FLUID VISCOSITY o Mathematically we express this as CIVIL EN 3130 LECTURE 26 DIMENSIONAL ANALYSIS 4 Ap D p u constant AP 11 p u constant L L U D Ap D U u constant Ap D p v constant L L p M 0 To attempt to determine the speci c form of the function f we might run CHANGE one of the vari CONSTANT a series of experiments in which we ables or parameters while holding the others o In doing so we would end up with a series of plots similar to what is shown above 0 While this approach seems logical in concept there are lDISADVANTAGES to doing things this way For ex ample 1 It would be dif cult if not impossible to COMBINE all the data from the various curves to obtain the GENERAL FUNCTION relating ApL to all the variables and factors 2 Obtaining some of the curves would be dif cult in practice For exam ple how would one vary uid DENSITY while holding VISCOSITY constant 0 However there is a better approach that avoids these dif culties CIVIL EN 3130 LECTURE 26 DIMENSIONAL ANALYSIS 5 0 Although not obvious at rst it is simpler to variables of the problem into two va2 COLLECT the groups DIMENSIONLESS of variables 0 We can then write a new functional relationship the in terms of DIMENSIONLESS variables as opposed to the ORIGINAL variables ie 0 There are several 1 2 p p021 g ADVANTAG ES to this approach TWO We are now working with only variables instead of FIVE the original This means the EXPERIMENTS would now just con sist of varying vau and determining the corresponding DAp p021 value The results of the experiment could then be represented by a SINGLE UNIVERSAL curve as illustrated above that VALID would be for any combination of smooth walled pipe and incompressible Newtonian uid Furthermore since we are now working with D I M E N 8 IO N L E SS variables the results obtained would be independent of the UNITS we are using o In the following exercise we verify that these new variables are in fact dimensionless CIVIL EN 3130 LECTURE 26 DIMENSIONAL ANALYSIS 6 EXERCISE Verify that the two variable groups vau and DAp p021 of the previous page are in fee dimensionless SOLUTION CIVIL EN 3130 LECTURE 26 THE BUCKINGHAM H THEOREM 7 3 The Buckingham H theorem 0 Two fundamental questions that arise from the previous consideration are i how does one determines the DIMENSIONLESS products for a given problem and ii how MANY of these products are required to REPLACE the original list of variables 0 An answer to the second question is provided by the so called BUCkingham H theorem named in honor of Edgar Buck ingham o This theorem can be stated as follows Buckingham H theorem If an equation involving k variables is dimensionally homo geneous it can be reduced to a relationship among k 7 independent dimensionless products where 7 is the minimum number of of reference dimensions required to describe the vari ables o The proof of this theorem is rather technical and is not included here 0 What the theorem tells us is that a physically meaningful equation in volving k variables such as 1 fu27 3739 16 can be written in terms of a set of k 7 dimensionless products 1 1 H27 H37 quot397 His 7 where 7 is determined by the minimum number of reference dimensions required to describe the original list of variables 0 The DIMENSIONLESS products H are often referred to as quotPI TERMSquot CIVIL EN 3130 LECTURE 26 THE BUCKINGHAM H THEOREM 8 0 Although the use of the 1391 THEOREM may initially seem to be a little esoteric and complicated several simple and systematic procedures for determining the H terms for a given problem have been developed 0 We outline one such procedure below Method of repeating variables APL D V List all the VARIABLES that are involved K5 in the problem K VARIABLES 2 Express each of the variables in terms of BASIC F I T DIMENSIONS R R OF REF DIM Determine the required number of 1391 TERMS KR 2 Q1 Select a number of REPEATING variables 3 REPEATING Where the number required is equal to the number of reference di mensions v 5 Form a H term by multiplying one of the 2 NON39REP39 NON39REPEATING variables by the product of OK F b L the repeating variables each raised to an exponent that Will make T the combination dimensionless Repeat step 5 for each of the REMAINING non repeating variables Check all the resulting H terms to make sure they are DIMENSIONLESS Express the nal form as a RELATIONSHIP among the H terms and think about What it means CIVIL EN 3130 LECTURE 26 THE BUCKINGHAM H THEOREM 9 EXAMPLE PROBLEM Use the method of repeating variables to deter mine the H terms for the pipe ow problem discussed previously SOLUTION 0 HDmav a K5 if 1 13 Di Ll39 iFL39qTa394 FLQT39 5L and H TTennsz kf39 5 3 9 ChocCg DIV and f or A triadREP A a b TE T D V prismwm aS LAE 0 lo lt For Tr 5 DWENSIonLESS L7 nquot 4 FL3L LT 91301471 F Hm 4 F Lquot T 39 J IN ORDER FOR THIS TO BE TRUE YOU MUST HAVE gt39lt 1 F 0 9 3JrorHo9L20 For a a bac 0 2m63 1 L AP I 1 ma EDT 21 97 TD 2 quot 2 Q o I A A H 7 l l fv CIVIL EN 3130 LECTURE 26 THE BUCKINGHAM H THEOREM 10 6 rst DWV 7T1FL39 TgtltLQLT397bltFL39qTC PD 7quot j 4 0 F9 F LI QJO Llo O Karl 6 1bag 0 For T 6 ALREADY CHECKED DIMENSIONS IN CLASS em CIVIL EN 3130 LECTURE 26 COMMON DIMENSIONLESS GROUPS IN FLUID MECHANICS 11 4 Common dimensionless groups in uid mechanics In the table below we list several dimensionless groups that commonly arise in uid mechanics See your text book for more information concerning some of these groups Dimensionless Types of Groups Name Interpretation Applications pvl inertia force Reynolds number Re Generally 1mportant 1n u V1scous force all types of ow Froude number Fr inertia force gravitational force Flow with a free surface ressure force L Euler number Eu Flows 1n wh1ch pressure p122 1nert1a force differences are of interest p122 inertia force Cauchy number Ca Flows 1n wh1ch u1d com K compress1b1l1ty force pressibility is important 1 inertia force Mach number Ma Flows in which uid com 0 compress1b1l1ty force pressibility is important wl inertia local force Strouhal number St Unsteady ow W1th a 39v 1nert1a convectlve force characteristic frequency of oscillation 2 1 l 1nert1a force p Weber number We Problems in which sur 0 surface tens1on force face tension is important Variables p density 39v Velocity l length chviscosity 9 gravity p pressure K bulk modulus c speed of sound w frequency a surface tension CIVIL EN 3130 SPRING 2015 LECTURE 27 Contents 1 Model studies and similitude 2 Reading Section 56 of the text book Homework POSTED PROBLEMS ON CARMEN om the text 5001 CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 2 1 Model studies and similitude o MODELS are frequently used as part of the DESIGN process in engineering o In the present context we de ne a model as follows De nition of a model A MODEL is a representation of a PHYSICAL SYSTEM that can be used to help PREDICT the behavior of the system in some desired respect o The physical system that we are considering is called the PROTOTYPE o In most cases models that are SMALLER than the prototype are used in the design process because they are LESS EXPENSIVE to construct and EASIER to handle in a laboratory 0 Occasionally the use of a model that is LARGER than the prototype may be advantageous if the prototype itself is too SMALL to be easily studied using standard lab equipment 0 Of course in order for a MODEL to be able to successfully predict the behavior of a prototype it is imperative that it be properly DESIGNED and TESTED and that the results from the testing be INTERPRETED correctly 0 The principles of DIMENSIONAL ANALYSIS that we cov ered in the previous lecture can be used to help insure that the model and prototype will behave in a SIMILAR fashion or have SIMILARITY CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 3 0 Recall that the BUCKINGHAM 1391 THEOREM tells us that a physically meaningful equation involving k variables U1 fu27 3739 16 can be written in terms of a set of n k 7 dimensionless products l39lltlgtnnn 1 0 To formulate this relationship we only need to know the VARIABLES involved for a particular problem 0 Speci c VALUES or UNITS for the variables are not required 0 Thus Eq 1 applies to any SYSTEM that is governed by the same variables 0 For example suppose Eq 1 describes the behavior of a particular PROTOTYPE A similar relationship can be used to de scribe the behavior of a MODEL of that prototype ie Hlm H2m7 H3m7 39 39 397 Hnm where the subscript m is used to indicate the model 0 Now suppose that 13911 is the variable that is to be predicted from the model study If the model is OPERATED and DESIGNED under the following set of conditions H2m H2 1 3m 2 1 3 3 Hnm Hn then it follows assuming that f has the same form for the model and the prototype that n1n1M 4 CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 4 0 Eq 4 of the previous page is sometimes referred to as the PREDICTION EQUATION o The conditions speci ed by Eqs 3 are the SIMILARITY REQUIREMENTS or model design conditions 0 These are the conditions that need to exist between the MODEL and the PROTOTYPE in order for the model to properly simulate the behavior of the prototype ie in order for the prediction equation to give useful results 0 In one of the dimensional analysis examples we considered the prob lem of investigating the DRAG FORCE D on a thin RECTANGULAR PLATE of size h x w placed normal to a uid of viscosity u and density p owing with a velocity v o The DEPENDENCE of the drag force D on these VARIABLES can be expressed mathematically as D fltw7 h M p7 v K6 R 3 K R 3 0 Applying the Buckingham H theorem to this problem gave the following dimensionless relationship D m h 2 0 We note that the third H term in this relationship very frequently in uid mechanics problems and is referred to as the 79w M 1 a w I appears REYNOLD39S NUMBER RN the ratio of the VISCOUS interpreted as of the uid to the o Physically it can be FORCE o In the following exercise we look at how we might design a model for this problem CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 5 EXERCISE Investigate how a model would be designed for the drag force problem discussed on the previous page Speci cally determine and discuss the similarity requirements and the prediction equation SOLUTION DRAG FORCE FROM DIM ANALYSIS SIMILARITY REQUIREMENTS Tra Tram Tr 1an FW 772 i w h h g39 m 1 MODEL MOST BE SCALED PROPERTY GEOMETRIC SIMILITUDE For 171 39 7 T VIM Mm W Kc ZL39M AliV MHa 777 r 777 CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 7 0 Generally as discussed in the previous example all the corresponding H terms must be EQUATED between the model and the prototype in order to have SIMILARITY 0 Typically one or more of the H terms will involve ratios of LENGTH o Equating these for the model and the prototype means that the model must be a SCAI ED VERSION of the prototype o This is known as GEOMETRIC SIMILARITY between the model and the prototype o In addition to this one of the H terms will typically involve ratios of FORCE o This indicates that the ratio of LIKE FORCES for the model and the prototype must be the SAME o For example in the previous exercise the REYNOLDS NUMBER RE was the relevant force ra tio but in other scenarios a different force ratio may be of importance o For example in ows with a FREE SURFACE the ratio of the inertia force to the GRAVITATIONAI FORCE is impor tant o This force ratio or DIMENSIONI ESS VARIABLE is known as the Froude number FR v SQRTCGL L DEPTH OF CHANNEL 0 When these types of H terms are equal we have what s called DYNAMIC SIMILARITY between the model and the pro totype 0 When both geometric similarity and dynamic similarity exist then it follows that VELOCITY and ACCELERATION ratios of the model and prototype are the same throughout the ow eld 0 This is known as KINEMATIC SIMILARITY o All three similarities geometric kinematic and dynamic must ex ist between a model and a prototype in order to have what s called COMPLETE SIMILARITY 0 We conclude with an example CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 8 EXAMPLE PROBLEM Model tests are to be performed to study the ow through a large valve having a inlet and carrying water at a The working between model and prototype and the model inlet diameter is 3 in I that should be used in the model SOLUTION DIMENSIONLESS ANALYSIS SHOWS THAT THE RE IS ONE OF THE RELEVANT n TERMS THUS REM RE DYNAMIC SIMILITUDE fm Vm Um f mE FL VIII 504quot AJM Vmo 7 pm CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 9 HOMEWORK PROBLEM E CAN CHOOSE A or 11gt 9 mm A C u v W 9CO qb3ltO 9 a LOquot D L 1T 7 Duquotfquot 7 oLr a b 3 C Mn T f a m REPEA7 FOZ 7E 39 hDJOWbc 1T MLTLT 9 MLquotL m L T Forw0 L CA Q3 RECTANGULAR PLATE WIDTH W HEIGHT H DRAG D FCWH I d h 9 TF1 J a 90 gt IQb 3L 5 77 73 0 az I bC Q L UJU CIVIL EN 3130 LECTURE 27 MODEL STUDIES AND SIMILITUDE 6 CIVIL EN 3130 AUTUMN 2014 LECTURE 29 Contents 1 Introduction 2 2 Laminar and turbulent ows 2 Reading NA Homework NA CIVIL EN 3130 LECTURE 29 INTRODUCTION 2 1 Introduction 0 In our previous lectures covering FLUID DYNAMICS we considered a number of topics concerning the FLOW of uids 0 Simpli ed equations describing the MASS MOMENTUM and ENERGY Of uid in motion were derived and applied to a number of different ow situations DIMENSIONAL ANALYSIS 0 The concepts of were also intro duced to aid in looking at problems that cannot be investigated through analysis alone 0 In our concluding chapter we will apply these various principles and concepts to the speci c and important topic of the ow of VISCOUS incompressible uids in PIPES o In particular we will take a closer look at the nature of the LOSSES due to friction that occur in the ow of real u ids in pipe networks FRICTION FACTORS that can be used in our equations knowledge of the type of ow with which 0 In order to describe appropriate we are dealing is required 0 To this end we will begin our discussion of VISCOUS ow in pipes by looking at the important classi cation of ow as either LAMINAR or TURBULENT 2 Laminar and turbulent ows 0 In the late 1800 s name of Osborne Reynolds made an important experimental obser PIPES a British scientist and mathematician by the vation in the uid ow of that would lead to one of the most important concepts in the general study of the ow of VIscous FLUIDS Osborne Reynolds August 23 1842 February 21 1912 was a British scientist and mathematician who made a number of important discoveries in the eld of uid mechanics He was one of the rst professors in British history to hold a title of Professor of Engineering CIVIL EN 3130 LECTURE 29 LAMINAR AND TURBULENT FLOWS 3 8 gt Laminar Dye Q 11A Dye streak I Ill Transitional INCREASING VELOCITY fr Pipe W gt Turbulent o A sketch of the experimental apparatus used by Reynolds is shown above 0 It consists of a PIPE of diameter D with water owing through it at an average velocity v and a small NOZZLE that injects a neutrally buoyant dye into the ow 0 From a series of experiments Reynolds observed the following 6 For quotSMALL FLOWRATESquot the dye streak re mains in a well de ned line with only slight visual blurring of the line due to molecular diffusion of the dye into the surrounding water For Slightly larger quotINTERMEDIATE FLOWRATESquot 7 the dye streak uctuates mildly in time and space with occasional bursts of irregular ow patterns appearing For quotLARGE FLOWRATESquot upon exiting the nozzle the dye streak almost immediately lLUCTUATES wildly and moves across the pipe in what appears to be a random fashion 0 These three different ow regimes that are observed are classi ed as 1 LAMINAR 2 TRANSITIONAL and 3 TURBULENT ows CIVIL EN 3130 LECTURE 29 LAMINAR AND TURBULENT FLOWS 4 o The term LAMINAR FLOW comes from the word lamina which means a thin plate or layer 0 That is the uid can be thought of as moving in thin layers or LAMINA one on top of the other that do not mix o The term TURBULENT has its roots in the Latin word turba meaning confusion 0 Indeed turbulence is a very confusing subject o In fact a complete theory describing the nature of TU R B U L E N CE is still lacking after all these years and remains one of the greatest UNSOLVED problems of physics and mathematics Interesting Digression According to one story when the German physicist Werner Heisenberg for which the Heisenberg Uncertainty Principle is named was asked If given the opportunity what would you ask God he replied When I meet God I am going to ask him two questions Why relativity And why turbulence I really believe he will have an answer for the rst 0 However what we do know is that the FUNCTIONAL de pendence of various parameters important in pipe ow is usually different for laminar and turbulent ow 0 Thus it is important to be able to DISTINGUISH between these two different FLOW REGIMES CIVIL EN 3130 LECTURE 29 LAMINAR AND TURBULENT FLOWS 5 0 To this end we note that our previous descriptions of SMALL and LARGE owrates upon which we classi ed LAMINAR and TURBULENT ow are inadequate in that they need to be de ned relative to some reference quantity 0 This is where the concept of the REYNOLDS NUMBER introduced in the previous chapter comes into play 0 Recall that Reynolds number is the dimensionless quantity de ned as l D IF D 15 HIGH BOUNDARY EFFECTS Low R L A 7IF VISCOSITY 15 HIGH SHEAR DOMINATES AND RE GOES DOWN which is the ratio of the inertia and viscous forces of the ow 0 Thus the ow in a pipe is classi ed as laminar transitional or turbulent based on the size of the Reynolds number 0 In other words it is not only the velocity of the ow that determines the ow classi cation but the DENSITY 7 VISCOSITY and pipe size as well 0 Now the RANGES of Reynolds numbers for which the ows are laminar transitional or turbulent cannot be PRECISELY DEFINED 0 However for general engineering purposes the following values are typi cally used for round pipes CD The ow is LAMINAR if RE lt 2100 Q The ow is TURBULENT if RE gt 4000 0 Between these two limits TRANSITIONAL FLOW the ow may switch between laminar and turbulent in what appears to be a RANDOM fashion o In the next couple lectures we will investigate laminar and turbulent ow in pipe systems in more detail CIVIL EN 3130 AUTUMN 2014 LECTURE 30 Contents 1 Energy considerations in pipe ow 2 2 Friction losses in laminar ow 3 Reading NA Homework NA CIVIL EN 3130 LECTURE 30 ENERGY CONSIDERATIONS IN PIPE FLOW 2 1 Energy considerations in pipe ow 0 As mentioned in the previous lecture in this nal chapter we will take a FRICTIONAL closer look at the energy losses that occur for the ow of real uids through PIPELINES The term quotREAL FLUIDSquot has been used here to highlight the fact that unlike the conceptual notion of IDEAL FLUIDS which have zero viscosity all uids possess some VISCOSITY From Newton s law of viscosity I L d d that viscosity gives rise to SHEAR STRESSES conditions you may recall under ow These shear stresses in turn create FRICTIONAL FORCES that transform the useful energy terms potential pressure and kinetic J into thermal energy heat This action is what produces what we have referred to as the HEAD LOSS IA In our previous considerations we touched brie y on these losses but did not characterize them in great depth Thus one of our primary goals for this nal chapter is to establish use ful relationships for computing the FRICTIONAL LOSSES encountered in pipe ow As a rst step to this end consider the ENERGY between and for the pipe shown EQUATION above CIVIL EN 3130 LECTURE 30 FRICTION LOSSES IN LAMINAR FLOW 3 0 Given that the pipe is horizontal 21 22 and has a constant diame ter D 3901 02 the PRESSURE DROP AP HEAD LOSS HL is simply equal to the between G and C2 divided by the speci c weight of the uid ie p1p2amp 7 7 hr o For LAMINAR FLOW the pressure drop and thus in can be explicitly determined from analysis to be a function of the uid and p1pe propert1es We w1ll look at th1s 1n the next sectlon 39 2 o For TURBULENT FLOW dimensional analysis and experi mental results must be used to formulate a frictional relationship 2 Friction losses in laminar ow 0 To examine the frictional losses in laminar ow consider the summation of forces in the c direction on a small element of uid within the pipe as shown in the gure below A EEC we T27T dw pdp7r7 2 0 where we have assumed steady ow acceleration a 0 o This expression can be solved in terms of the shear stress to give C0LP ft 5 dx 1 0 We also know from Newton s law of viscosity that the shear stress is equal CIVIL EN 3130 LECTURE 30 FRICTION LOSSES IN LAMINAR FLOW 4 E vmax 7 0 L Velocity pro le 717 0 Setting 1 and 2 equal and rearranging we have 9 r fquot 1 dp l d P d L d 7 d7 J r K 2M d3 A d D a pJ 4quot I FL 09 where we have used the fact that dy 2 d7 A amp o This expression can be integrated we will skip the details to obtain the VELOCITY PROFILE as a function of the radial distance 7 from the center of the pipe 1 dp 7 2 D2 7 7 2 u da 2 8 o This indicates that the velocity pro le 717 is PARABOLIC O The MAXIMUM velocity vmax occurs at 7 0 the center line of the pipe that is by setting 7 0 in Eq 3 we obtain D2 dp vmax 1617 d3 3 O The AVERAGE velocity 77 in the pipe is related to this max imum velocity by the expression 0 Combining Eqs 3 and 4 and solving for dp we obtain 3227 Jl dpDld 7 L CIVIL EN 3130 LECTURE 30 FRICTION LOSSES IN LAMINAR FLOW 5 0 Integrating this expression between points and that we used for the ENERGY EQUATION we obtain C fir HT 7 A 321le 5 p D2 A C L where again Ap 2 p1 p2 and L 32 31 is the length of the pipe 0 Recall from our earlier consideration of the energy equation that the HEAD L088 in the pipe was given by hL 6 0 Thus substituting Eq 5 into 6 we obtain an expression for the head loss in pipes for LAMINAR FLOW V 321le l V 2 3 Vma hL 702 where we have dropped the bar over 0 for notational convenience AVERAGE VELOCITY o By algebraic Inanipulations this equation can be written in the form 321le quotr192 hL DINJ XQXEX g 0 641 L 112 X X va D 29 where we have used the fact that y 2 pg V J 0 Since A e the head loss can be written as L m VLIaLJI39x Hem0L L AMINA L CIVIL EN 3130 LECTURE 30 FRICTION LOSSES IN LAMINAR FLOW o The dimensionless factor 64Re from the last equation on the pre Vious page is often replaced by the symbol f which is called the FRICTIONAL FACTOR 7 that is L 2 L a 114 L LTD C13 where again f 64Re o This equation is known as the DARCY WEISBACH equa tion 0 Although we derived the expression for the head loss in a horizontal pipe with laminar flow it is also applicable to NONHORIZONTAL pipes with laminar flow as well 0 Furthermore in the next lecture we will explain how the Darcy Weisbach equation is also applicable for computing head loss for TURBULENT FLOW provided a modi ed far more com plex friction factor is used in place of the LAMINAR FLOW friction factor of f 64Re

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