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# Fluids Final SG&NOTES CIVILEN 3130

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CE 3130 FLUID MECHANICS SP 2015 FINAL FINAL EXAM FLUID DYNAMICS II STUDY GUIDE CONTENTS Fluid Definition 1 Newtonian vs NonNewtonian Lecture 3 1 Pressure 1 Linear Uniform Acceleration 3 Steady Flow vs Unsteady Flow 6 System vs CV 6 Viscous Pipe Flow 6 Solution Strategy 6 Uniform OpenChannel Flow Problems 7 Hydraulic Jump Problems openchannel flow 10 Manometer Problems Chapter 2 12 Forces of Plane Area Problems statics chapter 2 13 Special Problems 16 Forces on Plane Areas with Manometer Problem 16 BONUS Momentum Problem 18 PIPELINES 18 FREE JETS 18 Fluid Definition A fluid is a substance that cannot support a nonzero shear stress while remaining at rest Lecture 2 Newtonian vs NonNewtonian Lecture 3 I FLU l l I 39u NEWTONIAN NONNEWTONIAN iship 39onship bet ween applied shear c ween applied shear stress and rate of deformation stress and rate of dcfonnat ion Newton s Law of Viscosity iiiu F u T T Adm 32H Pressure Compressible vs Incompressible Liquids are nearly incompressible while gases change in volume under compression p change of pressure needed to Bulk Modulus used to characterize K dVV change some volume Pascal Pa Cavitation occurs at Vapor Pressure EXAMPLE PROBLENI A standard atmospheric greasy and a temper ature of 60 F how mu uld be required to cmnprcaa a unit volume of water 1 SOLUTION V dVV 00 T 42039 TaUL C 2 K67 O39F quotELIIDOO 213 De ilion of Ik Modulus 7 in Z K Dlt WM 7 310 V a if 3uooooo ZJIO gsii Var HISJ Fo 100 Comf 5ton p w N 1NODNWamp5JIBLE 00 N A Linear Uniform Acceleration N o w cm mum r mm It MuttMn quot ed A Fvut d W X at y 839 L IN 65 0F oN MT run C PAAMLEL 139 F296 susFAte K UduFoIvA A accel tanw V y Rum 939quot nd 1 P 7 X fl 1 a k 3 P quot0 9 FREE n A Q f SoLVE m FaunaI 0P Y J h j G Y A 4 am IS VKO 5 quot 3 rlki V L06 39 9quotquot ut39 ANGs lAT u39 quot pwamw x J i 5 AHGL CATW 839 7940311051 39 Example Problem 2128 A rulricnl hux l m m an utlgv upvn at 11139 up and 1m lllfll with water is plural on nu imlimvi plnm quottakingquot 30 mnglc39 wiih tlu lmri39luulnl 39l lu lmx ulum39 weighs 500 N mul ms u rm irimu uf f1 Minn with Hue plum uf 030 lnt 1uim39 Ilu39 ann39ulumliun uf Hue lmx maul Ilu39 mmlv llw 39rWwmw39 mrfmv nmkm with tlw hm39immnl SOLUTION v1 90 Y M 85011 SOLUTION 21 amp LVN 1 VJ pl3V39IouL 1 Voox V ho H 00 4 t and 19306 rJV gt v ogt 39 X 715 quot J soul 4 05 a2gt391 NJ 403 N 0 an Kan 0 V Jig D v u o o FF a a 3 F41 J 139 ff chu i 03057144 9F 403 ms 3o39 jq F FNQ539J N b N zm J V51 30 1 m CHOg K W m 3quot39679IN 3903 4 W A 393 Q 135 quot53 5442va 1110 MNNGW L O A a o 05 3039 t 1355 5009 0 3 10 51 quotHquot w mm 30 pmc give For rm swr 0F TH F1155 Sultan s t 6 04 41031 to Q I 3 ml 6 6mquotozu 913 Steady Flow vs Unsteady Flow Unsteady Flow Lagrangian Description Analyzing the properties associated with the change of motion in a system of particles as a function of time Steady Flow Eulerian Description Observing the properties of fluid change at fixed spatial points Steady Flow occurs when conditions do not change with time at any goint System vs CV System a collection of unchanging contentsmass particles 0 The mass of a system does not change with respect to time o The time rate of change of the linear momentum of a system EQUALS ZERO Control Volume fixed space Viscous Pipe Flow 3 Different Flow Regimes Determined by Velocity Density Viscosity and Pipe Size 9Reynolds Number Re 2 Laminar Re lt 2100 For quotSmall flowrates dye remains in a well defined line Transitional 2100 S Re S 4000 For quotIntermediate Flowrates dye fluctuates mildly in time and space occasional irregularity bursts Turbulent Re gt 4000 For quotLarge Flowrates dye immediately fluctuates wildly and at random Solution Strategy 1 Energy Eq 2 Major amp Minor Losses 3 Find Reynolds Re 4 Find f Moody Diagram P0 for a free jet Laminar if l g lr g 39Tmmitiuml ll 5 WW Willy Tiub i em I l yquot V0 p0 for large reservoirs open to atm Uniform Open Channel Flow Problems APTEILNOmV PDILIE 03 I RESRVOILSgt Q 7 4347 s IraCLUOEZ MmeL brick4 395 F 2quot 0390 S AJD L 900 9 MM quot5 2 K v FIND 22 L ao g MInor LoSM i H goo a P39ampI 0 IE jg L iiiL Closed G DVIVCI E 4 kW E E 41 In 4r 72 45 V y 3 wwa hy r Lovae Benveins o 0 owl PL 3 0 M4 V M01 V2 3 0 L923 00Fgt yL amp V V 1 3 Zing 3 Vtoss s NATOL pawn PW 1 7 s CDMTLMUILV asv 7 V A quot74 0M1 S a Hi kl 0015W 2 A 367 1557 W quot 5 5392 W Z lop f39 L 2 91 Pk 1 Example Problem I PE EXAM AFTERNOON PROBLEM 505 A long concretelined drainage channel has a slope of 0001 ftft a bottom Width of 10 ft and side slopes The water depth is 5 ft and the Mappings roughpess goef fie39n t is 0012 Th Froude numberXor the channel is most nearly I FIND f f A 008 B 018 C 066 D 080 6E0 Enrle 39 1 lo 7 7 la F l 1 GIVEN So 000 00 quot o N v 39 5 DH 1 27 v2 SOLUTION fa 001 g MaulVINGS 3 3 Sn n A f r 5 A Dh A l 7 DH Aw 2 Mome 3 f 3 33 10quot Law and DA f I 1 F mas739 glp L S quot r 5 222 971 2353 550 Hydraulic Jump Problems open channel flow 7 i f5 a R FIDLY V lfco aquot g H WWquot I 3W GWEN 139 me at our 4quotf gtFmrv quot1r P n H ip uLl J39U WTquot AW THE EASE M A S lLI mm t i F A DA 5 Even TINAquot THE bE rTlH UPSTREAM Amp wnWH E h a1 fHE Juiwa AWE i J mesMHWELT E i F THE SWFILL WM 115 gm DVE I L SF iLL39wam f WHAT 0ng HATED ET TH E Ju W1quot Hm a gamma MEAM NIN Z S El SDM WIDE Y2 ELTAMAW r H ANNE L HYDMLIL 33mquot Ea 2 Jim07 1 3quot f H k 2 Fr A 9 quot I E q Q93 V 10 M w W 1amp1 Q V 3 Hi55 Samgt0 gag 7 36933 Du Er33 7 v2 mtgW a g E Y 13 m g V1291 1quot L A Vega D 1quot 2 r i 1 4 2 Q 94W 41 69 M 217 41 L F 9 Pat YGAL 9906 Wm336 735 90 em O O 7 11 Manometer Problems Chapter 2 227 In Fig 244 4 contains water and he umuonwlcr uid has a specific gravity of 294 When he left minimum is at zero on the scale pA 100mm H20 Find the wading of the right uwnizscus for L4 8 kPu with no adjustuwnl of the U tube or scale Figure 244 Problem 227 SOLUTION 12 Forces of Plane Area Problems statics chapter 2 0 Understand the prob Conceptually 1 Draw FBD 2 Find pressure forces of Areas a By Equation gammahbar A hbardistance from free surface to the centroid of the area of the surface the force is acting b By Pressure Prism 3 Write Appropriate Force andor Moment Eqs 4 Solve for Final Answer 13 Fole s 0 PLANE AIMEA39S mLmaxn rwmuncumcs 4 EXAMPL IS 7 Question Set 2 12 pts total 9 Pulley Square Gate 1 ft width P o NotmT henmopentobmlumhtm The square gate shown above is hinged at O and attached to a spherical weight 139 048 ft via a cable in tension that peace over a frictionless pulley The uid isoilwith aspeci cweightof lbft 21 12 pts Determine the weight W of the sphere that will keep the square gate mechanism in the position shown W 39 00 59 ANS 14 SOLUTION l u T r Hx 4 H a ZJ 3 39 m lt FWMM f T L L4 0 I It F80 j Faom139gt2 3 GEV7070 FaeJ FTP3 quot T 0 gt 7 7a I VE2 0 Frow Przrsssurze FILI r Ravi Fa MIN Y f v W 2 T4 YeasL V M 09347 I 1 75 an 439 Sq 39bF3 I 5 LWquot loo0 ml Special Problems Forces on Plane Areas with Manometer Problem 272 Find the moment M iil 0 Fig 263 to hold the gale closed 16 SOLUTION HW 97 NA Of lE 731 PT 1 YA 7 0 Q i quot 4 PM W m q 06quot E L siws39 39 95 i6 FTC PT Ya P6 ng Yu 0 1 0 quot Tquot L quot 3 quot 39 L35 F w U Pugh a 5 5quot L7 33 o 6 PfL LSh MGTHoD 39 x i guess I 3quot h w 3 57 R 3 3YXQXQ J gym E x W 11mm 3 Eli109 r FAQOA FB vimzp 1 FbKiQ FL4 in 7 3 L135 y5 3951 Y my bW 3J7 H QB HW 154 k 5 51 H 17 BONUS Momentum Problem Flow of Free Jets Neglect CV Momentum Eq Identify unknowns Energy PIPELINES 1 ControlVqume 2 Consider Forces weight of fluid 2 W pressure forces 2 F p A 3 MOMENTUM Eo 2 Fa Zoutmv Emmy 2F Z V 2 quotV V 39 M 0quot M gt pzAz cos02 p1A1 cos01 Z mv Z mv f out in PaA g as Pl A go 9 6 4 Va 5quotquot quM V cos 4 S 1 V FREEJETS OUT 3 ASSOCIATED PROBLEMS 315 339 362 0 Neglect 0 Elevation changes gt P2 P1 500712 1722 gt P3 P1 500712 1732 0 Surface Tension gt Pjet PATM 0 Weight of Liquid 0 Losses Due to Impact Draw Control Volumeinletoutlet Draw Forces on CV momentum fluxes gt 17m vout 170 Momentum Eq dentify Unknowns gt 2 Fa Zoutmv 2mm 12 Energy Equation PWN I WILL POST A COMPLEX EXAMPLE BONUS PROBLEM FOR FREE JETS LATER 18 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 CIVIL EN 3130 SPRING 2015 LECTURE 33 Contents 1 Brief summary of previous lecture 2 2 Friction losses in turbulent ow 4 21 Theoretical analysis of turbulent pipe ow 4 22 Dimensional analysis of turbulent pipe ow 6 3 Minor losses 9 Reading NA Homework NA CIVIL EN 3130 LECTURE 33 BRIEF SUMMARY OF PREVIOUS LECTURE 2 1 Brief summary of previous lecture o In the previous lecture we began our investigation of Friction losses that occur in the ow of real u ids in pipes o All real uids have some ViSCOSitY which give rise to Shear stresses within the uid 0 These shear stresses in turn create Friction forces that are responsible for energy losses 0 Our analytical investigation of this process for laminar ow proceeded along the following lines Laminar pipe ow analysis From energy considerations it was rst shown that the Head loss in a horizontal pipe of constant diameter with steady ow is simply given by K F1 quot F l A P v l L Y 1 where Ap is the pressure drop that occurs across the pipe 2a0 2 By considering the balance of forces Newton s second law on a uid el ement within the pipe a relationship between the Pressure and Shear forces within the pipe was obtained 3 Making use of Newton s law of viscosity 739 advdy the expres sion for the balance of forces was then integrated to obtain the vertical VelOCitY profile of the ow within the pipe which was shown to be Par abOl ic with the maximum velocity occurring along the centerline of the pipe CIVIL EN 3130 LECTURE 33 BRIEF SUMMARY OF PREVIOUS LECTURE 3 Laminar pipe ow analysis cont d A second integration of the force equation gave us an explicit expression for the pressure drop Ap as a function of the uid viscosity u the wge uid ow velocity v and the length L and diameter D of the pipe namely 2AM v L V 39 2 Ca Finally by substituting 2 into 1 and through a series of algebraic ma nipulations the head loss in a pipe for laminar ow was written in the form of the so called Darcy Weisbach equation V441 7 ll39IcQDl where f 64Re is referred to as the Friction factor o The expression above is applicable for determining the Friction loss associated with Laminar ow in pipes 0 You may recall that in a previous lecture we de ned D Rb ltC1OO 7 AMWM Flow CIVIL EN 3130 LECTURE 33 FRICTION LOSSES IN TURBULENT FLOW 4 2 Friction losses in turbulent ow F 4 l V to 21 Theoretical analysis of turbulent pipe ow 0 In most practical situations TurbUlent flOW is actu ally more likely to occur in pipes than laminar ow 0 Thus there is an obvious need to be able to characterize the Friction losses in turbulent ow as well 0 However as mentioned previously turbulent ow is a very complex process 0 Recall that the fundamental difference between laminar and turbulent ow is the seemingly Random behavior of various uid properties in turbulent ow 0 For example as shown in the gure above consider the Velocity vVt measured over a period of time at a given location in a pipe with turbulent ow 0 The distinguishing feature of turbulent ow is the irregular Fluctuations that occur 0 Over some period of time T as illustrated in the gure above these turbulent uctuations average out to give a steady Time average value of the velocity 17 ie V p 27 t0Tvtdt CIVIL EN 3130 LECTURE 33 21 Theoretical analysis of turbulent pipe ow 5 0 Given that the uctuations average out over a period of time the Time Average uid ow properties can often be used in turbulent ow analysis in place of the true Fluctuating properties 0 Speci cally steps 6 and Q of the analysis for Laminar ow in pipes can also be applied to the analysis of Turbu l ent ow in pipes with the understanding that relevant variables are replaced by their ounterparts 0 However the real dif culty of the turbulent ow analysis comes in ap plying step when Newton s law of viscosity was used to express the Shear Stresses in the force equation in terms of the Flu id Viscosity u and the uid ow velocity v 0 At this point in the analysis it is tempting to try ex press the shear stresses in turbulent ow by simply using the Time Averaged values of the velocity 17 in place d Of the actual Fluctuating velocity Q 39I M 0 That is we might attempt to express the shear stress in turbulent ow using Newton s law of viscosity as d1 7turb 0 However numerous Experimental and Theoretical studies have shown that this approach leads to a completely Incor r eCt characterization of shear stresses in turbulent ow that is Tturb 7E ud39Ddy o This is due to the fact that the turbulent uctuations of the velocity also make a Significant contribution to the shear stresses o The shear stresses that arise from these turbulent uctuations are termed Reynolds Stresses 0 Thus to proceed with a purely Theoretical analysis the question of how one computes the Reynolds stresses must be answered 0 To date despite a considerable amount of research in this area there is NO general theory to answer this question CIVIL EN 3130 LECTURE 33 22 Dimensional analysis of turbulent pipe ow 6 o The net result of this is that we cannot proceed with a purely theoretical analysis of turbulent flow as was done for the laminar flow case 0 That is it impossible to to integrate the force balance equation to obtain the turbulent Velocity PFOfile in the pipe and other useful information to explicitly compute the friction losses in turbulent flow 0 This is the reason why most turbulent flow analyses are based on DimenSional Analysis experimental data and Semi Empirical mmum 22 Dimensional analysis of turbulent pipe ow o The Pressu re D rop Ap for steady incompressible tur bulent flow in a horizontal round pipe can be written in functional form as Apfv7D7L7 7M7 Y where v is the average velocity D is the diameter of the pipe L is the length of the pipe 6 is a measure of the Roughnes S of the interior pipe wall u is the uid viscosity and p is the uid density 0 One important thing to note is that pressure drop for turbulent flow is Dependent on m Roughness mmhempe This is NOt the case for laminar flow o The seven variables k 7 of the functional relationship given above can be written in terms of the three reference dimensions M LT 7 3 0 Thus from the BUCkingham n Theorem we know that k 7 4 dimensional groups are required to write the function in a dimensional form R 9 Method of re eatin 0 One such representation is given by TI TI J I TI39V va riables p g I Ap 3va L e p39v2 u D D CIVIL EN 3130 LECTURE 33 22 Dimensional analysis of turbulent pipe ow 7 0 We can pull the LD factor out of the function E if we assume that the P rGSSU re d FOP is directly proportional to the Length of the pipe a reasonable assumption which al lows us to re write the expression as i A P l3 D Ap L e 2 R where we have also used the fact that the rst independent dimensionless group is the Reynolds Number o This expression can be solved for Ap and then substituted into the pre vious expression derived for the head loss in Apy to give TIRoughness factor quot friction factor 0 Note that if we call the function o the F riCt ion factor f then this expression is in the form of the Darcy Weisbach equation we saw previously in the case of laminar ow 0 In order to nish characterizing the friction losses in turbulent flows it remains to specify the Functional Dependence of Re and ED the friction factor on 0 Much of this functional dependence information was obtained as a result ofnuumunm Experiments o This information is typically displayed graphically in what is called the Moody chart see supplemental hand out and Figure 621 of your textbook which can be used to determine a value of the friction factor f for a given Re and e D CIVIL EN 3130 LECTURE 33 22 Dimensional analysis of turbulent pipe ow 8 0 Three distinct regions can be observed in the Moody chart G For Laminar ow f 64Re which is independent of the relative roughness of the pipe 6 D For what is called Completely or wholly turbulent flow very large Re the friction factor is independent of the Reynolds number and solely depends on the relative roughness of the pipe ie f Me D 3 In between these two regions for flows with MOderate Re the friction factor is dependent on both the Reynolds number and the rel ative roughness of the pipe ie f 5Re eD o The nonlaminar part of the Moody chart can actually be expressed in equation form by the following 1 eD 251 W 2010g37 Re o This equation is an empirical t of the experimental pipe flow data and is called the C013br00k formula 0 The dif culty with using it is its ImpliCit dependence on f 0 That is it is not possible to solve the equation Explicit 1y in terms of f therefore an IteratiVe scheme must be used to solve it o In summary the head losses due to friction for both Lamina r and TU rbUlent flow in pipes can be succinctly written in the form 2 f For Re lt 2100 L1 h here gt L fD29 W f L 2Olog63l7 251 For Regt4000 7 Rev Major LOSSES 7 gt 7 2 M CIVIL EN 3130 LECTURE 33 MINOR LOSSES 9 3 Minor losses 0 Most pipeline systems consist of more than just St raight pipes j 0 Additional components such as va IVES bends elbows tees Mino F 105535 etc are also typically present J0 fl l 0 These additional components contribute to the overall head loss in the system and are typically called the head losses that we have looked at up to this point that occur in straight sections of pipes are sometimes called the major losses 0 Minor head losses are typically given in terms of dimensionless loss coef cients K L that is a bb KI 13 o In almost all cases these LOSS COEffiCientS are deter mined by experiment 0 See the supplemental hand out and Table 62 of your textbook for values of loss coef cients K L associated with various components 0 It will be important to consider these losses when we look at pipe ow examples in the next lecture M h T m l l L minor va Colebrook 2 L a FOFMUlC x MOODY C HART Re 01 000 39 000 00 0206quot I 5 I Wholly turbm39etntflow quot Only39depends on roughness 39 W39WquotEL05 5 0 0 3 0504 quot A A f quot 2quotT quotquotquoti it 002 0015 7 0501 39V 39quot 39 1 L008 f 003 v k i j I i 0000 i 0020 0 0 ix j 1 0 i 0 a 7 r 00002 002 y i 70001 39 La lml infar I 7 r a 39 39 000005 005 004 quot Tra wait 60 n nra image 001 00093 0000 i w i 5 0H 0 2003 0 039 0 103 1 H I I l I I I f 000001 6 839 200 4 6 0 I 2mm 4 6 SI 2005 4 gal 2000 4 04 105 106 107 108 I 88 CEIHILLOEI I 30818 NCEI IIAIO 1Jeua Apoow SEISSO I HONIW OI 5L aLJ 1 KL 395 K h 2 0 p x l X I r I II I 1 g M u 77 bu r u n v I I I n I 7 I II Y I 7 I I I r d H I 1 I 771 I I W 1 I LILL U Q I L u a 39u39 DI 3 n l I II I I 39 I 39I I l r r In n E I n r E39 En E Q an Fl E39 u n J E i 5 253 Ea F rquot ll 1 j quot 17 I r 1 r 1 I i L F c 391 1 m J n I e 1 39 J IE1 5 I I 0 II 1 1 4 I 1 t J 39 l a w i I 1 E IJquot Id 3 I E j j q i awn J 3 1 l I x a u I I n I I 1 1 I I r 39quot414 39J I j j r M 394 7L41 I I E 795 7M k JL M I I1 I I J3 39T r r r m 1 Ifquot 7 i I L L I L D quotr 1 i a Fr 7 E m i it I D a In mg 39 D 5 1amp gq m D 4 H um um 225 as d I511 D lmEV E E m A n n E E a n 3 D n5 1 a 11 I I p I3 C mafia D I n l 9 L I T u L E J I F ur39 I r IIquot I 4quot quot lquot i 39 a I I E I I 39 I 39 J u 39 I 39 quot 39 7 I D D I I f I I39 l39 I I 39 l I39 II II I u I I I 1 I I F E j n I I I I I39 a 39 f 39 II I 39 39 I I I J I 39I I 39l39 I 39I I I I I I 39 I I39 r I L 39 I 39 I I i I c I I I r I L 39 I I I 39 I 1 I I quot39 I I I I if I I 39l I I I I L r 1 I I I L I I I39 I I 1 1 39 I t 39 quot I a 1 u a I I II 1 I I 1 I 7 I39 II 39 I I I III i I m r 39339 I I I I I 1 II I I p I I I I I l39 I I II I I I I n IlJ I I I 39 I I r I I 39l39 I I 1 l I r I I I I I 39I h 39L l 1 39 39 39 39 r I I39 39 39 quot 39 a n 139 I 1 lI 39 I I I I J Iquot Iquot I I gh q I I r l J u ll I I I I I I I 1 u D 1 I I If I I I I 39 I 39 39 I I J quot 39r 139 39 I 39 I l I I I I I39 I I I II 3 q I 39 I I I I r quot I l I I J l h r 1 139 I I a p I quot I I T I I itquot u E i I r III I I l I I I I I Il all i I I I I I in If D P P J 11 I I I 39l r 1 I I i 39 I 39 r I lI r I I P I quot quot r 39l39 39 I I I I I I 1 q I 1 r J I F h I i a I J I I39 I I I II F I39 II I ll39l I r I i 39I I I i I I l l f I II 299 The cylinder gate of Fig 180 is made from a Circular cylinder and a plate hinged at the dam The gate pasitinn is controlled by pumping water into 0r out of the cylinder The center of gravity of the empty gate is on the line of symmetry 4 ft 39em the hinge it is at equilibrium when it is empty in the position shown How many enhie feet if water must he added per feet of cylinder t0 hold the gate in its quotposition when the water surface raised 3 ft Figure 280 Problem 299 Qqq FIG 3 30 lt19 07 W 270J FUyrLI4FLa VVGATIS FHY Hng zFuy39fr FLLVGMEFHY yv K J i VH to an t g Mame wsv U 0111 33377le 5 CD YV YBWFar rxf 7 mama 339 3a FL ESW UY 9 FL F14 PnA z 3 yngl b FL39 FLA quotX W SUM MomW3 93 1 GENO 0 Venn lt3 FL1 FHVSgtO FH39y 395 J uwrrls F1400 ggegg um l QSrrY qY H F CD 7 1 Vmo L KL 315 W Sw lx g jSf g j Has 4 3 2 4 2 72739 43

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