FLUID MECHANICS ENSC 3233
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Date Created: 11/01/15
ENSC 3233 Chapter 12 Lecture 2 Hint for today Useful conveisions are 1 ft3 748 gal lftSs l cfs 4488 gpm 1 lbinz 144 lbftz 2208 ft of H20 550 ft lbs 1 hp Today we will cover System curves Pump selection using Specific Speed Cavitation and NPSH Energy Equation V2 p V2 p i71 zlha iizzhl 2g pg 2g pg note V1 V2 0 P1 P2 0 ha 22 Z1 hL The head loss will be a function of the flow rate Q h pipefriction minor losses chmm mm Example 1 Pumping between two reservoirs Given piping design andpump curve Find Q System measures Z2 21 10ft D 6 05ft galv iron e 00005ft L 200ft mum mmmmusty Substituting ields pumping between two reservoirs Solve for head teims eD 0 00505 000 f 002 rough pipe high Re approximation 002200ft D 05ft 2K entrance elbow exit 05 15 1 3 Human mm s 800 Williams mmmmusty Pump Curvefor example 1 Human mm 2 Continuity V QA QIr0524 Substitution yields ha 10 443 Q2 Q in Ss ha 10 220x 10395 Q2 Q in gpm Note at Q 039 V 0andhL 0 ha 22 21 10ft noflow shut o head We just computed the system curve Human mm a mmms mmxsmm usw System Curve may compute the required ha That is the system curve The pump operates at the intersection of the pump Q vs mm mm ha and system 7 curves Q 1600 gpm 112 665 ft swam zine 4a mmms mmxsmm usw Example 2 What type of pump is best for an application with ItZ 169 ft Rotation speed 600 rpm and Q 20 ft3s Q 20 ft3s 449 gpm ft3s 8980 gpm N quotrpmQgpm 1160 N LO 389480 1214 169 From the gure use a radial ow pump mum in Fromm uuhmxmmusw Specific Speed Pumps can be characterized at their operating conditions by their s eci c seed in English units Given a pump s desired operating condition s Q ha and a you select the type of pump by NS cm 12 Luna 2 x uuhmxmmusw What ifyou had a motor speed of 720 N 172018380 3480 From the gure use a mixed ow pump cm 12 Luna 2 u mmm mummm any Speci c Speed Pump Selection Diagram um 12 mm 2 y mmm mummm any NPSH A common problem with pumps occurs when the absolute system pressure drops below the vapor pressure of the uid When that occurs the uid boils and vapor bubbles form When the system pressure rises above the vapor pressure at the pump impeller the vapor bubbles collapse This is called CA WTA TION Cavitation is a big problem to be avoided To do so we check the NetPositive Suction Head NPSH In any system the actual NPSHA must exceed the required NPSHR um 12 mm 2 12 mmm uuhmxsuuuuusw 2 V NPSHA ampi7amp2NPSHR g Where 175 and VS are measured or determined on the suction intake of the pump and 17V is the uid vapor pressure Note 75 must be in absolute pressure units 17 17 a 2 17am NPSHR is speci ed for the pump by its manufacturer chum 12 m 2 13 Find the pressure on the suction intake side using the energy equation 2 2 V1 171 Vs I75 772177z hL1S 2g pg 2g pg S 2 V 75 S iziz 77711 1 5 2g L1 S Why doesn t ha appear oValues for 17V as a function of temperature are in Tables B1 amp B2 for water and Tables 15 amp 16 for other uids oValues for pm as a function of elevation are in Table C1 amp C2 mm 12 141 2 1 Ihmxsmlh any hm 2K V22g 0515 182225 103 ft 2 75ft 7 g 7103ft 7204ftgage pg 2g 75 204ft621 113113 1270 113122 75 88 psigage Psmbs Psgage palm p5abs 88 122 40 psia 576 lbftz mm 12 141 2 17 Example 3 What is the NPSH for the pump in Example 1 assuming Q 1600 gpm ha 665 it 21 2pmp 5 Elevation 5000 ft and Temperature 90 F Neglect pipe friction f 0 17am 122 psi pV 070 psi v 621 lbft3 Continuity V QA 1600 gpm449 gpmftBs7I 0524 V 182 fts mum 15 2 mm mm NPSHAampVii 7 2g 7 2 2 576 182 7071757144171 NPSHA7 128ft 621 2g 621 2 mum 1x Example 4 The pump in the previous example has 2 required NPSHR 10 11 Will it cavitate From before NPSHA 128 ft For Safe operation NPSHA 3 NPSHR 128 gt 10 this is safe Thus the system will not cavitate chm u m 2 ENSC 3233 Chapter 8 Lecture 3 Hint for period For laminar ow Re lt 2100 the DarcyWeisbach friction factor is given b f Re This period we will Examples of head loss calculations in pipes mm km 2 x mmm mammam By the DW equation h if L f D 2g Set the two equations eual Solve for f 7 pVDm a which is the equation for the quotlaminar linequot on the Moody diagram chmux km 2 Compare Poiseuille and D W equations For laminar ow Poiseuille s law from before Q w horizontal amp laminar lzsyd or AP Q1284il 7ZD Note Continuity Q VA D2 VA128ul WTIZW AP 74 4 7ZD 7ZD Ehmkamz 2 mam mummsm um my Example A horizontal tube 39 I D D 1 cm 001 m u 00012 Nsm2 o 999 kgm3 What is the highest ow for laminar conditions Reynolds Number Re 2100 maximum u Ehmkamz s mmm mm m um my horizontal laminar friction loss From Energy Equation hL M 21 7 22 horizontal pg Pg Ap 32ylV Pg ogD2 Elmmkamz z mmm mm m um my Solve for V 7 Rei 7 210000012 oD 999001 027ms Apply Continuity 2 2210 5m3s QVA Elmmkamz a Example What is the pressure drop in this pipe per meter of length l 1 In Energy Equation 2 2 Vi z1 ViJr Jr z2 hL 2g pg 2g pg 1 V2 7 h 7 77 p1 172 pg f pgf D 2g Laminarf 64 64 i 7 0030 mm be are Re 2100 Re 2100 f f Reynolds Number Re1000kgm3 011ms 001m0001Nsm21100 Note Re lt 2100 we have laminar ow Therefore f 64Re 641100 0058 DW Equation 10m 011ms2 0107m 2981ms2 hf 033 m 2 h fLL 0058 D2g chm rm 2 m Substituting kg 1m 027ms2 7 999700307 71 72 m3 001m 2 21171171 p2 110 Nm2 What is the wall shear stress T 7 All 7110Nm2 001m w 21 21m 2 2W 028 Nm2 Ehmkamz x Example Same pipe and conditions as before except Q 011s 10394 m3s Find hf Continuity V QA 1049065 11ms Reynolds Number Re 999kgm3 11ms 001m0001Nsm211000 Note Re gt 2100 we have turbulent ow use Moody Calculate sD sD 00046 cm 107 cm 00044 Get f from Moody diagram Re 11000 WK km 11 Example 10 In of 38quot steel Sch 80 pipe water at 20 C and Q 001 US Find In 0 1000 kgm3 H 0001 Nsm2 ID107 cm A 906X105 m2 6 00045 cm and Q 0011s 105 m3s 1000 liters in 1 m3 Continuity V QA 1059065 011 mS Ehmmeammz y f 0036 DW Eq 2 2 hL 0036 10m 11ms D2g 0107m2981ms2 hL21m mg m z 12 mmm numummam FindhL with Q andD known 1 Calculate V V QiZD24 2 Calculate Re Re pVDy VDV 3a IfRe lt 2300 use f 64Re 3b If Re gt 4000 nd f on Moody diagram with gD 3c If 2100 lt Re lt 4000 estimate f from Moody 4 Use DW formula for 11 1 V2 hL f D 2g note V1 V2 hL plipz21 zz 7 2 hL 10000 1100NSI1 007 mm 1000kgm 98ms hL 79 m DW Equation 1 V2 hL f D 2g Solve for V chmux 1m 2 WWW mummsm um ash Example A simple pipe system Find Q Z L 38quot steel Sch 80 pipe Water at 20 C 21 10 m p 10000 Nm2 22 2 m 72 11000 Nm2 Ehmkamz hLDZg fl Looks like trouble we have one equation but two unknowns Vand f V Lets guess a f value First calculate the relative roughness SD 00046 cm 107 cm 00044 Use Moody chart to estimate f based on gD Assumef 0028 this assumes complete turbulence or large Re Ehmkamz Look up parameters 0 1000 kgm3 u 0001 Nsmz ID 107 cm 00107 In Handout A 0906 cmz 906x10395 m2 600046 cm Table 61 Energy Eq head form 2 I72 zlh 72722hL P 2g pg 1 2g Solve for 11 Ehmmeammz V hLDZg fl V 7 79m 00107m 2 981ms2 0028100m V 077 ms Check Re on M00031 diagram Re 7 71000 077 00107 u 0001 Re 7700 Check Moody chart f doesn t match Ehmmeammz Try again assumef 0036 V 068 ms Re 6773 OK Q VA 068 x 906x105 Q 60x105 m3s m hm Calculate Q or Vwith known h and D oCalculate rD oAssume afbased on aD using Moody Diagram oCalculate be hL V g l oCalculate Re and checkfon Moody oRepeat iffdoes not match And don39t forget for laminar flow Re lt 2100 f 64Re m hm One last concept from pipe friction Why does a smooth pipe behave differently from a rough one turbulent ow W m i mu m hm The thickness of the viscous sublayer is about l 55 i u Where 14 is the friction or shear velocity u Lw p From the first lecture I LAP W 41 Solving for 5 55mm limlOJ L DAP pDAp m hm ENSC 3233 mw Chapter 8 Lecture 4 This period Minor losses Noncircular conduits m L There is no adequate theory to predict these losnses use the empirical relation hL KL 1 minor losses 2g for a complex pipe systems with several fittings and various length of pipe we would sum them up 1 V2 V2 h 77 K i L 2 D 2g 2 L 2g m L Minor Los s Minor losses are all the frictional energy losses that come from fittings entrances and exits Many times ey are the la ext cause ffr39 t39 39 ipe system source WWW stp com m L Inlets Friction is a li5 g amp strong function 1 F of the inlet square reentrant rounded well sha e ounded se Fiure 822 ae 438 Outlets LL lg lg 1390J Friction is not a function of outlet shape rounded well square reentrant rounded l or all outlets KL 10 m L inlets and outlets LL j elbows and tees 1 E V valves o gt expansions and contractions W m L Elbows and Tees HH Function offitting shape and connection Flanged connections have less friction than threaded ones se Table 8 445 Elbows Long radius elbows have less loss than 90 stande Tees Loss is afunction of line straight or branch turning ow m L brangh out um 1 Valves Valves are used to both shutoff ow The loss is a inction ofthe valve type and its setting There are several types of valves Globe ate SW39ng check Ball Butter y Example A pipe system is made of4 pipe Find KL for Sharp edge inlet Figure 822 KL Well rounded outlet Cormant for all outletr KL 90 degree regular elbow Table 82 KL 03 T branch l39low KL 10 T line How KL 02 Gate value 12 closed KL 2 Contractions Use Fiure 82 Note the velocity used is the smaller pipe V 2 hL KL Vi 2g mm 4quot to 6quot sudden expansion uxe equation FDl quot2 2 2 417472 031 IrDz 6 4 A12 K1i 1 L A2 Note that there is no need to convert inches to h 4quot to 6quot sudden contraction use Figure 826 v2 Ai 4a40444 A1 6 4 From Figure 826 KL 035 mm Expansions Use the equation z KL O AZ Don t uxe gure 827 Note the velocity used is the rmaller pipe V1 z u K a 2g Example A pipe system What are the minor lossesl 4 nominal Schedule 80 steel threaded connections ID4 972 cmA4 742 cm2 Q 00002 m3s V4 QAw 00002742 X106 27 ms ID6 146 cm A522 168 cm2 V5 QA rr 0002168 X 106 119 ms mm Lama 12 Example Determine pipe diameter required to obtain a given ow between two reservoirs Given Q amp hL nd D V2 V2 i z i zzhl 2g pg 2g pg hL21722 h ZELZJrZK V L D2g L2g mm Lemn m mmms mummsmmn any Entrance K L 004 well rounded rd 02 Elbows KL 064 regular KL 023 long Expansion use equation not gure KL1A1A22 056 Exit KL 10 WK ham 1 WWW mummsmmn any Note D is constant so V constant L V2 h 7 K 7 L 2 D Z L 2g Continuity V QA 4Q7ID2 Put it all together hL E Minor losses ZKL KL entrance KL exit 2KLelbow KL ZKL 0810 203015 255 There are two unknowns f and D but f D WK km 17 fL 8Q2 7 K 72 72 D Z L 2D4g 2 1 valve mmm cummmmuw Sum minor losses 2 z 2 h ZKLLZKLEZKLV6 2g 4quot 2g 6quot 2g 2 2 046402356LS21S2 2981ms 2931mS hL 055 m 006 In 061 In M x m 5 WWW cummmmuw FindD with known Q and In 1 Assume a value of D Usually make a guess so that V 1 ms or 3 s for liquids 3X more for gasses Calculate Vand Re Determine f from Moody diagram Compute 11 from fL hL E D Compare computed 11 with know value Jaws 8Q2 ZK 7 L 2D4g U 6 If computed 11 not equal to true value repeat mm x m xx Noncircular conduits Many conduits are square or rectangular Friction losses can be computed in noncircular ducts by use of the hydraulic diameter P wetted perimeter 2 4A 47iD 4D CircleDh 7 P 7ZD 2 Square D 5 P 45 Rectangle D i P 2a b Reynolds Number use Dh Reh amp chmux Lem mmm uuhmxsuuuuusw For noncircular conduits 1 Calculate hydraulic diameter D 4AP 2 Use Dh for D in Re DW and f calculations 3 Note that continuity remains Q VA where A is the true area of the conduit Example What would be the maximum laminar ow and friction loss in the proceeding duct Reh pVDh 2100 u chmux Lem Example A plastic 02 X 05 m air duct with S i as Q 03m3s What is h per meter of length 4A 4ab 40205 Dh 7 P 2ab 20205 029m V QA 03m3s02m05m 30 ms 123 kg 3E 029m 3 S Re LVD I m N 60000 179x10 5 m aDh 0 9f 0013from Mooaj diagram mmms mummsmmn any V 7 2100u 7 2100179x10 5 010ms pDh 123029 f L4 1025 Reh 2100 2 2 hiLiLV 70025 010 45xloesmm L DhE 029 2981 Clams hum mmm Dhlumhklhluil 2 2 LL 0013 3ms Om L Dth 029m2981ms2 m in pressure uriits ATP hprg 0021123981 025 Pam mg m ENSC 3233 I 5 x a Chapter 6 Lecture 2 Homework 83 85 88 812 814 818 827 831 835 This lecture we will PoiseuilleHagen Equation DarcyWeisbach Eq Moody Diagram mm mm 2 x mm Enrme mu usw From before for laminar ow 2 u M 1 1 2 R u 1 1w R Integrate u over the pipe area to obtain Q Looking down pipe 1 u dA u 27Wdr volume ow in donut mm mm 2 Hint for the period 5 i as In a complex system of pipes and ttings the total head loss is computed from the sum of the loss from each part IV2 V2 h if K7 f ZfD2gZ 2g The energy equation becomes mm mm 2 mam mummsmmusw R R Q I udA qur IuZ rdr Area 0 Wote dA 27Wdr R 2 2 Q ludA APR 171 2 rdr A 0 41 R Integration yields the Paiseuille Hugen Equation 7 7134M 7 7ID4Ap 821 1282 laminar ow amp horizontal laminar general case mmm ummmuw 2 2 E 171 h V2 172 pg 77zzh 2g P 2g pg f 2 2 2 V1 zlhPVi zzZfLLZKL 2g pg 2g pg DZg 2g Example How does the average velocity relate to the maximum Average velocity 4 2 VQA7zRApiRAp 8M 7sz 821 From last lecture MJW 2y 4y max cumsme a mm Enrme mu usw The mean velocity is 12 the maximum mm mm 2 7 mm mmm m m Now if we ran pipes at various velocities we would nd h f cx Vquot n l for all pipe with laminar ow n m 17 smooth pipes with turbulent ow n 2 rough pipes at turbulent ow Since we operate mostly rough pipes in high turbulent ow say V2 hoc Lets put it all together mm mm 2 m mmms mummsmmusw Energy Equation on a pipe 2 Note point 2 is downstream from I Z Viz 22 hf nopumps amp turbines 2g pg 2g pg 2 2 hfV1 V2 171 17221722 2g pg 171 172 P3 21 7 22 uniform pipe 2 WWW mmmmw h f 0C D 2 g Introduce a proportionality parameter the Darcy friction factor f hf This is the Darcy Weisbach relation for hf There are other relations around but they have mostly fallen out of use with the introduction of calculators mm mm H mmm ummmuw Pipe Friction What form does energy head losses hf have If we ran experiments with different pipes of various lengths but at the same Vwe would nd h f i I7 h f 0c 1 If we ran experiments with the same kind of pipe but with different diameters we would nd h f 0c 1 D mama y mmm ummmuw Returning to the energy equation 2 2 VL zlhPVi zzh1hf 2g pg 2g pg 2 2 2 Vi zlhpVi zz 2g pg 2g pg DZg This is a form of the total energy equation where pipe friction is the only energy loss All we have to do is gure out f Emma m 2 u Lets do more experiments Now what happens if we change the Viscosity of the uid but not its density f lu for lowV f i forhighV Likewise compare different pipes by their relative roughness gD g f 8D for lowV f 8D for high V Elimm Luna 2 2 fhf D2g L BuckinghamPi n5VD84u0 j3 MyLiT H n j532 Let D be the repeating variable By inspection H1 SD H2 Re pVDp So take all the data andplot it up asfvs Re with lines of constant aD TAEILE 81 mmums mhhmxmm usw Equivalent Roughness i39oi New Pipes From Moody um 7 and euiiinnuk Ref 01 Pipe Rivelvd steel mu Drawn uuuug Plastic gins cums mm cums mm Equivalent Rnugiuw F Feel 0003 7003 10017001 000004 L003 000085 00005 000015 0000005 00 uuooiu Millimeters 00791 0045 000 i 5 00 nwmh we would n f l0 for low V f i 0 for highV Using dimensional analysis We know f VD8J0 Note that f is already a dimensionless H group Emma mm Example Calculate the head loss for 100 oSf 0lm commercial steel pipe owing water at 2 ms ipVDilOOOZOJ Re 7 u 000112 From Table 61 l8x105 s 0046 mm 46 x105 m sD 46 x 1039501 000046 From Moodyf 0019 7 V2 7 0019100 m 2ms2 h f ng Emma mm 0lm m 298172 s 39m If we changed 0 keeping the other parameters constant w ENSC 3233 I n M 39 Chapter 1 Lecture 3 Hint for today No class or discussion sections on Monday Today we will cover 0 Surface Tension oViscosity oReynolds Number 0 Speed of Sound amp Compressibility ofLiquids 11111 11 91111 1 111 111111 A capillary column surface tension holds the liquid above the free sur ace gt H 1739 o surface tension 19 contact angle 1 A force balance yields quot 2quot quot Wt Force u yV length of surface 17 cos 9 yrRZh 2rR 7 cos 9 h 2 7 cos SyR 11111 11 U111 1111111111 Example What is the weight of 1 00 ft3 ofwater at 32DF p 194 slugsft3 7 pg 1940 slugs 32174 1152 7 6242 113113 100 113 weighs 624 lb Note We will aslsiume that that specific weight ofwater is always 624 lb 3 11111 11 1 U111 1111111111 Variation in geometry How does this change for two parallel plates wunitlengthhwhL3 Length2unitlength2L 11111 11 U111 1111111111 Surface Tension Surface Tension is a indarnental property of all liquids If a liquid didn t have a surface tension it would be a gas The surface tension we observe results from the atomic scale forces on the liquid molecules Surface tension may be visualized as a force that acts normal to any section cutting through the liquid interface It will have units of force per length of surface FL The two most common expressions of surface tension are capillary rise and bubble formation 11111 11 Example A bubble in beer What is the gas pressure inside the bubble R10 mm o 007 Nm 2F 0 pressure area 7 Surface tensioncircumference 0 4 271111 szZ p erR 7 2 007N m 0001m Is a bubble in the foam di erem 140Nm2 2 2911113911Z 11111 11 mwxnvvuy Mm Viscosit Viscosity is a measure of the ability of a uid to transfer momentum within the ow eld Image dragging a plate across a surface with a film of uid in U velocity of plate A area ofplate u point velocity of uid b film thickness y coordinate system F force required to pull plate mm m 7 mw m Mn Mm Example A linearbearing Find F mm M in Dunn m Mn Mm Note no slip boundaries F Ce F F Aquot r r shear stress at thewall For anyplace in the ow eld y absolute viscosity At the wall mm an Example Aplate being pulling between fwo others i with a different liquid on each side Find F mm m u du 50 dy Ifwe ke b and Usmall laminar ow the velocity distribution will look like mm m Newtonian versus Newtonian Fluids mm m mu m Mn Mm alumniH mm sum mm m nl sitmwum 17 Reynolds Number A very useful parameter we will use frequently in this class is the Reynolds Number Re R a u u where Vis the velocity of ow and L is a characteristic length The Reynolds Number is function of both the uid properties and the ow cmme 1ch 3 13 Compressibility of Liquids m mm V a s Liquids have a small compressibility governed by their bulk modulus of elasticity E The bulk modulus relates the change in density to the change in pressure dp E v i dp 0 Over a small pressure change EV is assumed constant 2 d p2 Ev l p I dp p1 0 p1 Integrating Evlnampp2 Tpl where l and 2 indicate before and after cmme 1ch 3 15 Example What is the Reynolds number for a ow of water in a pipe with a diameter of 01 m temperature of 20 C and velocity Vof 2 ms L diameter 01 m Table B2 page 761 o 998 kgm3 u 0001003 Nsmz 7 pVL 7 998kgm3 Zms 01m Re 2 0001003Ns m 199000 chm 1 ham 3 1 Example What is the change in density of Water as its pressure changes from 100 x105 to 100 x 107 Nmz 1 to 100 atm Ev 111g 172 T 71 01 I72 171 CX P2 P1 P E 1 EV 215 x 109 Nm2 Table A3 01 999 kgm3 p 1000 kgm3 Only 01 increase This is why we normally assume liquids are incompressible chm 1 ham 3 17 Speed of Sound The speed of sound in a uid is of course an important property in high velocity ows For an ideal gas it is given by 12 cideal gays kR T where R is the ideal gas constant and T is the absolute temperature For exact values of real gases we look the value up Tables B3amp4 gives 0 for air at the standard atmospheric pressure chm 1 Lecture 3 13 ENGSC 3233 Chapter 5 Lecture 3 Hint for the period This is an important equation V2 V2 zlhpi zzhL 2g pg 2g pg This period we will cover The Energy Equation and Examples mimecm mmmmmm QM WWW g t epdV tepwo Wl CV cs Break W into two components W VVsha anmalstress Wsha Tsim a For normal stress from a complete analysis Wnor39mal stress 7 IpV quotdA CS work on the CS m MchokamsmeUmvmm Energy Equation The potential energy per unit mass of uid is an extensive property and may be de ned as V2 Eeppa7gz e total stored energy per unit mass M internal energy heat Substitution into the Reynolds Transport theory material derivative yields ma Mums okum Sm Umvm y Qnet in t Wsha in CV IPV quotdA cs 9 epdV epVo ndA at CV cs Move pressure integral over Qnet in Wsha in CV a V p V2 A 7 epdV n77gzoV0ndA atCV cs I0 2 For steady ow ena 0 and along a streamline ma Mechanms om m mummy M epdV epwomdA CS Di arm Energy Sources Time rate of change in CV Net energy uX out of CV Xe0 7 Qnet in t Wnet in CV Q heat transfer per unit mass W work per unit mass grnhM ma Mechanms om m mummy Qnet in t Wsha in CV 2 2 gzpQ 7aV7gzpQ out in 01 Qnet in t Wsha in CV 2 7 gz ml 2 m Vigz o 2 ou T his form is only used in heat transport applications Rearranging Qnet in 1070 17in Wsha in V2 V2 7gzpQ 7gzpQ p out p in In words heat uXchange in heat energy shaft work in hydrodynamic energy out hydrodynamic energy in With no heat sources or heat transport Qnet in 0 Example Given A pipe with a constant diameter Find The pressure drop along the pipe V2 p1 V 2 p Lizl hp iizz hL 2g pg 2g pg P1 172 Pgzz 20 hLl Can the pressure increase in the direction of ow m WWpMmMUWW The increase in heat storage will be that due only to uid friction out T in inlossout let out be position 2 in be position I and rearrange V2 W V2 mgzli iigzmzmz p 2 39 0 2 Note rn pQ and W gr nhp Then dividing by g yields an WWpMmMUWW Example A dam and hydroturbine are used to generate power Solve for the power generated as a function of Q 21 and 22 Neglect friction losses 2 2 zl hp Vi zz hL 2g pg 2g pg ma quot1 me m mummy 2 2 VT 71 h Vi 22hL 0 2g pg where ossg hL heudlassfrom I to 2 or multiple by pg 2 W 2 PViP1PgZ1PP2szzPghL SolVi for shaft work Wsha in png PgQZz 21 Question What sign will Wm m have What does that signi l mmmwummmm Example A jet of water strikes a wall and all kinetic energy is dissipated How much will the temperature of the water increase 2 2 VLgZI hp Vigzz llass2 p 2 p 2 MMW 1an mm Example Water is pumped over a levee Q Asuctmm A2 21 22 517151172 h are known Findpmxmm p m Mmommmu m y V 2 ialassz but with no heat sources or transport 172 7 171 llossz V V V i u 7 u 2 2 1 From Thermodynamics cPT cp speci c heat at constant pressure 4182 Jkg C at 20 C T absolute temperature Apply continuity V2 QA2 Apply Energy from 1 to 2 2 VL zlhPVi zzhL 2g pg 2g pg NoteV1ampp10 V2 hpi zzizlhl 2g pg ma Mechanms om m mummy V12 70pT2 cpT1 CPT2 T1 2 Tz Tl i 0P 2 Note from the SI de nitions Jkg Nmkg391 kgms392mkg391 mZs2 mmmmmmmm y Apply Energy from 1 to the suction side 2 E 21 h Vsuction pmction 22 hL 2g pg P 23 Pg Continuity Vsuctmn QAsuctmn Since we have not passed the pump 11 0 V 2 Fraction Pg21 M 22 hL 2g Question What sign does the 75mm have mm Manhunt 0mm Sm Um en mi ENGC 3233 Chapter 12 Lecture 3 Today we Will cover Pumps in parallel and series Variable speed pumps Reducing NPSHA 211me 12 hunk mm Manhunt 0mm Sm Um en mi 211me 12 hunk mm mum 0mm Um a w Pu1nping in Parallel and Series In many applications the required Q and ha cannot be achieved by any single pump Speci cally either Q or ha is too large This typically occurs for water treatment 5mm Lemuel In many cases the pumps are directly integrated 5mm Lemuel Pumps in Series Big ha When two identical pumps are placed in series their ow rate Will be equal and each Will provide lZ ofha The trick is to create a pump curve for the pumps operating in series Qtwo Qone Q ha two 2 ha one 771w0 77011201t Qone amp ha 0112 bhptwo 2 bhpone at Q amp ha 0112 Everything is the same but the head is doubled 5mm Latin 2 Parallel Pu1nps Big Q plumbed together When two identical pumps are placed in parallel With their suction and discharges connected they Will produce the same head and equal ow rates Qtwo 2 Qone ha two ha one ha 771w0 77one at Qone and ha bhptwo 2 bhpone T he ow for a given head is doubled 5mm Latin 2 171 diagram 1 000 Example Two pumps with the curve shown are operated in a system with 111 10 00001 Q2 gpm What Will Qsenes lt55 Qpanzllel 6D 5m An N ham serles W parallel o 200 400 600 300 1000 Q 9pm I diagram both lar e installations and in sma processin applications Variable Speed Pum s P Most pumps may be operated 71 lt quot1 lt quot2 k Increasing NPSHA Lecture 2 In any system the actual NPSHA must exceed the required NPSHR or cavitation Will occur 7 p in absolute pressure Wh p5 and V are measured or determined on the suction intake of the pump and IV is the uid vapor pressure When NPSHA is too small you have to get another pump With a smaller NPSHR or change the system Ef ciency of different types of pumps The different types ofpumps centrifugal mixed or axial ow are suited for different regions of speci c speed because of their best ef ciency range Speci c Speed Was de ne wxQ NS 7 Why 4 N nopqugpm Wm a for the Example How can you increase the NPS pump shown V2 NPSHA PJPJ 7 2g 7 Apply energy equation from reservoir to suction side to P V2 iizl isamp25hl1S 7 2g 7 2 V1 Patm 2g Solving for 25 V2 P Patm 7 21 2s i hlais Head losses hm fLD Vf2g Substitution forps yiel PSHA V2 V2 7212seise1lte ji 2 2g D 2g VJ Pv 7 2g 7 Patm mmms mmxmm any Simplifying 2 NPSHApatmzlizsiKe 7amp 7 D 2g 7 Continuity VS 4Q7ZD2 2 NPSHAp mzlizsiKe jig7amp 7 chum 12 m z w To increase NPSHA increase 21 raise the water level decrease 25 lower the pum increase D increase suction line diameter decrease K2 install bell tting decrease L mount pump neXt to the reservoir What can be done with pv mm 12 m 2 2n ENm sWC 3i 3 Chapter 2 Lecture 2 Today Hydraulic cylinders Manometry Hydrostatic pressure distributions mmmmm While we sometimes de ne the pressure of one uid in term of the length of another such as when atmospheric pressure is de ned in mm of Hg its more common to do it in terms of the uid itself the working uid Later we will use pressure heads in the energy equation and it will be important only to use values measured in terms of the working uid mmmmm unihme in my mmmu Hint for today Measuring Pressures in Then Heights Head h p 1pg pIy Last time we listed a pressure in mm of Hg So why Usual units are ft of water mm of Hg m of water These are called Pressure Heads or just Heads Note went using pressure heads you need to define the height and the uid If g and p are constant P2 quotP1 39Pg 22 21 Let p 0 ie atmospheric and note 2221 h head mamquot 2 mum Hydraulic like in oil cylinders Fbig Fsma p 7 Abig Asmall F big Abig F small Asmall Abig Asmall FbigpAbig Fbig Fsmall F small P Asma mamquot s mum Example Lifting a house p 14 x 10 2000 p31 Dummy 10 cm Dbig10 cm What is FM and Fbig chmnmm Piezometers are used to measure the difference in pressure between a liquid and the atmosphere p2 p1 pg 2221 note 22 21 h amp p2 Parm Ogage p1 pgh Pmm abs p1 pgh gage chmnmm mm W 2 2 001 Fsma pAsma 14x107Pa X 7239 7 1000N 220 lb 012m2 FINg pAblg 14x107Pagtlt 7239 100 000N 22 000 lb mm mm 2 x Example A simple piezometer is used to measure the pressure in a water pipeline Maximum pressure will be 30 psi What39s h h P1 W h 301bin2144in2ft26241b 3 h 69 9 ft do you have a long ladder mamquot u Manometers Manometers are the simplest and most of the time most accurate method to measure pressure in a uid There are two general types piezometers and Utubes mum y U tubes measure the difference in pressure between two uids by using a third You have to solve the hydrostatic equation in series for each uid or mum 12 P239P1 wag 2221 P3 39 P2 39pbg Z339Z p4p3 ng 2423 Adding these three Eqs Yields p1 p4 pa zz21 pi gem pc gZ423 l Note 1 At the interface between the uids we assume the pressure is constant That s only true chmmmm n For manometer problems and a lot more follow these steps 1 Start at a point of known p 2 Trace a path through uids to the point of interest 3 De ne points 123 etc You need apoint at every liquid interface 4 Determine 2 at each point 5 Apply the hydrostatic Eq from point to omt gtgtDon t confuse z with heads chmmmm m if there are no surface tension effects which is reasonable for large tubes Note 2 The densities of gasses are about 11000 of typical liquids Thus if one or more of the uids are gasses we can usually ignore their contribution to pressure changes If pc is small ie a gas PI39P4Pa zz21 pig23Zz pcgZ423 p1 p4 pa zz21 pi 2322 mm mm 2 1 Example A waste tank What is the pressure at the bottom PI39P4 pa zz21 pig23Zz pcgZ423 mm mm 2 n A word of caution You will see lots of formulas for the different variations of manometers Don39t bother to remember or use them You will probably make a mistake in their application It s easy to get a sign error mum is Hydrostatic Pressure Distribution p2 p1 pg 2221 mum xx mummsmmn my Hummus thummmusi l WWW Example Leth pm 0 gage pressure Note h 22 2 Drop quot2quot subscript p pgh gage chmnmm ENSC 3233 K 5 x a Chapter 7 Lecture 1 Hint for this period Section 73 8 steps and Table 71 are the most important information in Chapter 7 This period we will Start Chapter 7 Dimensional Analysis and Similarity chmn rm 1 x mmm uuhmxsuuuuusw One important restriction Each fundamental dimension must appear in at least two variables One Exception In some cases k r will form one valid H group chmm rm 1 Dimensional Analysis Procedures that use the dimensions of variables to form variable groupings H terms that are dimensionless Examples of variables we will use diameter D 0 L velocity V 0 LT gravity g 0 LT2 weight Wt e F M LT2 viscosity u 2 F TL2 M TL W7kmmi 2 WWW mummsmmn any Determining l39I groups 3 methods Inspection Just take a look at variables and trial and error a solution Rayleigh Method Not used here Method ofRepeating Variables An method to determine H groups usually called the Buckingham method W7kmmi s Buckingham Pi Theorem k variables with r fundamental dimensions will form n 39 H groups Where l l s n k r Example For the variables listed above determine the number of H groups F57 D Kg Wt ftU r 3 M L amp T H sn532 mama 2 Exam pl e Pipe ction was shown by Henry Darcy in 1857 be a function of the pipe diameter D and the height of the pipe roughness c Find a dimensionless group By Inspection D 0 L c 0 L HcD A ratio of two variables with the same 139 39 is always a valid PI group mama a Buckingham 11 Example Given a triceratops nd Ugroups lst We define the relevant variables V H amp g The number of variables k 3 2nd Define their dimensions H L g 0 LTZ Number of dimensions r 2 3rd Compute number of H groups needed by Buckingham 7 Theory n k r n I 4th Select repeating variables We have two reference dimensions L and T so We select two variables With at d T Rule of thumb Choose in order velocity length mass or Wei ht rotational spee Dimensional Equation The dimensional equation is a c H Lb L T T2 H LaTuLchTizc Each dimension must cancel out so Length Labc0 Time T a200 ote We have two equations and three unknowns We will have to assume the value of one variable Let a I It is traditional to let velocity be to the rst power Then from the time equation 20 12 C From the length equation b a c 112 b 12 We pickHand V Note since there is only orie 11 group We could have skipped this step 5th Form the H group With the repeating variables and one nonrepeating variable HV Hbg 01 mm I H 39 cancel The H roup is d39 The exponents a b and c are unknown at this point We must solve for their values mums HVaHb c VHrlZgrlZ 43H 6th Repeat step 5 for any additional l39l group Not needed for this example 7th Check the H groups to insure they are dimensionless H V LT W 1 LTZL mums n 0 dimensionless mammal 8th Express the relation if more than oneMHW grou H1 pa 12113 4 2 quotis a lnction ofquot Since we have only one H group we can say H constant For any triceratops Lets give this H group a name the Froude number Fr H Fr 7 J31 chmn rm 1 u mmm uuhmxsuuuuusw H VVJ P mHm Do we have a reasonable model today for a triceratops Given model Hm 5 ft Vm 30 mph 44 fts Triceratops prototype Hp 7 ft Solving for Vp VP 44ft s E 5ft 52 s 35 m chmn rm 1 m Modeling and Similitude The performance of a prototype may be determined from a model using the relevant H groups if similitude requirements are met Example Determine the speed of a triceratops Fr 7 1 gH For a model of the triceratops to predict the performance of the real prototype their Froude numbers must be equal W7kmmi 1 WWW mummsmmn any What has this exercise demonstrated 1 Dimensional analysis allows us to analyze systems we don t llly understand oIn this case we showed that the Froude number is a relevant parameter oIn other systems different numbers apply 2 Modeling allows us to predict a prototype from measurements on a model oThe model measures such as velocity or force are not necessarily scaled by the length ratio W7 mm n Frm Frp Vm VP 3 pHp lgmH m Since gravity will be constant unless we go elsewhere such as the moon gm gp g Vm VP gH m ng m7 mm is One final word on the method Dimensional analysis and modeling are Ar TS The answers are not unique We just as likely could have come up with H gHV2 The ReekerNumber However the Reeker Number is just a different way to express the Froude number Dimensional analysis does not provide single correct solutions Choosing different repeating variables or arbitrary exponent may produce different PI groups However all correct solutions will be equivalent m7 mm xx Multiple PI groups 5 m 5 For the triceratops a more accurate analysis would include the length of the animal s stride T quot L Now we have k4VHgampT r2LampT l lznkir422 numm mm m We now need to plot TH versus Fr for a range of similar animals or the same animal running at different speeds to determine the function Calculate THP for triceratops and estimate Fr from graph of the function Calculate VP from Fr VgH12 TM 4 VP Fr ngfZ f F Note other estimating methods exist for this problem that bring in other variables numm mm 22 WWW mm 51 Um my Again use H and V as the repeating variables From before we have H1 go H1 FrV1lgH For the second PI group use H Vand T H2VquTC my mm m In a39 o elephant o ostrich In b man and 2 children 0 on soft mud Q on sandy mud 0 man hard ground Girds are different gerbil species m7 rm 2 nm Mazhmxs um 51 Um my 1 H2 Ej LbL T By inspection a 0 and b c H TH 1Tl Ml 12 FI TH mm Laznml 21 ENSC 3233 Chapter 9 Lecture 1 Hint for today For most everything except flat surfaces use 12CL0V2A CDpV2A This week Chapter 9 Flow Over Immersed Bodies mm 1 Historically these equations were impossible to solve for most bodies of interest Dimensional analysis and experiments show CDpV2A 12CLpV2A whereA is the area of projection normal to the ow usually and C D and CL are empirical drag and lift coef cients respectively They are similar in concept to the DarcyWeisbach f They are computed by experimental measurements mm WWW msz my Flow Over Immersed Bodies Consider an object immersed in a uid V far upstream velocity of uid relative to the body If the uid is stationary V object velocity To determine the Lift and Drag 9 we need to determine the sum of the pressure and shear forces acting on the objects knuva 2 WWW mummsm um my D 912 0 VA L fIz 0 VA Like the DarcyWeisbach equation the key to using these two relationships is to keep track of the Reynolds Number Re Re 0 VLy where L is a characteristic length The length can be a sphere s diameter the cord on a Wing or various other measures knuva 5 WWW mm m um my 92Fx jpcos6dA IrwsinQaVI A A 13 ZFy Ipsin6d4 Irwcos6d4 A A Lem2M 2 WWW mm m um my In this Chapter we will norrnall use where 1 0u is the kinematic viscosity Drag Drag is caused by viscous friction and unequal pressure distributions Sometimes they are broken out 9 F f F F We will usually keep them together Lem2M a mmm numummam Viscous effects are always present Re ltlt 1 viscous effects dominate Re gtgt 1 inertial effects dominate Finally when Ma gt 03 compression effects are signi cant Thus we assume C D shape Re Ma sD C D is determined from tables or charts mm 7 Example Drag on a simple body using CD table Given A tractor trailer like a 18 wheeler Final The drag at 70 mph with a fairing From Table 930 CD 076 Estimate the frontal area 7 X 12 84 ft2 Convert V V 70 mph 5280 ftmile 3600 sh 103 fts Note Density p yg 0075 lbfff322 fts 70F Drag Equation 7 12 CD 0 V2 A 7 076 075 2 21032 84 78001bf mm m WWW msz m For common shapes CD has been plotted as a function of the Re and or Ma Fig 921 page 526 Smooth Sphere amp Cylinder low Ma Fig 922 page 527 2D shapes low Ma Fig 923 page 529 2D shapes high Ma Fig 924 page 530 3D bodies high Ma Fig 925 page 531 Sphere with eD low Ma knuva x mmms mummsm um ash Determination of Drag V given For compleX bodies usually only a single table value is available These values were obtained at high Re low Ma Table 928 page 536 2D shapes Tables 929 amp 30 page 5378 3D shapes Pay attention to the orientation oftheflow to the object Lem2M y mmm mm m um ash Power with V constant 1 Find appropriate C D graph or table 2 If using a graph calculate Re 3 Obtain C D 4 Determine A Most objects use frontal area SphereA 7ZDZ4 2D objects A D l Wings and flat plates use planform area Wing A cl 0 core width amp l wing length 5 Compute drag using 9 12 CD 0 VA 1 Compute 9 or obtain engine thrust T 2 Note 9 T 3 Compute P P V VT knuva u m m 12 Lift Lift is caused primarily by unequal pressure distributions resulting from the ow around the object C L shape angle of attack Re sD 1 12 C L 0 VA BEWARE for wings A is the planform area mm B mmm numummam Lift with anown 1 Find appropriate chart for surface Determine relevant parameters to use chart Airfoil type ap positions on 2 If a polar plot determine CD 3 Read CL from chart 4 Compute 4 12 CL 0 VA Remember for Level Flight 1 Wt mm m mmms mmummay mmm cummmmuqy For wings the C L is usually eXpressed as a function of C D andor the angle of attack Example Lift Drag Polar Li or kmmvl x Lem2M u ENSC 3233 5 i W Chapter 12 Lecture 1 Today we will cover Types of Turbomachines Pump Curves The objectives of this chapter are for you to 1 be able to use apump curve 2 be able to select a pump for an application 3 be able to determine if a pump Will cavitate and 4 be able to use pump similarity laws mum 1 mum mmmw Axial Flaw Turbines in unauka minimum Turbom achines are classified by the type of work they do pumps compressors and turbines and the direction of the fluid entering and leaving the machine axial or radial magnum 2 mmm ummm any Axial Flaw Compressors and Pumps Gas Turbine magnum 5 Axial Flaw Turbines Propeller Turbine mmm Wmmm mm mm Axial Flow Water mmme mm mm mmm Wmmm Radial Flow Compressors and Pumps Centrifugal pumps Air W112 Luna mmm mmmmasv We will limit our discussion to pumps Pump Curves The performance of a pump will be a function of the w system ow rate Q It must be measured for each type of pump and is usually presented as aPump Curve The quotpump curvequot plot the relationships between the ow Q the head applied by the pump ha the e ieiency 77 and the pump shaft brake horsepower W112 Luna Water chmm Lemuel Large Axial Flow Pump Curve Pump 1 chmm Lumer WWW mm st um 17 WWW mm st um 17 7 firm 7 vs 0 r Eltlclency 7 7S EEF Opnmum point at pzrauon a 396 Enmipncy quoto and shall nmszpowevx m 39 hail lmlsepnwev u450rpm 7 w o 50 mo 1 50 20a Capaciw L1 Dis mum mm a m m Positive MW H displacement VLWg pumps ammo M901 me J Pm momquot was mm vurmdc V mm 12 mm y Large Centrifugal Flow Pump Curve Pump 2 Hug V We i mm 12 mm 12 mmm uuhmxsuuuuusw Pump Energy and Ef ciency EnergyEq V2 V2 i zlha i zzhL 2g pg 2g pg Solving for ha Vzin 7 ha MZZ ZthL 2g pg Example Given ha 100 ft of H20 Q 200 gspm ands 77 07 nd bhp Convert Q to ftjs Q200im310409 3s 4488gpmft s7 Pump efficiency 77 7Qha 550 bhp bhp 7Qh n550 3 3 bhp 624lbft 0409ft s 100ft 661117 07 550ftlbshp chum 12 m x m m mummsmmn any On either side ofpump If V1 V2 21 22 amp hL 0 h Pz P1P2 P1 a Pg 7 Or between two reservoirs ha 22 21 h The power gained by the uid is In English horsepower units 9 in ftBs W 550 mm 12 km 1 1 Example The pressure increase across a pump is 30 psig What is Q 77 and bhp Pump curve follows Assume suction and discharge lines are equal in diameter EnergyEq 2 2 VL zlhaVi zzhL 2g pg 2g pg Solve for um head ha mm 12 km 1 n The power delivered to the pump shaft by the motor or engine must be greater than this VVsha T0 gt Pw The pump e iciency is thus In English units where bhp is the brake horsepower supplied by the motor or engine bhp Worksim 550 when Worksim ft lbs mum is h 7 p2 71717 301bin2 144in2 2 1 7 624lbft2 h 692 ft oszo Go topump curve tofind Q I and bhp Q 2500 gpm M 557ft3 s 4488gpmft s7 77 70 bhp 61 hp Pump curve for Pump 3 u n ma u EN SC 3233 Chapter 10 Lecture 3 This lecture we will cover More Normal Depth Hydraulic jumps Mxl kmmz Use Manning s equation in all normal depth V iRinSlZ Rh in meters w V leg35 Rh in feet Solving for S7 meters V 2 71 s 7 0 Rf3 or hL SOLL Rh Mxl kmmz From the previous lecture Normal depth is de ned as the uniform ow depth for a given Q in a channel ie The slope of the energy grade line and water surface Sfis equal to the slope of the bed SQ Energy Eq 2 V V 4liy121A LieyzetzzethL 2g 2g Elana m m 2 Example Find the normal depth for a rectangular channel given 11 Q 17 and S7 Manning s Equation SI form with continuity Q VA lARZ3s 2 I l Solve for A and Rh ARZB AA23 A53 quotQ 1 P23 P23 5372 Elana m m z Frict39on or Energy slope 2 2 hL V1 V2 S i 2 7 72 L f L 2g yi 1 yz 2 BedSlope So 217 22 L When the ow is at normal depth y 1 y2 I S7 S normal depth Elmmmkmmz For a rectangular channel A by P b 2y b 53 b2y23 Rectangular S1 The geometry prevents a direct solution for y A trial and error solution for y is required Elmmmkmmz De ne Error as Error Equot Guess a value for y and calculate Error If Error m 0 y is correct stop If Error gt 0 y is too small guess a bigger y return to step 2 If Error lt 0 y is too big guess a smaller y return to step 2 Note the calculated y yquot E Mxl kmmz 1 1 Epgbyf 7 Epgby pQltV2 7 V1 Note V Qby and rearrange Mxl kmmz Hydraulic Jumps A hydraulic jump occurs when a supercritical ow changes to subcritical ow It typically occurs below dams or obstructions The jump is a region of very highenergy loss that is di icult to predict Thus the energy equation has limited value and you must analyze jumps with momentum Elana m m z x 2 b 1 1 Pg y eyz QEEWJ 2 b y2 y1 Rearranging and solving the quadratic yields Note The I and 2 subscripts may be exchanged Solving the Energy Equation will yield the head loss in the jump mam z u Momentum ZFx lpidA ledl PQV2 V1 From hydrostatics p pgy For a rectangular section dA 7 9y 1 I pdA E Ogby2 Elmmmkmmz y V12 V22 7y1217y222hL 2g 2g Power loss in jump P nghL SI QhL P En 11sh h 550 g p Elmmmkmmz 12 The jump can be shown on the speci c energy curve Mxl kmmz 2 V2 uh 1 ipltm21 lthgt 2g 2g 7 7 NotepW 0 andzj ZW 11 And assume V1 m 0 WI 2541 2 H H Q J buhdh jb2gh12dh b2g12H32 0 0 Mxl kmmz Sharpcrested edge Weirs with free over ow At the crest the pressure in the ow is zero The velocity will not be constant It will be greatest near the crest The total discharge for a rectangular weir will be Elana m m z 1 This is based on ideal ow for a more accurate estimate CW23b212H3 Where CW is the weir discharge coef cient Other weir shapes will have different equations and d erertt coe iciertts For rectangular weirs with no end contraction CW 0611 0075 PW Elana m m z n H Q mod1 Imaging A 0 uh will be given by solving Bemoulli s Equation along a streamline from an upstream position V2 2 i Zl Lh 201 7 2g 7 2g Elmmmkmmz 15 Broad Crested Weirs critical ow On the weir the uid pressure will be hydrostatic and the ow will go to Fr l or V2Vc y2yc Elmmmkmmz xx Bernoulli Equation Note Mxl kmmz V V iy121LJc22 2g 2 2g V12m0 2221PW y1PwH Since the ow is critical Fr mm 1 V52 gyc 0 7PWH21ampyCzlPW 2g 2g or y 23 H From Fr l m gy 23 gH Continuity Q VA 23 gH 2Hb b23g12H52 Again we will use an empirical discharge coe icient 9 Cwbb 23 H3 CW 0 651 HPW 2 2n Elmmmkmmz 21 marqme may mmm ENSC 3233 Chapter 4 Lecture 1 This period we will cover oFlow Classi cation oVelocity Description 0 Acceleration and the Material Derivative chmnym x x mummsuu may mmm Classi cations of ow 1 2 and 3 dimensional Do the streamlines curve in space chmnym x unmsm mamaawn Hint for today mamme 2 ummmummm Steady or unsteady Does the average velocity at a point change with time mamme 5 mm m mmelmmms Three ways to visualized ow elds Path line Trace made by a single particle over a period of time particle track Streak line Position of all particles that have passed through a given point dye trace Streamline Line drawn tangent to the velocity vector at an instant of time Streamtube A closed surface made up of a set of mmme 2 Laminar and Turbulent Flow Is there small scale mixing of uid mmme a mmxmmvusumnlmm Fluid motion descriptions In uid mechanics there are two basic methods to describe uid motion Lagrangian Identify for a single particle its position velocity and density as function of time I To describe a ow eld you need to de ne many particles Eulerian Flow is described as a function of coordinates and time for entire owfield V Vx y z t We normally want to achieve an Eulerian description chmnym x 7 Example Given u x ms v y ms w 0 Steady 2D ow Find V Solution V ui vj xi yj note vector V x2 J12 amp 9 arctan vu chmnym x m unmsm mamaam Velocity In Cartesian coordinates the uid velocity is given by V ux zti Vx ztj wxyztk Were u is the x component v is the y component and w is the 2 component Note each velocity component may be a function of position and time t This implies a compleX velocity eld that changes in time and space mamqu x umszwmmm Stream Function A more complete analysis will show that along a streamline e e M V w f0r2 D 0w or mm m mmwjhdmms Example Given u 3 fts v 2 fts w l fts steady uniform ow Find the velocity magnitude speed Solution V 3i 2j 1k fts V1 V32 22 11 2 387fts MM y mum mammal Integration shows If constant M V or were zis the stream function The stream function provides a method toplot streamlines 51mm m x u minme may mmm Example Given u 1 v 2 ms Find Streamlines dx 0 If If u v In a 1 2 Z x 2 11 letz3 Thenatx71y7 4 x2y 2 chmnym x u mmmmummm This operation is called the Material or Substantial derivative or wwuvwwa Dt 7 at 0x ay 02 The partial with respect to tis the local acceleration While the three partials with respect to x y amp z are the convective acceleration chmnym x m Acceleration Consider a particle of uid uxyzti Vx ztj wxyztk Acceleration of the particle is a alValt However Vis a function of xyz ampt so we must apply the chain rule of differentiation 0 0V dx 0V a3 0V dz a7777777 at 0xdt aydt 02d but note it aSCalt chmme 1 Example Given V M 2xt One dimensional unsteady Find The acceleration Solution an Bu Bu an ax 7uiviw7 at 0x 6y 02 a we be 2x at at at e L2H he 2 0x 0x 0x o uo yo39Mo 2 0 a2x2xi2i 002xl2t2 chmme n 0V 0V 0V 0V aiuiviw7 at 0x 0y 02 In scalar terms x at 0x 0y 02 0v 0v 0v 0v a iiuiviwi y at 0x 6y 02 0w 0w 0w 0w a uiviwi 51mm 6 m x u WWW mmxmm any EN SC 3233 Chapter 8 Lecture 6 Hint This period Pipe Systems Examples mm mm a WWW mmxmm any Conservation of Energy The energy equation may be applied along both paths in a parallel pipe system mm mm a WWW thmxmmu l Pipe Systems In piping systems there are only two additional concepts to be applied which you know already kams 2 WWW thmxmmu l 2 2 VL zlhpl Vi zzhLla 2g pg 1 pg 2 2 Vi zlhplbVi zzhle 2g pg 2g pg Subtracting the second from the rst results in hLia pla hle pl With no pumps hL la 1quotle The head loss along arg path in a parallel pipe system must be equal to the head loss along the other paths kams 5 mm unihme m usw Continuity The mass ow rate in must equal the mass ow out at every point in the system Assuming steady state for variable density 2 pQ Z pQ in out for constant density 2 Q 2 Q in out WXLemm General Solution Method for pipe systems i as h L and D known Q1 unknown 1 Write energy equation for each path 2 2 VL zlhp Vi zzhL 2g pg 2g pg 2 EXplicitly de ne head loss in each branch Li D 2g pathi V h K7 Lpathi 2f 2 2g 3 Assume a ow in each branch Q t i i 4 Solve for the head loss along each path if equal stop if not guess again and repeat mmms mmummm mmm Dmxmmusw Example A parallel pipe system Example A dual exhaust chm mm 7 WE mm x mm Enrme mu usw ENSC 3233 Chapter 8 Lecture 2 This lecture we will PoiseuilleHagen Equation DarcyWeisbach Equation Moody Diagram Etamm km 2 mm Enrme mu usw From before for laminar ow um ilgjz u r 3 1w R Integrate u over the pipe area to obtain Q Looking down pipe 1 u dA u 27Wdr volume ow in donut Ehmrxx Laml a 2 R mam mummsmmusw Hint for the period In a complex system of pipes and ttings the total head loss is computed from the sum of the loss from each part 2 2 IV V h 77 K7 L Zszg 2 2g The energy equation becomes E11958 mm mam mummsmmusw QI Area R R udA qur IuZ rdr 0 Note dA 2727dr R 2 2 Q ludA APR 171 2 rdr A 0 41 R Integration yields the Paiseuille Hugen Equation 4 4 72R A 7ZD A Q p p laminar 0wamp horizontal 811 128g 4 ill 11 L7 laminar general case 128p mmm ummmuw 2 2 V V i zlhpl zzhL 2g pg 2g pg 2 2 2 2 V V V V i zlhpi zz2fiiZKL 2g pg 2g pg D2g Example How does the average velocity relate to the maximum Average velocity nah RzAp V A Q 8M 7sz 8y From last lecture MJW max 2 4y Elmmkamz mm Enrme mu usw The mean velocity is 12 the maximum in laminar ow umxx rm 2 7 mm mmm m m Now if we ran pipes at various velocities we would nd h L DC V n l for all pipe with laminar ow n m 17 smooth pipes with turbulent ow n 2 rough pipes at turbulent ow Since we operate mostly rough pipes in high turbulent ow say Lets put it all together umxx rm 2 in Energy Equation on a pipe 2 Note point 2 is downstream from I I Z 2 2 V V i 21 72 Q 22 hL no pumps amp turbines 2g pg 2g pg 2 2 V 7V 7 1171 27 more 2g pg hL M 21 7 22 uniform pipe Pg E11958 mm x 2 hL 0C L L D 2 g Introduce a proportionality parameter the Darcy friction factor 2 l 1 Pipe friction D 2 g ZDg h 7 f L W2 This is the Darey Weisbaeh relation for h There are other relations around but they have mostly fallen out of use with the introduction of calculators E11958 mm H mmm ummmuw Pipe Friction What form does energy head losses hL have If we ran experiments with different pipes of various lengths but at the same Vwe would nd hL i I7 L DC If we ran experiments with the same kind of pipe but with different diameters we would nd hLoclD Ehmmeammz y mmm mmmmuw Returning to the energy equation 2 2 V V i zlhpl zzhL 2g pg 2g pg 2 2 2 Vi zlhpVi zzfLVi2 2g pg 2g pg DZg This is a form of the total energy equation where pipe friction is the only energy loss All we have to do is gure out f mg m 2 u mem mummth mu any Lets do more experiments Now what happens if we change the Viscosity of the uid but not its density f lu for lowV f i i for highV Likewise compare different pipes by their relative roughness gD g f i 8D for lowV m f 8D for high V 39 Emma Luna 2 13 H1 VD8 H2 VD0 H3 VD f H1 gD H2 Re pVD Note that f is already dimensionless so it is a H 2Dg L LT2 f hL 72 72 V L LT H3 Plotting up experimental data yields By inspection Emma Luna 2 u mum mmmmw TA El L E a 1 Equivalent Roughness for New Pipes ll rnm Moody um 7 and Culrhrnnk Ref 01 Equivalent Roughness 4 I39ipe Feel Millimeters Rivclvd steel 0003700 004 Con 10017001 03730 Wood we 1000070003 010709 39 391 000085 016 G ilvnmcd iron 00005 015 Cnmnmu 0 chl or wrought iron 000015 0045 Drawn mung 0000005 00015 Plastic gum 00 smooth 00 mumh Ehwmx mm 1 Ehwmx mm 17 If we changed 0 keeping the other parameters constant we would n f l0 forlow V f p forhighV Using dimensional analysis We know f VD 8 144 0 Buckingham Pi k5f VD map r 3 M L T l ln k r633 Let V D amp ube the repeating variables Elmpax mm 15 Example Calculate the head loss for 100 oSf 0lm 5 commercial steel pipe owing water at 2 ms Re pVD100020118X105 u 000112 Get 8 from table 5 0046 mm 46 x10quot5 m gD 46 x 1039501 000046 From Moodyf 0019 7 V2 7 0019100 m 2ms2 DZg 01 m quot1 29817 S2 hL 39m Elmpax mm m ENSC 3233 x m Chapter 8 Lecture 1 Hints for the period The two most important equations in Chapter 6 are Darcy Weisbach pipe friction head loss hL f g 2 Minor ttings loss hL KL 7 g chm rm 1 1 Friction Free body inside a horizontal constant diameter pipe Momentum Eq ZFX 711471721472wa OQV2 V1 With V1 V2 TW AI71 172 Mm 172 71D 41 mm mm 1 Read Chapter 6 Not responsible for Today we will Start Chapter 8 Viscous Flow in Pipes or why we have a Reynolds number Re pVDy kami 2 Elme m um ms KB or 4 71 p2 BTW So to predict pressure drop we need the shear stress kami 5 Review progress to date We now have four basic tools of uid mechanics gtgtEquations of State Hydrostatics Viscosity Gas LawCompressibility gtgtContinuity gtgtMomentum gtgtEnergy In other words you are about equal to an 1850 engineer What do we lack gtgtT he ability to predict friction momentum transport Mum 3 Reynolds39 Work Near the end of the nineteenth century Osbome Reynolds 18421912 was trying to gure out how to predict the pressure loss in pipes One of his experiments was to measure ow in a simple pipe system able injection Ehnmkami a mm msczm He ran the system for a lot of conditions he changed the valve setting the inside diameter D and even the liquid ie u For each run he would record D u and Q He could calculate the mean velocity Vin the pipe using continuity for incompressible uids VQA QEDZ As Hagen had before him if he used a glass tube and injected dye he found something curious chm rm 1 7 mumsmm um msczm The transition range was hard to predict F able injection chm rm 1 m In laminar Relt2300 ow there is no mixing of the uid and instantaneous point velocities are the same as long term averages 1 E 1 I time F aye injection Ehmkaml x um imam If you were to measure the long time point velocity ur you would nd laminar u 1 E u ES 2 ur 0c 17 VEumax turbulent r 17 uroclilie Va um In turbulent Regt4 000 ow the uid is well mixed and instantaneous point velocities are not equal to the long term averages aye injection ur Long time average W M time mama y uuhmxmmnmynsczm If we could look very closely near the wall during turbulent ow we could see this turbulent 5 laminar The turbulent ow has a laminar Viscous sublayer The laminar layer obeys Newton s law of viscosity u Tlui 0 mg mm 12 The turbulent regions obey a new Viscosity relationship T 7 1 where 17 is the turbulent eddy Viscosity It is not constant but is instead a function of the Re In general 17gt gt21 Another way this is described is to call the turbulent shear stress the Reynolds Stress rmu ipuv 0 chmux rm 1 Note at r 0 239 0 At r R 239 1W wall shear stress Return to chmux rm 1 Note Because turbulent ow is always three dimensional the Reynolds Stresses must be analyzed in 3D It gets pretty messy Return to the pipe with laminar ow Place a control volume at an arbitrary radius r Ehmkaml Newton s Viscosity Law in cylindrical coordinates alu 239 7 u 7 Substitute into shear stress relationship and rearrange alu 7EEerr 211 Integrate Idu 7 E I ralr 211 u i rz C 411 Flow area 7172 Shear surface area 27ml Momentum Eq in x ZFX plmzipzmz TZMaOQAVz Vl Note again that V1 V2 let pp1p2 E l r 01 A 21 19 21 uuhmxmmnmynsczm Note at r R u 0 Therefore C Ale 411l u 08 7 r2 411 Note from before A 239 7R W 21 solving for Ap 239 2l A L p R