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# BasicFluidMechanics MEM220

Drexel

GPA 3.65

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This 196 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM220 at Drexel University taught by AlisaClyne in Fall. Since its upload, it has received 159 views. For similar materials see /class/212391/mem220-drexel-university in Mechanical Engineering at Drexel University.

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Date Created: 09/23/15

Today s Agenda 12412 Calculate free surface slope and pressure distribution in fluid moving with rigid body motion De ne a streamline and the streamline coordinate system Understand the origin and restrictions ofthe Bernoulli equation Calculate static stagnation dynamic and total pressure using the Bernoulli equation Apply the Bernoulli equation and conservation of mass to solve inviscid fluid ow problems N A A 01 The pressure at a point in a static fluid depends on A Depth Depth and specific weight Depth and container shape Depth and surface pressure Depth surface pressure and secific weiht 20 20 20 20 20 03 0 0 F 1282012 Fluid moving with rigid body motion Fluid mow wr39i iiow skw S Mgs as Quit v1 931 mtg Avehl New39i39w39s 1quot IN 4 L kwd u b l elm k FSma yK L dungREC SD R 5 L 10 quot39Vurvfxu Wcssm A1640 k l a M943 393 315v Sum forces on fluid element in x y and z D Mew NM M 239 Law a y dimiqu 2 c c MBYL Sum LLS 25 3Ffsgt p 37sz izy ZIF 8 KF3SL quoti5 x Aie 4quot W 2 A M W Ix 2 SR quot9 555 2 S L SxSJSZ 7 was T 1282012 Combine forces using directional unit vectors A waiimi wu vwiars If vL 5 We quot3 We 3amp5ampgt2K38f wow Dumb iiMusk 93 E LM 6V oS NO HM iiy Fluxa MM wWwd SkwgLmss r 5 393 33 3 Q a Mm mass calamion YL SSVLN 3 quot Equation of motion for fluid without shear stress Rmnt VLS39IMB hAL KIMj ovum vLMv r dAVLcri h 0 Manny A A 3 V 3935 Ta Li39s39tD Lgk Kiwi EEK nu tr Msnka 519 I39K lt 1282012 Find slope of free surface of fluid moving with RBM Z Ww kb A UAW Mennmks a P an r gt 55 73349 e R v T E g n 0 S nu ax o X39 n K no A max JiVlL HM 23 Z 3 33 E a y mi A Smu a 0 2 h a 739 Kgmgg has 4 LAP m 2 Anna Express as total pressure change at free surface TutA enssrm Jung 3 ME Arriquot 39 39J7 V g J3Aa 7 a AP a3oa 7547 a A4 41m u mku ho SM 2 mm a3939 A PMS un t Swan X L AW SID 0 39l s K3 Slog 0 9 Wk elk J31 13 6 now 4 sur tu bus at 139ka swan v Wellh u Can 1 sum 1 u39A Mum 1282012 Rigid body motion example 2 An open cuntaincr of oil rests on the atbed of a truck lhm i Li zlvcling along a hm i 20111sz mad 3155 mi 39hr As the truck slows uni V formly In a complctc stup in 5 5 what will he ihc slaps 0 he 01 suri acc durng the period of Cun x izmi dccclcramm 1 at V quot h z glue 7 wkquot an M 3 75 y c lM W s Am quotI if 1 WW S r Mquot 8 s What is pressure difference from A to B a 43 0301 A ii 3 2m 4km 4 Lawn A E R347 2 75900 ORV A L VB APsA gang PK 239 1282012 1282012 How does fuel pressure during acceleration compare to the fuel pressure at terminal velocity A rocket accelerates upward at 20 msZ 33 33 33 J quot WNW A n Vwiaum B Smaller WWW CThe same 4 33 lvis u dy a How does the fuel pressure gage reading change v on m illmug ynur car at 50 mph in m x at hnvn in lhe diagram in x t x V m 7 direchunl You 50 mph Air t L m U mlle at a consmm mu m39cr 5 w w mmmlmgnnynurl 1k is bullnm lnl l nl39rhc Hulk Gasollne T e A B Decrease How does the fuel pressure gage reading change Vnu m h39n39lllg your car at 50 mph 12 quotw m htmn inihcdmgmmiin hcryduccuum m 50 mph Air Limitmu u u my 39msrmu mic um i mum win lhc mum on uur g T r m ihc 1mmquot Mi om mm Gasoline 0 20cm m 2 m v v oulsw w1 y 3 4e 5s Luau 311 mlh nLaMLycta dinedquotm 3 13 2 V harkvl 5H SMIA Ali3 4quot is k 35quot 4quot SM 391 ags d a Aguiuchmi u en gut VLKA M w Mow13L Types of hydrostatics problems Pressure variation with depth P2 3 pgh Manometers and barometers FR Pgh A Force on a submerged surface I n yR y A y ZMW 0 Buoyancy FE PgV for neutral buoyancy FEW WLngga dP 0 Rigid body motion 1282012 Today s Agenda 12412 1 N w A 01 Calculate free surface slope and pressure distribution in uid moving with rigid body motion Define a streamline and the streamline coordinate system Understand the origin and restrictions of the Bernoulli equation Calculate static stagnation dynamic and total pressure using the Bernoulli equation Apply the Bernoulli equation and conservation of mass to solve inviscid fluid ow problems Define streamline coordinate system Wmhui Luv quotDA vhf mswcmw Lu Vang13 Mu nquot 5 WM fwtdL g mgmkklomlvde rh wusth x mumLb lambdaKn W OK Whit qusthuw39M m39 3M3 Hm vialqu Wwdk KKM loud www Ittm L 4 1282012 Define particle motion in streamline coordinates PbsA IM 55 ousting aloha VL a3 1 1 qun IV at Accdwd u 25 I 195 2 isot c l3S b c h Ml l c an 73l Apply F ma to fluid particle along streamline l 251Equot SMQSRS3J 25 W3 MgQS nS K Ws My stK3 1VSM9 a m gsgg stSuS we 3 S V2573 5 RB oadLmlo valwx 2quot LA A Eyq s ne l x 5553 AM I 1 3351ae Vi 1Q As 1282012 Apply F ma to fluid particle along streamline A 1 We I A vid wlo envmu ABMA wequot a 4am iv the side A9 Lquot L Abe30 Violin us As ls 1Q As 523 mJMIsmbJWT 39MMML Whiz Sine vi lbms llsz quotbumlkmthj seHalsdl MS 4 r VL 3b Gals ham h 3 W A M J 4 14 has 4 Ml m 76 W Mam Bernoulli equation 51 How vs 0 Muqu Mr 0 S lunda no elm 3 wk A nidt n fwl 41 gkmmha Q luwrvcss Ll i ems P 2quot l 37 hashmi W W Wquot 3 u m 1 LL wh mal Lind quot quot 39 u 4quot M U WQ erul39m 5 Mata 1282012 1282012 What is the definition of steady flow A No fluid acceleration No change in fluid 25 25 25 25 properties at a given ocation overtime C No change in fluid properties at a given time over any location D No change in fluid particle properties as it moves through the flow field Today s Agenda 21612 Solve fluid flow problems using the Conservation of Energy in CV form Derive the differential form of the conservation of mass and the conservation of momentum Solve simple viscous flow problems using the Navier Stokes equations N A Course Logistics Project grades are posted on BbVista and graded projects will be emailed back to students later today Since many groups did not do well on the project you can correct your mistakes and resubmit for 72 credit Only resolve problems that were not correct the first time Resubmit new solutions electronically as PDF with explanation Maximum 12 additional credit eg if your original score was a 60 you can move up to an Must be submitted before the final exam Project 2 will be posted by Friday 21712 at 5 PM Due 3212 at 5 PM no Project 3 You must work in groups of 4 students Midterm grades will be posted by Friday 1 PM and exams will be returned and solutions reviewed next Tuesday If you have questions about dropping the course please email me or come see me on Friday Final Exam 8 AM Tuesday March 20 Location TBD 2202012 RTT applications Cons of mass momentum energy DB 6 A Dquot zchpdeJspbVn7 A hi1 O IpdVIpltV gt1A m CV CS T17 Z ZEIdeVIpVQf gtjA m atCV CS BE e 2 an Vzz Wnel aZCLWWipeV n2 A Common forms of Cons of mass momtum energy Conservation of Mass Simplifications Shdu lu Guam Quip tn at ads laxMU 2i Ewan CS fvt wl J J Bypu run JIM Conservation of Energy P V2 P V2 Simplifications 2 Z z h h II le Ululd 39UMIM pg g g m g lK 3J49 Z pVA 2 pVA out in Conservation of Momentum Simplifications 9 Leif 2202012 2202012 Cons of energy includes stored energy heat transfer and QM Wm I pedV WWWA CS CV 25 25 25 25 A Momentum B Pressure 0 D Friction The loss term in Cons of Energy is always gt0 50 50 B False 2202012 Conservation of Energy problem Water flows steadily in an inclined pipe with the static pressures atA and B as shown Which way is the water flowing Loss must in 30 Y 445 Mal w 1 us 0 HI I armLn 4 um MW stdMW Z imt SAW V EL Mt L 3 quot Ohmmustwlr p5DSl V Emu g h gt1l Ll Mud hr 4439s Determine the filter head loss The pump adds 20 kW to the owing water The only loss 0 05 ms w is across the filter A973 m B 0205 m 0331 m 2202012 Determine the filter head loss The pump adds 20 kW to the flowing water The only loss 03905 quotii is across the filter Cur 0 L vzi 1 4qu zl g39a LLdr 157 7a O 005 3A s 2 9 SN ilk v v PMIquot A Mr4 st I I z 5 L L m k 5 73 39 F r MW mam1 M k 40m f sf73m Huey lungn x Is 1113 k u 1L m l Today s Agenda 21612 1 Solve fluid ow problems using the Conservation of Energy in CV form 2 Derive the differential form of the conservation of mass and the conservation of momentum 3 Solve simple viscous flow problems using the Navier Stokes equations Why differential analysis AIMS w in solw Iv uid MPAL J unwindquot miv Fl cv uSL NHL h 15quot 3 CMI kw mmmohm 653 41 has WAR q inqu v a Lvu AL Lks i 1 mendchm 4M dangerm mu sx V954 uwru um 01W mm W m g in km 1 A Jaw I 2 quot l JulHQ Conservation of Mass via Divergence Theorem CV 51M 0 32va k 14mm JVMSLus Swine His unk VJNIM MIS I L Nan 50 W W 145 mind wants 39Dyubm Il W 51 b wi vm k av 639 am 0 3mm Spa 39 gt0 9 Sowu w u 39bh lt0 gt3 Sink II 2 2202012 Conservation of Mass via Divergence Theorem SuiMb Cu xiiMayan L Wan W 0 aw quot 53 4 1mm what 84 anJV cv 3 MN41 043315 5 m3 4 M1 Lama mama x23 Mikh 3quot6VO Differential form of Conservation of Mass G39Mual FVMquot E39TVquot ltI 1F Llwis Shad 6610 y t o Shde muva 6 lt1 0 g w 3 L W61 942 g 0 301 31 LA Gavdull J 5 1 D 2202012 Cons of Mass example problem m For a certain incompressible two dimensional ow eld the velocity componem in lxcydirection is given by the equation U 1w X39ly Determine the velocity component in hex direction so that he VOL umclric dilatation rate is zero 1 ahaKM CymuLMln t 2g 0 Mom 23 3 quot 1 lm 97 3 I 1 1 13 gl tbvrh a 4nd u Su S3 ily z 3 439 3 2202012 2212012 Today s Agenda 22112 1 Review midterm solutions 2 Derive the differential form ofthe conservation of mass and the conservation of momentum 3 Solve simple viscous flow problems using the Navier Stokes equations Midterm multiple choice 1 True or 1 laminar ow a fluid particle can cross a streamline 2 A cylindrical steel needle remains on the water surface due to The needle is 5 cm long and has a densin of 7700 kgm3 Assume 0 c0ntact angle between needle and water What 39 maximum diameter needle that remains on 1 1e water surface Neglect buoyancy A 15 mm i L aquot B 11mm av ZrL C 05 mm 1 D 49 1mn 392 b 1 gjuu Midterm multiple choice 3 Which of the following is a dimensionless parameter pdensity Vvelocity Llength gravity P n39essure i1Viscosity FSlu39f CS tension A pVZLil391 B VgquotL 1 39J 94 1 AP C AP r1V Zgt in 5 D pVLo l 4 What does it mean when a ow is described as inviscid A The ow is moving at negligible velocity B No acceleration is occurring C The uid is incompressible 39T he Viscosity of the uid is approximately equal to 0 Midterm multiple choice 5 How is the Reynolds Transport Theorem written if the uid ow is steady DB 7 inquot T 1 pbii iii A Dr CV J t if f7 PA 0 221 I201 2 2212012 Midterm multiple choice 6 Water nd an unknown liquid 36 15 are in an open I tank Ha 15 ft and b 1 ft what is the height h of the third liquid A 1 ft g 13L who PLGVUJ39lub39Q a glycerin water tit 56 1 s h C 075 ft km 39 l D 06ft Q G 7 In which of the following equations should absolute rather than gage pressru39e be used A deal gas law Conservation of mass Hydrostatics If Bernoulli equation Midterm multiple choice 8 Match the pressure definitions with the appropriate parts of the Bernoulli equation 11 Dynanuv pressure Pl pg Static pressure Stagnation pressure f l 39 1 Total pressure P 7 7 JV 2 9 A block of weight W slides down an inclined plane angle 6 with the horizontal in a thin hn ofoil Viscosity u The lm thickness is h and the contact area between block and oil is A Assume a linear velocity distribution in the oil What is the block s terminal velocity 7 A sinBu l Z F39 O A o M B tth391wsmequot Kr fins C ltVquotsinEil39Aquot r r 1 kf D 1139 39srneir Midterm multiple choice 10 If ow is steady what may change with time at a given location Circle all that apply A Velocity Temperature Density com None of the above Midterm Problem 1 SOLVE I You are going on a road trip in an old Hunda Civic A One cylinder is not working because the piston is stuck Gas enters the cylinder at 018 k in after combustion exhaust gas is released to PM at 590 C from a 6 c the in diametg 1 pg What is the exhaust gas velocity Assume exhaust is an ideal Uas composed primarily of car on dioxide in air enters at 0054 11135 and CN lquot39 Who S ta o I talk Kiwi 005 1 l 1263 a 091 A 2212012 Midterm Problem 1 B The speedometer is broken so you make an accelerometer from a Utube th 25 cm and L 12 em how fast are you accelerating R c 1 011 WI th A a L In 1 M E1cl 5 inks Midterm Problem 1 F39 viscosity is similar to water R if 5 Sn L 3 Q4 37010 v v H 3 You are drinking soda and wonder how fast bubbles rise in the uid The drag on each bubble as it rises is given by Stokes drag F SnilVD Ifa bubble has diameter D mm and bubbles are carbon dioxide what is bubble terminal velocity Vquot 135mg l 110 kg m and fa 2212012 Midterm Problem 2 H A circular gate on a river dam is hinged just below center 2140 inch b38 inch and held in place by a stop er A What is the resultant force from the water pressure on the gate When the water height h 25 ft W W Ea SiluA A 39 in Ell I wt v 6 M wate Lt hmge a A stopper b v Midterm Problem 2 B How far from 1 iiuge does this resultant force act Make sure to indicate whether the force acts above or below the hmge E4 5 1L 1 jg Mm k 3 4 M g V04 kg 16 x3 L lo Law 1 In T 2212012 Midterm Problem 2 C What is the resultant 19 mg on the stopper for this water 4 LA 2 IM o m gb l height v Wat 7 4 M g 574 K hinge stopper Midterm Problem 2 D When the water level gets too high the gate opens and water flows out into a it t e same shape as a te 391 that is I en to aLLLtos here at the to IS t 2 739 39 a a nt of the fully opened gate w t e I 1e river is 150 ft wide and ignore 6 open gate and stopper on t 1e water ow A Cm wuss VAA V545 l V AT 3mm 39ie ktt leiquot 14K V1 Q 01 kl 2212012 Midterm Bonus BONUS At what water height 11 will the gate open ioiun E Melts Maud KM Lung Today s Agenda 22112 Review midterm solutions Derive the differential form of the conservation of mass and the conservation of momentum Solve simple viscous flow problems using the Navier Stokes equations N 0 2212012 2212012 Differential form of Conservation of Mass a 95 o Stul 1310 Linnka MM 6270 Outta D Cons of Momentum via differential control mass 5 SJ Lu 5mg 5852 A 1 a h v Av 3quot it twiij 517 5 f a a w nfrv w a DV a Di low Linda A quotX r aquot r if 59 539ELI 3 ERA 5355 5 163 1 3 8ka SE94 39 KFV i ar SF 39 r 2212012 Cons of Momentum via differential control mass in y dmdw But Ir tr Nr39mol amp quotny 5455 V rxg krrl a 3 40 h A J w I SW fitif 39wa 5 I g is L Wzgguyxrz 0 r l 397 Express stress in terms of strain rate P Z 35 swimw as 3 JV Tu rrg A W km 3 7 beAS 34 quot1 a 9 q w W Substitute into Conservation of Momentum N r it Dr 2543a it 13quot m r and M31 Wuumfu r or vim ii HM u 1 it 1 9 DH 3 9394 J KE 239 593 43 2 04 he at 2d ohmtrio m 3 T 3 9 A3345 39lw x iwesr39 7 73 9 1k to was gm Navier Stokes equation in x direction l x h 9 A J 3 3 Ag L J vi u imuwtcdhn n ISMu 9 M F ungiulj h ML MMI 2quot 0rd Burt gal L3 Mini 3 Luke Law 5 2212012 Today s Agenda 11912 Pressure variation with depth l Pressure measurement barometer manometer 3 Force on a submerged surface 4 Buoyancy U39l Rigid body motion The center of pressure on a submerged hinged gate is A Above the gate centroid 63 B Below the gate centroid CAt the gate centroid D Not enough I information 1192012 Summary Force on a submered surface ZW o k W wait Er FNA W H3361 3 tr 4 Example car in a lake 95 4 19 quotWk 4 139 makul 4n oPlM 3911 1w hinge 01 4 tgglmA 1w 7 I Imo m mymxmau handle nnpoord l 3 Ix 411 30 1 3 TM 3 I hula 3L le qb 33 95 WK 3 1M 39 g 1192012 1192012 Example car in a lake ZMeO R Emd at W4 4 badv ll h an d Ie W Equot oopo XE ghb Urn 5quot 910309 gram L Wquot W Example midterm problem SOLVE I The gate on u mmll reseruur I designed in open when m water luvc high thereby r 39ngwulcr inlu the ocean scawnler H Ihc quotm mm 39339 wm s gmu i5 8 m wide and 55 m lull39 quotWEE WITW77 A What is the resultant force of the seawater on the gate and how far from the hinge along the length of the gate does tth force act 5 kcA 7 Ul liloquot Uh tei ka T l 111 jg o 3 34 1 W 3 391 CM Us 30139 M 5quot L R I 303 Example midterm problem SOLVE I llic gnu an a smali reservoir IS designed in open when the war reicasiug wuler mm ihe ocean scawaier H the N of height lisw is cnnsizmi at 40 m and the 8 m wide and 53 m tall n um tcr ievel gets ion hlgh thereby 5mm 31 191 1 in 394 3 2 Bl 4 15M 2 ug Es sw w W LKon39Nn my i B Ifthe water height hW is 30 m what is the net moment at the hinge mm Today s Agenda 11912 N 9 5 01 Pressure variation with depth Pressure measurement barometer manometer Force on a submerged surface Buoyancy Rigid body motion 1192012 Buoyancy a La mid M A W A 21 oLLwn hwy mssuc ua HMX Mac43 Wu V Bunnar VON A 39Equot EA39FW w rm VIMA 4325quot Kim quot31 quotEMMquot 3 7 5 131 1amp4qu J 3 WW Archimedes Principle WAR m1 Lamammdloe Ii 40 tuxV 4 4L Msfbui Hod Eamm wtr ufml 4M7 Au my 4 u snafu maven 4 0 50le Majmoa WNW39Mwmg wlk IJ My mam H gsw 9amp5 L g L W szw s T R gt1 xl 1192012 Wrecking ball example mrr armn w T An EWC VL 53 this Nquot lller Nmr raw E g i Vlln39 hos 39 e T 10839 w WV ML wowvb WWquot What is the net lifting force of the balloon L A 5m diameter balloon contains helium at kPa abs and 15 C R 2077 m2SZK in sea level standard air Neglect the balloon material weight lt4IL395HA3 6 Q 125le Qlikrzm VD A 67 N ww ggim B 134 N gig 0522 N Qv rgna vt D654 N m E 787 N 1192012 Helium balloon example 1112le A um lull Lfswsm z 1 w W 3 l Midterm problem example 1 You an m clung 0mm landing sul39rty m m quotcw Bocmg 7S7 Dreamllnex Aswmc lhe unmn is a cylmder um l mo 1 long and w n m mammal mm 1 mkeol39l39muss of 16925 slug A Ifllh39mn n lands m I quot mman olumc will be submergch walnut3 my 3 w Hm Hmvh i 6 Z fmw 9124 S W ml 31 mm wlm pcxccmugc ol39llu 1192012 1192012 Stability A floating body will always be stable when its CG is above CB 7 CB is below the waterline s a CB is above its metacenter Metacenter is above CB WUQWgt Metacenter is above CG 1192012 Today s Agenda 11912 Pressure variation with depth l Pressure measurement barometer manometer 3 Force on a submerged surface 4 Buoyancy 0 Rigid body motion What are the primary force that can act on a fluid if no shear stresses are present A Weight and pressure B Weight pressure and friction C Only weight D Only pressure E Only friction Today s Agenda 22812 1 Review common project questions 2 Solve cylindrical viscous ow problem using the NavierStokes equations Obtain nondimensional parameters using the method of repeating variables 9 Problem 1 Volume flow rate to achieve 700 MW Powergeneration The hydroelectric power plant at the dam has main generators turbines each wi 0 MW capacity and 95 f ciency The turbines are the water enters the turbine a la y lame e height 4 9 m and exits axially diameter96 m The water in the reservoir in 110 m above the downstream river The tur ine outlet is to the river bottom assume a river depth of65 m nd assume there are no other losses in the besides turbine efficiency What volume flow rate is needed to achieve 700 MW output from the turbine W nmrbine 2282012 Problem 2A Vertical force at turbine bearings Turbine installation The Francis turbines used in the Three Gorges power plant each weigh 450 tons The turbines are mounted vertically and use thrust bearings to withstand he forces from the turbine weight and the water ow What is the vertical force on the turbine thrust bearings Problem 2B Horizontal force on housing What are the horizontal and vertical forces applied to the entire turbine housing if water enters the structure at a 45 angle assume 15 m diameter pipe and exrts horizontally assume 96 m diameter pipe 2282012 2282012 Problem 3 Downstream river height Flood control The dam is also intended to minimize downstream damage from oods by controlling water flow along the Yangtze River At its absolute maximum the reservoir water level can rise to 1804 m with a maximum discharge ow rate of 102500 m3s This discharge ow does not pass through the turbines and water can be assumed to be an inviscid uid Nearthe downstream town of Wchang the river is only 100 m wide What is the s mom and height at Yichang during a maximum flood discharge Problem 4 Flow rate to empty lock Shipping Two ship locks were installed at the dam site to increase the shipping capacity along the Yangtze River Each lock is 280 x 34 x 5 m and there are 5 locks in a staircase con guration If a ship traverses all 5 locks in less than 4 hours what is the volume flow rate from one lock to the next Project submission Due by Friday 322012 at 5 PM Submit as a PDF to BbVista 1 submission per group Include all group members names on the rst page Use project problem format see syllabus Equations can be handwritten or typed Iftyped use an equation editor Show all the relevant work include diagrams and explain clearly what you are doing Remember signi cant figures gt Any questions See me in 0H Wed 911 AM Today s Agenda 22812 Review common project questions Solve cylindrical viscous flow problem using the NavierStokes equations Obtain nondimensional parameters using the method of repeating variables N A 2282012 2282012 Navier Stokes equations In x direction au au 6 au 6P 6 azu E3qu p u v w 7 pgxy Z a 6x by 62 6x 6x2 by 622 quot2 F In y direction av av av av p u v w 6 6x ay 62 In 2 direction 6P E 7 pgy u 62v 62v 6 2v 6y 6x2 ayz 622 iwwimwiw E j pat ax 6y az angz ax2 6y2 622 Which of the following assumptions is true 59 16 c ddy 0 11 14 DatybuU1 AA 3 y Ni Cylindrical coordinate system Poiseuille Flow b c an E m 0 avzv 2v 9 zv I7E T r 2 2 z p 61 V66 I 62 6z 39u r6r 6r r 662 22 CM r at LED 3 539 L a V 39 e 7quot 77 erL G ASSUMde A lawnf C lb vulva 39b All xo 39D A nmdn39c 1i Joo 3 N1 39J39S D 7rquotJ v a 31 C ldublt quotW I 0 1 I LP 8 3v1 d cc ah au r lu t quott v Tv quotJ Clhv C1 w t silt Jt Cylindrical coordinate system Poiseuille Flow l 0 IL 039 33914 CJvv J 6 A lj BC 4 LA CC N K quot o A wave quot 77 1 L m s wa Cz w av 12 c 4 9 Vznt39 nk1 vdnd W al C 030 t 0 2282012 2282012 Cylindrical coordinate system Poiseuille Flow Inlt mlow Q Q SVLZ39irVJr Today s Agenda 22812 Review common project questions Solve cylindrical viscous flow problem using the NavierStokes equations Obtain nondimensional parameters using the method of repeating variables NA w 2282012 Dimensional analysis uni in s wrl TwLIUM Mat ml in 11 50le L3 km a Sub mu M jddwmw ivwls V 39Dw 53E HAMY film 4 W Rquot M o vdau V 34 MpUun w ham or kwwon CowLin OIMIJ39US 4 Ovut A m m s ms Defining physical parameters with basic dimensions NLT F LT Wi mus 4F 399 L W L Tam T innu EC F 1 1 W E L T1 1 F1 4 VA 1quot U Q L e a M quot AT L Use dimensionless parameters to reduce of ind variables Crud A39M lhmllss W 10m 39atam rs M l39 Huasfwhoa VhohlA dimmmlm 3 QNCD E39s 170223 l Mutbe equal lonJ s 41 bwhafdlmdln u wqu um L39u39ljh39k andIn 1 nub quotmarchb qp39nlnlu JIMmulls Propeller example Find T for given V Imfwbnf uMLLks T 1 Mpw AWW AAfaint Even Aquot VMlIehs Ivi39l um vi lam Minnieu mmquot 2282012 Propeller example Find T for given V Mmue l vi dlimis lmlq s A 39E Athhal 39HAuruM k 395 VFiVbn39nuu midi r quot 39ll diMuASmAr umtll l 3 bars SAD MS 1 firms Stink Vlfu39l39ms Halls mHIndify Ts glt39ln iii M Q 9L Myst L D WW 6 V Propeller example Find T for given V g m ll 71 JWMASmlkil arm s T zT gADIVt 39 L dwmims Ey mSS w hams cl Last A 5 c L o v L m EVAN Pum h L 39M Lfr39m g2115 M D A A39 l i L ol5ABc 39Z llKW F D lC c z 2282012 Propeller example Find T for given V 9 W 4 wt WT TWWQ Wu Ag bexf 1T 55 c l WPva 73914 n 73 Eg mss Tam Asa 4 an 1Tquot39F 71213 T H4 3 Km V V 2282012 Today s Agenda 3112 1 Obtain nondimensional parameters using the method of repeating variables 2 Use dimensionless parameters to analyze a model system A Determine ifinternal pipe ow is laminar or turbulent The Pi terms must always be A Negative B Positive 0 D Equal to the number of original variables 94 312012 Wing lift example 7 M L V39L Iu 7 4E QMpClV Wyn 34 Wing lift example l 1 m aw mL rlt239gtYL39gt g f 0 zac 051 1r mg LJ v US33 1T5 CK LL V TE 6 TV TV QLW1T3 hgt4 3139 if g 312012 What is the relationship among AP R and o 1 Pouiside 1 M Ki 3 fps P k 3 3 o u 398 szi K K j i APR Km if A I S 7 1 05 B HIZAPRUx AP C H1 R m AP D H1 Ra Soap bubble example A K i W Pom E L L gm LL L i R i 6 an L if A K x in A 6 lt quotgt F quot 5 F L 30 L 495 M o 6 6 4 ll 5 17 WW igt l Lx o 1A A Ar ghw 312012 Common dimensionless groups in fluid mechanics mm 7 mm mm m mm mm 1H Rcnol numbcr Rr inch km a y 0 impmmncc m 7 mm mm nll lylm m llmnl llyimmu x pmhltm y I39luwx Ill l cli llw uumn cxsvlnlin of III llmd l nupnrmm If the Reynolds number is large pVl 79 Re A Inertial forces dominate B Viscous forces dominate 15 C Flow can be 3 4 considered inviscid DBoth A and c 312012 Today s Agenda 3112 Obtain nondimensional parameters using the method of repeating variables Use dimensionless parameters to analyze a model system Determine ifinternal pipe ow is laminar or turbulent N A Model and object similarity CD Glamtn SmMy mu 40L km Mm 8W m LL LL 3539 a at in ti2 Kmmulxi Similarla vdm l Academia al wluhl 1i 1 s MediaRad PFothd Dammit L494 mnwa MI fartlama Lwi m kc olv39ocol 24M 312012 Example problem Car An automobile is designed to travel at 65 kmhr A model of this design is tested in a wind tunnel with identical 1 rd sealevel air properties at a 15 scale The measured model drag i nf ing Determine the drag force on the automobile an e r 39 come this A m at CDoww DV x Emu e7zzow Example problem Fuel pipe line The pressure drop per unit length in a 025 inch diameter gasoline fuel line is to be determined from a laboratort usinghe sa yiubing but ith v vFa er as the fluid 2 Q E 0 f ts ig t isl th39e water velocity to use in the test LT A 529 fts 5M5 B C041 ftS D 1 89 ftS v111 We go lsI WR A t M 15411quot 1 A a LS IE If gasoline velocity should b 63 312012 312012 Example problem Fuel pipe line The pressure drop per unit length in a 025 inch diameter gasoline fuel line is to be determined from a laboratory test using the same tubing but with water as the fluid f gasoline velocity should be 10 fts what is the water velocity to use in the test AP 9 E 15 v25 3 t r 7 3 37 V f c Example problem Fuel pipe line The pressure drop per unit length in a 025 inch diameter gasoline fuel line is to be determined from a laboratory test using the same tubing but with water as the fluid At this water velocity the pressure drop per unit length was found in the test to be 045 lbft3 What is the predicted pressure drop per unit length for the gasoline line 57 A 0125 Ibft3 B 0252 Ibft3 C 196 Ibft3 0051 Ibft gl l 1 39 131 W Example problem Fuel pipe line The pressure drop per unit length in a 025 inch diameter gasoline fuel line is to be determined from a laboratory test using the same tubing but with water as the fluid At this water velocity the pressure drop per unit length was found in the test to be 045 lbft3 What is the predicted pressure drop per unit length for the gasoline line if ta 6t 00510 39IH Example problem experimental data The pressure drop due to friction for flow in a long smooth pipe is a function of average flow velocity density viscosity and pipe length and diameter AP fVpLD Rewrite this function in terms of dimensionless parameters 312012 312012 Example problem experimental data 2 Plot this function using the followmg data Appl2 pm L U D Example problem experimental data Plot this function using the following data Appl2 pm L 2 D 100E11 y 14289x15474 O R2 0712 0 100910 O 100E09 100E08 100E04 100E05 Pl2Re 312012 Example problem experimental data Plot this function using the following data APpDz D pm 2 X L 100E09 y 00691x1E266 R2 09983 100E08 100E07 1 83 AP D3 VD 00691p i L k J 100E06 100E04 100E05 m2 Re 2232012 Today s Agenda 22312 1 Solve simple viscous flow problems using the NavierStokes equations The NavierStokes equations are derived from A Consenation of Mass 47 B C Consenation of Energy D 2nd Law of Thermodynamics 19 2232012 Navier Stokes equations In x direction 6M 6M Bu 6M 6P 62M azu azu 0 u v w xu at 6x 6y 62 6x 6x2 6 yz 622 In y direction av av av av 6P 2 p u v w u 2 kat 6x 6y 62 6y y In 2 direction Assumptions needed to solve NS equations A Muftika FlnI quotlons39lln1 3 Shall e 0 P 31 um I dIuJ rm Igolmp IA In L C W 399 w 3 vwo quotD How CAM was imaw h meS k39tnknlm nbs klK 55 mild in 11M ho y 539 I wwn i n b 0 V D lwldhw Wilma M 60 aka Cl 2232012 Boundary conditions that apply to NS equations D Naif alumni1w um UN N mil l6 Ml bum AV 17 B 1 Wuquot 3 TA Tn 3 H hle 345 Ml rt na kt golfm fa o r GD N duhhm 0 salutary a Nibj 1 230 Which of the following is not true A Assumptions apply everywhere in the flow B Boundary conditions apply only at boundaries C Incompressible flow is an assumtion D The noslip condition is an assumtion 3 types of NS problems to solve by hand Problem type Assumptions 39 39 Diagram Atm IUA 44 no 03 M Llnm39wquot a I I 1k 1 ful ll ML cm ullen l 3mm s AL 1qu W I M I cwiit SUN 0 Daryl in 0 B it Q 393 u h E 11 N Fu Sim as 3 in 339 39 w W W39 1 Fixed parallel plates Iscous incompressible uid ows between two in nite D VEE39M 0 OW vertical parallel plates Determine an expression for the pressure gradient in the ow direction in terms of the mean velocity Assume ow is laminar steady and uniform Q A o E a E 7 pyMILVLVAJLPg azaxayazay axzzz C A G C 4M x 33635 0431 W1 5qu AW l39Diu h 2 4744 MW mum 2 0 VM 7 n 39S Bmlkswmvionzl jm quot39 aw Df 334391 13quot 23 A quot 2232012 What is the boundary condition at x 0 Directmn uf How HM A v O B u w O Cdvdx O Ddudy O 1 Fixed parallel plates M1741 7K 397 jg3 4quot quot513 Dreuion al low At W m E ff 8 2 1143 I39m3quot g L 973 6amp2 Q39CI rhodi vi hg9 I 11 9 W4 CL Al39mmo i atec W l at My 0 V ampS mc1 FEE ameu 2232012 1 Fixed parallel plates OIW M V13 1quot R 9quot l l Direclloil ol flow w 2 4 h L If S VJ Z u2 hg k h zilw 3 fair 41m 9 skim 3 5 411m JE quotl 7lt33quot3 In aw1 ltnol ohr l V 9 id 1 w u 7 2 Couette Flow Two horizontal in nite parallel plates are spaced a bottom plate is xed and velocity U Determ39 ulem The the upper plate moves at me the velocity distribution between the plates and the llowrate per unit width in terms of b a 399 N39S M x du hnx p 8 C i 14 vw bar i fayaz 9 Can A WW 9 MY IM GD SINK MS MEI4 1 Ruuunf D WV 2 72 8282 ax x 14 X By 439 39 id Fixed L I 1 at 74 Ward a ID v13 T W D 1 MM m 7 ln 0 5 2232012 2 Couette Flow u M1124 4 J12 r39Hr A 5 35 Ag53 W W Mw C a Caw C Am 3 41 c cI Q1 Ina 9149 s W J vdudi nd G dirt 13 0 ourCL u HLcn c I CL O a L 392 q E u f Lj 1 1 s M at Star Wk i iu v 2232012 382012 Today s Agenda 3812 1 Solve internal flow problems with major and minor losses 2 Calculate lift and drag for objects ofvarious shapes Fully developed flow means that A The uid velocity is constant acrossthe pipe cross section B The uid velocity pro le 70 U The uid is accelerating at a constant rate 382012 Minor losses from geometry and components 1 v1 Lani1 g5 lhlmhr KI 38 AP lug law 7 314 a atquot L 31quot 9f Il39l nugt Minor losses from geometry and components um I 2 382012 Equivalent length FPan m 1n nudd lofnaaumm cm L 14141 lasscs uuuld Lt floriou 13 quotKM WM Lulmajr Lm Solu any Mark W 11 n D D 3 It Example problem major vs minor losses Water ows steadily through a 075 inch diameter galvanized iron pipe system at 0020 cfs Your boss suggests that major losses in the straight pipe section are negligible compare to minor losses us to elbows and ttings Do you agree 1quot l D 7r 1 warm rimJ mama 9 l 39 quot W 139 i213 MM 9 ll 1 1 r 2 v1 an MW 33181 a 1 a 7 43 Ml Hem om f mt LiarW3 382012 The Moody Chart n m mm mun Example problem major vs minor loLsslegs K Water ows steadily through a 075 am Wm mm mm inch diameter galvanized iron pipe system at 02 c our oss suggests that major losses in the straight pipe section are negligible ared to minor losses due to elbows and ttings Do you agree VI In t Mil If i L1 105 1 mgr k 51 3 Luv 09 ml as 45 um The pressure drop in a pipe is greater if the flow is A Laminar B C It depends D Pressure does not drop in the pipe due to conservation of mass Type of internal flow problems Fur kHn 51quot Await 43 1391 END Solut hrde Sm new Lawrhs M ma a 382012 Example problem Find AP Water ows at a 39 39 1 inch uialllelel Find the pressure drop per unit length along the pipe 1 I Q 9 5quotquot V f 411 gr NH MUM VJ so a lvlo 1f Ilqunnhui lenlunM n m I39 mm 7 Anti i uhIvnmk 61593 W 395 ovwlt 10076 3 Km 9 mwmm mm 1 Uhm uHH lixniunilli V A L234 1 ii ln quot The Moody Chart 382012 What if the fluid is glycerin Glycerin flows at a rate of 20 gallonsminute in a 1 inch diameter stainless steel pipe Find the pressure drop per unit length along the pipe I 1 rb Kam rlw A 313 RES 5 d Kquot k c 231 lo 165 K 300 IF 007 Q P 1 10 ac 4 AT 9 What if the fluid is air Air flows at a rate of 20 gallonsminute in a 1 inch diameter stainless steel pipe Find the pressure drop per unit length along the pipe 0 R39 B tlo39 Eh u 1TH 157 K H313 P O HS 4 at ALE owl km s w if 103 p ll 1 81 lm hr with Tm Me Wto 382012 Example problem Find D quot fasteelr39r quot 39 F pressure drop of 5 psi er 100 R of horizontal pIpe Izqummm lenlvmM n m I39 1mm qu r uhIvnmk l A De 39 wfz w L f l l g W y q rwawugvg Gum 05 9 05quot KL urn 65 Moons 39 km W k oom 61M9 091 1 JW 4quot J e The Moody Chart 382012 Example problem parallel pipes ll 8110 The llowrate between tank A and tank B is to be 39 creas 7 by adding a second pipe 39om node Cto tank B lfthe elevation oft e 39ee surface in tank Ais 25 ll above that in tank B determine e dia e new plpe Neglect minor losses and assume the 39iction factor for each plpe Is 0 VFW 935 40 5 v1 amp V I 1 w 4 v K Z 1125 RITE 1 SH NM V LoSHs 19 2 0 aka zbn39 Q m 97 I 1r 939 Q5 R39ws I51 2 Example problem parallel pipes Amtquot 8110 The llowrate between tank A and tank B is to be increased by 30 by adding a second pipe 39om node Cto tank B lfthe elevation ofthe 39ee b l sur ce mm B determine the diameter D ofthe new pipe Neglect minor losses and assume the 39iction 0 m factor for each pipe is 002 a V u E Q Q HQS LEI LA Ql ls l 191 1 11 quotm4 2 l V lioVs I ls l E 3 11 quot N 5 2 3339 figs Q 1034 2411 vyksquot I V x 1 Draw 5quot f i 382012 Exam example problem A pump transfers water from a pool 6 m deep to a slide which is at a 30 angle to the horizontal The total length of pipe is 35 m and the pipe diameter is 15 cm The pipe is made of plastic with surface roughness a 000005 m The water volume flow rate through ump is 03 m3s KLfor each rightangle pipe bend i and the pump is 70 efficient What power is required to run thetpump a v J v Cms39laun 23 2quot 1 tin l quot39L Q h l E 3 7quot 000 005 R12 R 2 Z39H 1 ZD a 00003 4 oolS r L wl E liq it g 301 if Iz ln 382012 Today s Agenda 13112 1 Transform a Lagrangian property particle to an Eulerian property field using the Material derivative 2 Use Reynolds Transport Theorem RTT to express Conservation of Mass in control volume CV form Choose appropriate CV to create simplest solution A Bernoulli Equation For ow that is Inviscid 2 Steady 3 Along a streamline 4 Incompressible P0V2 pgz consl 2 E IOV Perpendicular to a an ER streamline 212012 The exit pressure for an incompressible fluid jet is A Static pressure B Dynamic pressure C Stanation ressure D tmosheric ressur 25 25 25 25 Midterm example 3 an ul39hnl mm 5 t x n cm crcd by A lhm L1ch ml uni Huck in u m llir1vrgc m w mmmphm m an mulur ul lum I in Am h 5 m Inn is lhcuxlumc lhm rm mug mu m uk ml Bernoulli Gnu A L 11 7 IA at AL 0 A39 l39 Tlvt L9A lKXfll4 i quotPgD 2c IA UL gQ31 w We 39 VL 71239 quotIs Q V L1nLIquot oz l 0 0111 quot7 i m at am cur 212012 212012 Midterm example Bernoulli and cavitation x iquot k 111 i lhtmullmumJmm rlul 3m Q 5 iv o qPl aV m iv er wri R two 9 Wyamow 21 quot mas Fwy 0 m V V A VLAL Vb VI T Vl I L V5 9 n0 m w senPo mm M Sire an h Stine Pressure outside a tornado is pressure inside a tornado A Greater than B Less than C Equal to 33 33 33 v a 391 35 a 9 1 WW Fluid kinematics study of fluid particle motion Phbhvv ws M OMLAUMH n w 9 km A Ann L th faw txJu w M Puk 0 MW Mk WM W m Raw ww b 41M laws Was gor39t w MT Define Lagrangian and Eulerian fields Like 23 Cum Hutifcwl uo 13W AWL MLasucl Jg We Mr39hm n Z Em Ta 9mm FeTar auv whvuqs 13 x quot quot MW Mat Wham1 Edy 04 WVqu M a 39VM ota hv Oilr 39hw39u lvu39hov TERI 1qu Car47039 Vo w attl dm meumd 39 lb 0 loudlyquot 212012 212012 Define Eulerian field properties SCA at 1 PipL k QLl 139 M mm 0 KL J 1 f i 39VGHKU OSM 421317 k a 3 Chi 3 1 M M 2 A W W 54m gnaw17o Havana a Quaemh asit my 48 p911 ch 1 Eulerian acceleration vector Weigh MAM TLtmJ 1a wrikvugahiut 14de all 9 am 3 4 NA iquw 31 3 21 m jg A 017 what its nv thw av 25 as 3 quot aL39LI I J 39 7E4 79 Va fr w 27 3 396gtI Method2 DMV h quot 1 V 212012 Define Material derivative 3L 2i n39 E ILxiVaa39twil at i memukdwl ML illwat wwt Mee c AS VVLNLt point 3733 unsimaa Wernmi39 0 F rrhaj39 u 5 395 L unwam we o is wattage Unsteady vs Convective terms MW Wu Hake 5443 sid Va o 3390 3 E E T39 3T 3139 13439quot H DD u VTjw r L39 4 HERT a writ Lea I Today s Agenda 13112 Transform a Lagrangian property particle to an Eulerian property eld using the Material derivative 2 Use Reynolds Transport Theorem RTT to express Conservation of Mass in control volume CV form Choose appropriate CV to create simplest solution A Reynolds Transport Theorem RTT Ahmfwm plum WM 1W8 ML My 5 835 Lin in axial valnu u v rm Nui Atplua JA m upjjlgvk 61151quot 75 mm Mame Twyla 5 an wdem PM E wt 5 M a quot CV 2 at t gem 1quot quot Bu 5 GWEN5 wa39OX39t at fe gone 51 393 441 quot 3154 39r 5am 212012 Reynolds Transport Theorem RTT Swat 4 nfv hwl t 32mmquot Lutrle 4am bun 359v DJ W 39 2 Vb cg a QVASLTIQVJA V So be W T39 r WA VMJ W L f s rh SW quotD 6quot L d v gas b 9quot m SL9 A 7 V 39 39 01 lea a gul dyL ruff ampw Lb 14 4 U A ATEnu y a Reynolds Transport Theorem RTT 1 asen l CINE mmlwwo Mk I quotIrma Mm wad Mn iSML clva and wept 12 Park Mu MW Vinma vcdvugrmlnl V Sumerian C63 quotHusk Cs MMT P55 A A n h divL0K 1 5 as Vw V MJRM ll hr 3 a A VM M a qux g a gut39 f Eme SLS 34 4 n 212012 Reynolds Transport Theorem RTT b5 9 S q 39 N L v a A A m 3 u Q t inu di X39 hi rth art hvt Hm M wf CM CV 01005 CS ampu 1 has a 7 Mk CD 9 ampI D Vt gguequot L 3 w W t W D A A WH a Egygxmtgrk 4 Mass Lack w lw wt 9 RTT Application 1 Conservation of Mass may 4 b 1 3amp9st Mt 95 DMM a quotK O D Fgcvqb f E s bl gt014 L Mwl v chive cash1391 o ugLVR3M a o r vem m e m a UA e w 39 gm gm 212012 212012 Conservation of mass example 1 A rocket motor operates steadilz as shown The combustion products approximate an 9 quuld nxygcn ideal gas with R 1775 11252 R Calculate l quot5 M gt AWE l gt noon V2 ln 115 on PM at out lamMU i m 238w 1 15 llllll7 of 5 5m lllIlIquIKl39 T llIslugt w 399 s W 95 5 4 MA3A KLV A A 1 Vs an I MW If E sa WW aha93 4w V2 1 1102012 Welcome to MEM 220 Basic Fluid Mechanics January 10 2012 Professor Alisa Clyne Today s Agenda 1 Introductions 2 Syllabus overview 3 How to do well in MEM 220 4 Survey 5 Fluid Properties Today s Agenda Introductions Syllabus oveniew How to do well in MEM 220 Survey Fluid Properties CHLOONA Introductions Professor Alisa Clyne asm67drexeledu 2158952366 AEL 1700 Office hours Wednesday 911 AM TA recitations Danielle Jacobson d93934drexeledu Office hours Monday and Wednesday by appointment in MEM main office Randell 115 TA grading Miao Yu 1102012 1102012 Today s Agenda N Introductions Syllabus overview A Logistics B Grading C Objectives D Schedule E Problem format 3 How to do well in MEM 220 4 Survey Fluid Properties 01 Syllabus logistics Overview Mechanical Engineering requirement in the thermofluid sciences fluid mechanics M EM 220 thermodynamics M EM 310 and heat transfer MEM 345 Class Tuesday and Thursday 200 320 Randell 121 Recitation Wednesday 100 Curtis 353A 200 Curtis 353 300 Curtis 353 400 Curtis 258 Textbook Fundamentals of Fluid Mechanics 6th Edition Munson Young Okiishi Huebsch Website WebCTBlackboard for course materials Syllabus grading Participation 10 Recitation quizzes 20 Projects 20 Midterm Exam 25 Final Exam 25 1102012 Syllabus grading Participation Each student will be assigned a clicker which will be Quizzes Projects Exams to answer inclass questions based on reading and simple problems Percentage answered and correct will both be used to calculate grade 8 quizzes will be given in the last 15 minutes of recitation based on the weekly recommended problems see schedule closed book closed notes no makeup quizzes but lowest score will be dropped Each course module will have a project which is an openended real world problem Use fluid mechanics concepts from class with outside research to solve Teams of 4 students Midterm exam February 9 Final exam week of rch 9 Raise any grading questions in writing to professor Syllabus objectives N 01 N5 Explain the critical properties of a fluid State Newton s Laws of Motion Conservation of Mass and Conservation of Energy 1st Law of Thermodynamics as they apply to fluid mechanics Understand how each term of important fluid mechanics equations relates to these fundamental laws Classify fluid mechanics problems into several basic types Solve the basic types of fluid mechanics problems including statics and dynamics inviscidviscous internalexternal incompressiblecompressible Justify the validity of assumptions necessary to simplify problems Predict the qualitative outcome of a given set of fluid conditions based on an intuitive understanding of fluid mechanics and dimensional analysis Develop a demonstration to explain a fluid mechanics phenomenon and explain it effectively 1102012 Syllabus schedule Week I Date I Subject Homework HYDROSTATICS Jan 1 0 Introduction Fluid properties 1 Jan Wreka 1119 C111 16 2732 40 ReCQI units 2126 45 65 95 102 Jan 12 Hvdrn tatlc manometer Jan 17 Fluid force on a surface buoyancy Ch 2 2 18 27 30 41 2 Jan 18 Hydrwtaticx 28412 45 56 59 72 97 98 Jan 1 9 RecQZ uid properties 110 115 Ri id body uid motion Project 1 Due Syllabus sched u le Week Date Subject Homework FLUID DYNAMIC Fma andBemouIli Jan 24 Bernoulli 3 Jan 25 ReeQ3 hydrostatic 3 13 8 E 256333119532 13 Jan 26 Bernoulli normal to streamline 39 39 39 39 39 line Jan 31 Control volumes KIT 4 1 74 4 4 Feb 1 ReeQ4 Bernoulli Ch 5 5152131 Feb 2 39 5 1 Feb 7 Conservation ofmomentum 5 Feb 8 Midterm ReView 5 2 Feb 9 MmTFRM FXAM Feb 14 Conservation ofmomentum 6 F217 15 Conservation ofenergy 5 25 3 Ch 5 45 56576696 ReeQ5 Conserv ofmoss momentum 101110112124 Feb 16 Conservation of energy Feb 21 NaVi 39swkes ammoquot Ch 6 13 84 90 92 95 7 F217 22 NHVIZVVSIG es 616 3 100 Feb 23 ieegoConservotion ofenergy 6 86 9 Pmde 2 Due 1102012 Syllabus schedule Week Date S Homework DIMEN S ubj ect IONAL ANALYSIS AND MODELING Dimensional analysis Ch 7 12 14 18 41 8 Feb 29 fimej gfal aquot if 77 3977 46 52 60 Marchl Q qwer 0 quotV 39 39 ProjectSDue Modelin Syllabus schedule Week Date Homework March 6 Ch 8 2 5 27 28 42 P e ow RecQ8 Dimensional drabsir 7 53 60 69 77 83 93 March 8 13 12 1338 43 44 14 FinalExam Ikview 64 72 92 95 98 101 1102012 Syllabus project format 1 Problem description Diagram the problem with all variables 2 Given parameters and axeslabele I velocity v 5 ms 3 Researched parameters including reference This section should also include a free body control volume or control mass diagram whichever is appropriate 4 Parameter to solve for pressure P Nm2 5 Assumptions inviscid ow 6 Solution 7 Answer Syllabus problem solution suggestions Describe the fundamental concepts underlying the question that is being asked Consider which physical principles are involved Begin with a basic relationship or equation that can be used to find the unknown Make additional diagrams as needed remembering to label with all variables used in the solution Work your problems algebraically to the end Substitute numbers and units into the algebraic formula only when calculating the final answer Rememberto consider significant figures This is especially true in Excel worksheets Write all numbers that are not dimensionless with units Box the final solution ofthe problem If a problem has more than one part box the solution for each part Check your result Do the units match Is the answer reasonable Is the plus or minus sign proper or meaningful Include any explanations necessary for the grader to understand your thought process This is critical to partial credit 1102012 Today s Agenda Introductions Syllabus overview How to do well in MEM 220 Survey Fluid Properties 01wa4 How to do well in MEM 220 AQNA 0701 Keep up to date do reading and problems each week Participate actively in class and recitation Come to office hours Be courteous to the professor A Arrive on time and do not pack up materials before the end B Focus on fluid mechanics during class C Try not to distract others from teachinglearning Focus on learning the material Contribute your best effort to the class 1102012 Today s Agenda U PPONT Introductions Syllabus oveniew How to do well in MEM 220 Survey Fluid Properties Today s Agenda 9900 Introductions Syllabus oveniew How to do well in MEM 220 Survey Fluid Properties A Define a fluid Understand and use dimensions and units Measure fluid mass and weight Define pressure and viscosity 00w 1102012 1102012 States of Matter ELIMD Srle Linub AS ihtu wenssx bh quotl hwmev ssiLh Cm stink km 4 degrvn WA dlLvm Alrg39m M9LCVLLS Ma lmiu ScarsJ7 you 2ch Fluid Definition ARMJI o quoS39Vanu 3939Aa ll Cums Laknumluk m dA shu EVLSS Simu S Vu39tSSG ska Am in 11quotij Qrco Dimensions and units D iVNMSa M 14st arch WA M VL 791m MU 4 i Lt Umzk1 yomkiwiwc discuth 293WS Dimensions and units Tables 11 13 14 1102012 Properties of common liquids Table 1516 name 15 w mu m m Speci c Dynamic Dullbil Weight iu u Tammiaunt p 7 4 Liquid 1 C lkum l lkNnr l 1N smll Waicr lib 9W 1Nll Ll F i I Today s Agenda Introductions Syllabus overview How to do well in MEM 220 S urvey Fluid Properties De ne a uid Understand and use dimensions and units Measure uid mass and weight De ne pressure and viscosity mewwe P 90 Quiz 1 on Wednesday will be on the units sheet 1102012 Today s Agenda 11212 1 Define a fluid 2 Measure fluid mass and weight 3 Define pressure and viscosity 4 Calculate the effects of surface tension What is meant by defining a fluid as a continuum 03 Fluid parameters vary continuously Fluid parameters are discontinuous Fluid parameters are the same throughout the whol fluid Fluids deform continuous with shear stress 1122012 1122012 Fluid Definition 80L 1 D Alum 5 F W1 rm W m Today s Agenda 11212 1 Define afluid 2 Measure fluid mass and weight 3 Define pressure and viscosity 4 Calculate the effects of surface tension Fluid Mass and Weight Measures 1 01 mags CW1 Valium er 1 we pt I at 397quot a quot 7 awr quot49 4 owT ML MAM Mug AFW LJ Fluid Mass and Weight Measures ms LV M 5 gunL1H DWS sc guuglmhM 1 Setting v31lnuU Uquot i F Nu H 894 W3 A L3 K V 8 ka rm1364 NMENSIOQLESI huMWlSVUL Hails Q Consvkayd39 quot 1122012 Properties of common liquids Table 1516 name 15 slimim Duuiiu Weight Tammiaunt 7 4 Liquid 1 C 1k39nr39i ikNnr i 1N smli wim 150 mm mm Li 139 7 1 Fluid mass and weight problem A person weighs 180 lb on Earth 1 What is the person s mass in slugs and kg l lb W 5 51 I M2 39 r 8 Sim 3Uqu 7 k 1lt39 i3912 L t 8H 3 quot l l k I 1122012 Fluid mass and weight problem A person weighs 180 lb on Earth 2 Assume the person is a cylinder that is 6 ft 183 m tall and 08 ft 0244 m in diameter Find the person s density and speci c weight K Welt1iowoow witM3 v V 39rR B W g What is the person s specific gravity of A 0954 kgm3 B 1048 kgm3 C 1048 D0954 1122012 Today s Agenda 11212 1 Define a fluid 2 Measure fluid mass and weight 3 Define pressure and viscosity 4 Calculate the effects of surface tension Pressure definition is WMalv39u WL UL14 HNJPIMA IM 394 it i A intIE h gv u lm ln I L39ifplomsl v 1 1122012 2 ways of defining pressure value A10Solut NASqu fola hvivlv APwai Luum Almjs 20 2 6113 fussm withquot 4v Iota WOkablic gussm Cumin 03 in 1mm w m 5 mtss M 8013 mm You measure a negative gage pressure This means J cor gt Your gage is broken Gage pressure can t be negative The pressure is greater than Patm The pressure is less than Patm You are at high altitude 03 O U 1122012 Viscosity definition quot fLSis39iuMkaQ A wt 1 Mo l39lm Muron 14m mm 911 l I JEN r E M W7 A esiw SW 39 3 bur 61X L V M Mb 21 3 swim NW Wait Types of viscous fluids Nariqu Hm M Cmsimd39wwnll J Jr islimfln rdchu 9 Skin 11 th a gas fakvd39 am 6quot V Q gfggxdw 2 we gum m X 41 ii 1122012 Shear stress problem M m Jannme 35 f ZPO M Am W5Zo c 41 0 1quot 630 I in 100 My A 871 L I 00W339 s Gas compressibility l 604 LAM M M 1 RR T e mwpudm Hi j T L 13A wsiml Huqu iv ado 52 Ami wt ask an 9 In Link a Mob P Q T 1quot CoM Msstc jam 1122012 3303 02320 bi FEE L its Lass Er JE lt51 rig b9 tr twiy 15 552 E lt11 325 it An 92 42859 3 Fa mi 49 a 225 Agwr v 02320 1122012 Today s Agenda 11212 1 Define a fluid 2 Measure fluid mass and weight 3 Define pressure and viscosity 4 Calculate the effects of surface tension Surface tension 54 m VLer lujik at 4113 nthEU 4341s and minimum M Mn 1 1ch3 Sole vw olms Mal3 Agra galMQ F tb Kiwi 39 iszbcm Classic surface tension problems W MK 2amp0 A f ilk AW 1 Pressure difference dwell it Art luar 2 Capillary effect K2 k o k Q15 mewsefanzk tr 39 Plug 31 Today s Agenda 11212 1 Define a fluid 2 Measure fluid mass and weight 3 Define pressure and viscosity 4 Calculate the effects of surface tension 1122012 Today s Agenda 31512 hook d rag Explain how winglets increase lift and decrease Calculate boundary layer transition point and thickness Calculate speed of sound and Mach number Course review Lift occurs because gt 093 Pressure on the airfoil bottom is less than ressure on the to Fluid particles moving over the top and bottom ofthe airfoil must meet at the trailing edge 77 3152012 3152012 Circulation and topspin s xmnuj No sq Lon WM V Vhl w Y gt 39Phuu I h J Fr my Vb u w Winglets L 46 a J W Winglets I mDurm DR Daemon nf mgr 17Mmmet stuN mm Winglets REHUHION m INDUCED nRM SPANWXSE urr DETRIBUHON hunky wmmm m rLbwymrvrwvmn m v w v Wm m Basblin g mam121mm oral wmg chum I valevenre 1 II J M W quotMW 3 1mm 8 0393 Macnzom lvkwdhmenl g mu cum n mos g 07 E or 137 mu Mann 018 Im Hemmer v50 wun wungm 050 Basehua 1 17mmmet stuN mm 3152012 3152012 Today s Agenda 31512 Explain how winglets increase lift and decrease drag Calculate boundary layer transition point and thickness Calculate speed of sound and Mach number Course review N A00 Boundary Layers 399 060w din Ito Mun 1M uncut MauiI l fElquot ltUeO F A 011 3 I gt H ill CM W d w 3951 RM 08 mus FI H Mull 3 wqu quot5 W WM 5 Macaw mlh y mi altoLuz ml 1107 Laminar vs Turbulent Boundary Layer BL ball Wx u 43m W Av39kkvbk lvd th orst a RU HDE eS lfzs L 5 9L E Q dab re 39 K 134 a gt3 1 93M 3lag IQ4 Boundary layer thicknesses WM 3 9 y memj ax muss k Jun d 101 MJJUU Swsegwggaa a 6 mm r WW 3152012 Boundary layer calculations Laminar Turbulent I S Jquot mu Thickness a p 5 x 4 x 11 Displacement l m E 0123 thickness 739 tr x 4quot Momentum Q 5 am 9 z 9 01 thickness x EM 1 9 0021 Skin friction a 39 T coef cient C4 KT CF Rt 1 Example problem Al 0000 H a 35mmquot W i1swoquot MW que RL U 6HOS Mv ni 339 event If 5 v01 out 7 mb f gyfzml Aquot izbrlo 4 00 irks H11 2quot Sifts l 3152012 How far along the hood does the BL turn turbulent Approximate the hood of your car as a flat plate 35 ft long moving at 50 mph p 000238 slugsft3 p 374 E7 Ibsftz Requotans 5 E5 hagzx M l9 39l l39 31 A B 00088 ft C 157 ft D 23 ft 55 Enos 00013 9 10 371 loquot What is the BL thickness at the end of the hood Approximate the hood of your car as a flat plate 35 ft long moving at 50 mph p 000238 slugsft3 p 374 E7 Ibsftz Requotans 5 E5 gillml A 00859 r 60 B 00247 ft 39 0 00137 ft 16 13 3152012 Today s Agenda 31512 Explain how winglets increase lift and decrease drag Calculate boundary layer transition point and thickness Calculate speed of sound and Mach number Course review AWNA Compressible flow ma 43mm m G Wu at which Yum um k MM valid P l g 95 In 9L1 W Ailud39mnlh 4J1 Gas PW bahsold 1 FE 3152012 3152012 Types of Compressible flow l quot quot quot39 lt mm a as m 9 Wsth 9pr 03 A g L I 0 aw k 5mm Wu 1 o bem quotWquot Mach and c calculations Cm m3 at 58 Hard 174 I NA 239 LESS 00112 01qu as m mm amo39Vs Q 3152012 Shock wave 39Msw nuzg h whko Novamoo Ob A N4 quotl Acmss SWILWV H n 739T7 7 le m4 Oblique shock wave development ts S at Supersonic pressure source Today s Agenda 31512 Explain how winglets increase lift and decrease drag Calculate boundary layer transition point and thickness Calculate speed of sound and Mach number Course review 3152012 Topics covered before the midterm Fluid properties measures ideal gas cavitation surface tension 6 F 1u P 2 RT E 6y p U L Hydrostatics pressure vs depth manometers force on submerged surfaces buoyancy rigid body motion Pz131pgz1zzpgh ngpgv Bernoulli inviscid incompressible steady streamline ow E PVZ 8n 7 SR P 0VZ pgz C Conservation of mass 03J39pdVfpI7 1A ZpVAZpVA 81 CV CS m Topics covered since the midterm Conservation of momentum vector equation a a A 0 I pdV IpVn1A atCV CS Differential analysis Navier Stokes 61 au 6 aul 8P all 8214 8211 p u v w 7 Xu Z Z Z 8 8x 8y 82 8x 8x 8y 82 Dimensional analysis nd Pi terms H1 1601an Internal ow calculate major and minor losses power 2 Z Z Z 1V Jr V z 7hLhW hLmajfLV h minKLV g 2g m g 2g d2g 2g External flow lift and drag boundary layers L CD1 CL1 7 UZA 7 U34 2p 2p 3152012 What you should remember in 10 years l A A 01 Hydrostatics pressure increases as uid depth increases Buoyancy related to weight of displaced uid objects are stable ifcenter of mass is low Conservation of mass as crosssectional area decreases velocity increases Bernoulli as velocity increases pressure decreases Conservation of momentum uid needs to be driven by a force gravity pressure friction What you should remember in 10 years O l G 9 Differential analysis Navier Stokes equations are the basis for computational fluid dynamics software Dimensional analysis Reynolds number relates inertial to viscous forces and indicates if flow is laminar or turbulent Internal owenergy equation system losses come from both friction and geometry External ow lift occurs because airfoils turn the air ow 10 Projects You can use simple uid mechanics concepts to analyze real world problems with some assumptions 3152012 Final reminders Final exam CAT 61 Tuesday March 20th at 8 AM Bring something to write with and a calculator BRING YOUR CLICKER NO CLICKER NO FINAL Fill out the online evaluation forms especially about the projects 3152012 362012 Today s Agenda 3612 Determine if internal pipe flow is laminar or turbulent l Calculate entrance length for fully developed ow Calculate losses major and minor as input to energy equation A Internal Flow Cmiwk mu MPH13 U whim uSL mun rst rnsslm lAuA 3 a a v ekf 23 139 13 lg mluAa11 Mus m L ml Lab M wywds Malay Van s G lw c laww w Jug Laminar vs Turbulent Flow W LWW F P v gt leO 55 3 777 I 1 2 I I Milan lTthuk I a KO 000 WWW 7777 77177117 Example problem laminar vs turbulent flow Carbon dioxide at 20 C and 550 kPa abs flows in a pipe at a mass flow rate of 0004 kgs Determine the maximum diameter allowed if the flow is to be turbulent M Rug tggvA saw v v 2 Mi 44A alaw lsl 4 U ifml D inno azym 000 D39 oot gt 1M 362012 Hydraulic diameter and transition region H barium 13 ml nud uu kylmhe Landry ml39zd g LIAt UL Autumnal 1 nu F P Ll pk 5quot mm D H int Haw wheat mm M Wt Q H vimi Wis le 5 km vlvMS Flf W541i wusu 91149 Today s Agenda 3612 1 Determine ifinternal pipe ow is laminar or turbulent 2 Calculate entrance length for fully developed flow 3 Calculate losses major and minor as input to energy equation 362012 Fully Developed Flow All1 W l4 w Urlvmu ld It Emlrmu l XL MIW 00lo tb XL Jmmm 44 R39squotb Fully Developed Flow 362012 362012 Is the flow laminar or turbulent 001 Air flows through a rectangular galvanized iron duct 030 X 015 m m3s 123 kgm3 alr179 x105 Nsm2 Pair LIA Vb on Due R x 68 A Laminar 3 le Kim H B Turbulent lidW quot 1 Jawl C don t kno WE mg 23 K glut LL13 04123 Pa lV Uloquot h 3 13 4 a oquot at f 08gt 99 s Is the entrance length larger for laminar or turbulent flow a 39 M tomb 9H l lkE b 73 A B Turbulent Is the entrance length larger for laminar or turbulent flow 0quot Air flows through a rectangular galvanized iron duct 030 X 015 m at m3s pa r 123 kgm3 alr179 x105 Nsm2 Rt c 30155 It l M r 099111 0ob801901n 3L3vu Bulwmu 4Hwa WY YDM a 33m Today s Agenda 3612 Determine ifinternal pipe ow is laminar or turbulent Calculate entrance length for fully developed ow Calculate losses major and minor as input to energy equation N4 362012 Major losses from friction Mfuss A K or A k 11 1 A 11 i quot 3 quot D 2 WW quot1 W4 L L b Jinnah LI RAMk mwa 9 wind 1cm Who13 M Turbulent friction factor For a smooth walled pipe Prandtl 1935 1 l 1 2010gRed f2 08 f NI Red 4000 10quot 105 106 107 108 f 00399 00309 00180 00116 00081 00059 362012 Data show a dependence of f on pipe roughness r sand grain size k pipe diameter V V Nikuradse 1933 L 1 l as 39 32 34 35 35 4 42 44 16 43 M 52 5 1124 50 Hme was temeen lozuoou uni ma a Accounting for pipe wall roughness Colebrook 1938 f1 g 2010g 2511 Red fE The Moody Chart 362012 362012 Example problem friction factor Water ows through a 2 inch diameter stainless steel pipe at 02 i13s Find the head loss and pressure drop over a 200 it long pipe section mm slugsit3 We 234 x 105 Ibsit2 4M W m M Pip u umm mmm v e quot6 mumm m L n I quot 39 s 039 Nm I39m an L D I 1 7n 1 me W rmm mun ii 0011 nwm mm W wti hf you LaniMi m um mus M5 H i M1 c 2153 IR The Moody Chart Wm Is the friction factor larger for laminar or turbulent flow 74 A Laminar B Is the friction factor larger for laminar or turbulent flow Air ows through a rectangular galvan39zgg iron duct 030x 015 m at m3s p kgm3 pa 179 x105 Nslm2 ipmx rimm mm E 315 5 001 6 K J a quot cams mini mm llwuu 4W1va 4mm Mm ELL m lumnmlii mammal 0 Mv dad 40qu 6 1 362012 362012 The Moody Chart Laminar vs Turbulent Flow 39LM39M r 4 de a quotL 9n f39mu m r 739quot HM 362012 Minor losses from geometry and components A kL g K Lh39hu v1 hL MV kL Minor losses from geometry and components Outlets W a n lets Mott J 9 dU i I 4 Q I x Minor losses from geometry and components um I 2 362012 1302012 Today s Agenda 12612 1 Calculate static stagnation dynamic and total pressure using the Bernoulli equation 2 Apply the Bernoulli equation and conservation of mass to solve inviscid fluid flow problems 3 Understand how the Bernoulli equation can be applied perpendicular to a streamline Bernoulli Equation For ow that is Inviscid M t 0 Steady I java Ma a no oter MWI IM Along a streamline Mann Incompressible lt W M PWN 1 2 VEERV pgzconsl wtLan Aw FIMJ 31 Mm wk n The Bernoulli equation is not valid for flow through a pump A B False Types of Inviscid Flow Problems 1 Static stagnation dynamic and total pressure often for velocity measurement l Internal ow with an area change A Splitting pipe 4 Free jet 5 Open channel ow 1302012 Pressure definitions using Bernoulli 1 P EQV 2 pgz cons 5quot 1 sink AvL m t W WM 941314 ffl luf 95h w4wml asvfnsvu an Mun3 wrkluFim l di cl W K visa 1 PP F MinaMi bib Jay45 umber 103 J nude 99le it 4 NF ilkLt 11a ys hw HM Mquot 3 What is the pressure on the front of a car at 60 mph A922 Si 9 whey L B0064 si 0 052 psi 9 quotEmmum Mama D 105 psi 9 ewe3 v i x a E 428 pSi 9 31 chairAu 9wquot FVL 39cPx 7 i vf 39sz 1n 0 3 1302012 Pitotstatic tube M11mmw 1 b ltP i l a 7 611324 V mun 0P6 J vi ML L 5W 39 l e awed m I Z3w39k 4 T HTLEVKH Types of Inviscid Flow Problems 1 Static stagnation dynamic and total pressure often for velocity measurement N Internal flow with an area chane 0 Splitting pipe 4 Freejet 5 Open channel ow 1302012 1302012 Conservation of mass Control Volume m gmaSS Wk 4 WA 97 ma numb 56424 Qo39M39I 3 vdeuvdL Js Q VA Cmsbrvukul aquot wuss E ln AM L mm mm Amw o i39g m 279 quot 7quot viLWJW ak31hwm wuanc v lt Wm w olt I 39 0 2 QVABW 2UAur V VI Internal flow with area change P2 P1 Mullquot Q DL 5 1 1 l t wdw 3L Lf l2 RLa 2 YaJ J Tm Cmuvxkd n M 3quot A i V 1 Wshut Arr avug v VF VLTVLW z MV5Aamp MMWCM 21 7 1 2 I lsa g LV1 tL igV 5 b 39739 K 1 SAVII VT cmy mu u rr sunall lt 39z Internal flow with area change example problem 0 A JLAL D A I V 153 vi D Tquot Q nnuL u 39b wvllC 3 139 VtV15 L21z 7 1 Q2 Y Vt 3 2 43 009315 his 2 LE 0 3 6395 j numb Q c M75 ankA VLAI l Types of Inviscid Flow Problems V lStatic stagnation dynamic and total pressure often for velocity measurement 2 Internal flow with an area change 3 Splitting pipe 4 Freejet 5 Open channel flow 1302012 1302012 Splitting pipe 1 inlet multiple outlets Draw S39ktamlmi e qc9cpy 9 N wfwtrw 9 Draw CV V Cons Mass dvaw CV Swatt J 40 voiced 39 va WL quot54 Splitting pipe example problem quotZL quot a A 3372 Wider lows through a horizons hl anchizn g pipe as Shi qu in 1 0 0 4 03 d lug MHZ Henmine the pressure 1 sec in 3 1 5 Cam Was 0 o Me My wmu QM CD 4 Wag f TH ie tttq i vl 4m M 5 84 quot L Vquot v1 long A gamma Qu V A a C M ll U l n r M ZLSKIS V3 A3 gunvibe WGkQ P iKVJ P A I 393 1w k 1302012 Types of Inviscid Flow Problems 31 Static stagnation dynamic and total pressure often for velocity measurement 2lnternal ow with an area change iSplitting pipe 4 Free jet 5 Open channel ow Freejet fluid discharge to atmosphere Qx Hm wllwdwucrwaIIANJ i r g w ma a rah WLIM Mantra jali 4 oquot Ll V s zlo DB 20quot 3 3M 339 a u 1 s g s 2a Q V39k V 741 m 2 i 2 05 m bLM Autumn M ht39eMnU 7 I39 L I7quot39 I 2quotI 939 M 21quot 1302012 Free jet example problem 3 72 1139 viscnus sl lecrx are neglected and Ihe mail is hung lev m ermine 1hr liuwrilu from 1hr rank shown in Fig P372 Q 7 i l M a 39 ha u Bumm vclm BlQ 4le m FW W ww wwrn a Eifmo shrub 1 vLjW39 N no 4F 9 Mkab hu a Q i 59 021 39l 7 3km quot 3139 i ge39vLLL4 z39 In at 3 61 A quot 001 3INs r 39 3 Types of Inviscid Flow Problems Static stagnation dynamic and total pressure often for velocity measurement N Internal ow with an area change A Splitting pipe A Free jet 5 Open channel ow Open channel flow 34 13mm 2 Causal Wm VlAlr39Vl A M EM g 32x quot y igvfd 3 2a m if 4 1 kmquot 3311 23 21 LLLLHL n Open channel flow example llOD Water How in a recmn ularc nut that is 20 m wide as shown in Fxg PBJOLThe upsmzam depth is 70 mm The water surface rises 40 mm us it pusaes over 1 portion where the chan nal bottom rises 10mm t39vismuse 39ecu are negligible vth is the owrzuc39 l W 439 0 I I gt 7 we deurma Mm W4 M 3 I 016 A h ONl V A3939391AL Jl IL A1 W I 139 mm bad mi Maui 1i 3 MUS i L1L z w L711 Mls 0 V A 039 FM quotaquot 2quot 39 21 7amn 1302012 Solving inviscid flow problems using the Bernoulli Equation 1 2 For flow that is P 3V pgz consl 1 Inviscid ap Steady Along a streamline lncompressible at PQ N Hints Choose 2 points along a streamline and apply Bernoulli equation between them gt Pick l point where you know fluid properties and 1 point where you want to know fluid properties In a large tank at a point far from the outlet V e 0 Pressure at the free surface or in a eELiet is Pam Can use either absolute or gage pressure be consistent Bernoulli equation normal to a streamline lauiri su Zr gg ys 3 at vquot 7 a 5 13quot tum MF PP E 1quot P l as NM W5 tacky 5L wvdA Iqu M 5439 l 1302012 1302012 Bernoulli equation normal to a streamline V1 VigtVo Maknuuu of L o 394 0139 qu 7 Bernoulli normal to streamline example problem 311 4 u at t ninumnprcssmtv inviscid uid ows Stead 12h ular slreatulines around u 7 39 radial variation of 3quot cnzunml Mud m hmvn in 15 Waf e is given hwltetc V is the f v s radius r r t f velocity at the msxdu of the bcn r Dcierminc the pressure variation across the hand in emu of V ru p4 r and p WllCl39L pH is Lhn pressure al r rm Plol the vhf pressure distribution p pm il ro 12 m r 18 m V quot gt quot l1 ill4170 u mu inquot mm mm nuiu i mun lL Lu imam Lquot aw TM39H LM n M e l 1 fqu 77 r a JP g 941m V 39 quot V F l a g aquot r we J 1 V4 9 K r r f u 90 mamath 5 Kr V P vcv T Saint L 9 Q 813 i v39 03939Y y N I 7 9 Today s Agenda 12612 1 Calculate static stagnation dynamic and total pressure using the Bernoulli equation 2 Apply the Bernoulli equation and conservation of mass to solve inviscid fluid ow problems 3 Understand how the Bernoulli equation can be applied perpendicular to a streamline 1302012 Today s Agenda 2212 Solve fluid flow problems using the Conservation of Mass in CV form Choose appropriate CV to create simplest solution Use Reynolds Transport Theorem RTT to express Conservation of Momentum in CV form Solve uid ow problems using the Conservation of Momentum in CV form JON A Midterm Thursday February 9th 2012 1 week from today In class Covers material up to and including today Conservation of Mass but not Conservation of Momentum Closed book closed notes Equation sheet provided Example midterm and equation sheet are on WebCT Midterm will be reviewed in next week s recitation Check online for review lectures 222012 222012 Reynolds Transport Theorem 1 Transforms fundamental laws from control mass to control volume form gchpbdw IpbI7 2 A 0ch 2 Dr S Hurtful f wt 9 m M 439 B M Vj oh 5 w labgal 40quot CS item 8 Simpli cations DB Steady ow m IpbV iA Dt CS 2 Constant properties DB 6 acrosseach m de b VA b VA inletoutlet Dt exclj p p 2 39 39 The RTT transforms fundamental equations from A Control mass to control volume form 87 B Eulerian to Lagrangian form CUnsteady to steady 8 a form 4quot 1 DCompressibe to V f A incompressible form f E gs 3 9 km a a e 0 5 f Conservation of mass example 2 A syringe contains a liquid 8G 105 lfthe liquid is to be injected steadily at 6 cm3s how fast iii should the plunger be advanced quotDIEWM CV it M exAL 441k 7 lg at 0 an QM VQL 1 All a v a r v URI quot5 Conservation of mass example 3 An incompressible uid ows past an impermeable at plate with uniform inlet velocity pro le and a cubic polynom39 l 39t locity pro le x will plan mnlllv 1 mm Mm w a a 735 n0 SI S I ve t m la 53911 ginJ Compute the volume ow rate Q across the top sur ce of the CV owl u St og oi a Q 1m i quot39 uni 5W3 ll 222012 222012 Conservation of mass example 4 For the open tank of water derive an expression r 7 i 7 b for how water height h changes with time CDM ungtul d o LV o 5 5 li 95ml SM 5 mm mm a TLL 74 39 Ellg Z LAz KV39AquotW3AS 4 Li a c quot511 m N34 Allah r Today s Agenda 2212 1 Solve uid ow problems using the Conservation of Mass in CV form Choose appropriate CV to create simplest solution Use Reynolds Transport Theorem RTT to express Conservation of Momentum in CV form Solve uid ow problems using the Conservation of Momentum in CV form JON A How to select a control volume CLWSL CVCS lkak fussu Mus Mum wfwrhq pro kWiW or nish 79 mm 040W rudhlw FlruAlwlaV 139 0 er Va as k 9 CW Whirl musl oL douJ E K le39wl voluw can Wu walk I an Moving and distorting control volumes Mania 0 D39 O d M 9 max I vu39v lNk u avS LAV Joaccwm39l39 394 2 V155 AVquot 1 0 ALLM BM 339 n1de v10 S 9 av Vid n V V a quot7 Vrd39u Vd tquot 222012 Find v2 C 107 ms D392 ms V girl 13 Vl vl 11 Today s Agenda 2212 wrv 5 Solve fluid flow problems using the Conservation of Mass in CV form Choose appropriate CV to create simplest solution Use Reynolds Transport Theorem RTT to express Conservation of Momentum in CV form Solve fluid flow problems using the Conservation of Momentum in CV form 222012 RTT Application 2 Conservation of Momentum quot a quotJquot 9amp5 T mv 9 V J quot 7211 Emmi Am B Jo RTT v g 2 a 9 7 quot 2F ESQH Sax v M V quot39 vir z V Di 9 d w Mam hh HW hub on WWW MLquot u wk awn CS CV wf quot quotquot Types of Forces acting on a CV G GVNAHqu 3M 1 7 Ru Aw erwi 8 1 m gm Awajs hwmaleh CS lwbc 39439 MAP CV SI 30x 222012 Conservation of Momentum is a vector equation 5 SDINL d m39 l allmlw MA 33 a ConSI39dlr 114060 of U S V WMA WTquot33A Ti lat ET MW M x 3 ouTLET AM 6 mevwr v x The standard unit normal vector on a CV is A Tangent to the control surface pointing upwards B Tangent to the control surface ointin downwards C Normal to the control surface ointin outwards D Normal to the control surface pointing inwards 95 0 0 1753 c 222012 Today s Agenda 2212 Solve uid ow problems using the Conservation of Mass in CV form Choose appropriate CV to create simplest solution Use Reynolds Transport Theorem RTT to express Conservation of Momentum in CV form Solve fluid flow problems using the Conservation of Momentum in CV form JON h CoMm example Force on a flange chtSaxlm mu Th Drb gt00L l w 1m v6 1mm P90 mMm 3 Cans mp4 M 1 2g igftv39aA M RR R VlAu QAlvlV Ez 71lb CM WW u 7 E w a gillW a In l5 1 222012 Today s Agenda 31312 Calculate lift and drag for objects of various shapes Understand the physical basis for lift Explain how winglets increase lift and decrease drag Review common Project 3 questions ham x External flow Stalwa tibia Muua ma M K Mu In m mmqu m5 may or Tff fm tk mum We all MintyJ 4 m 44 Jr 3ka MD 413me caulk faWW kg 3132012 3132012 Lift and drag If a z M 5 00 SAP 0 CL it lww l tk tract 3 umsuhku atquot 1 M 4 MM 1quot 01 1 39 W c Q U A Ed mlua M 1 W Example wind to blow over a truck I9 2m 0 w 2QJSM l L M Div 1 A A 2 h w m dd Cuquot v39cl 6 gnu L u mismva a Hawk 3132012 Example parachutist terminal velocity PW 1 bu3 ago 90 I M W CH A JWMH quot wl olh wDNl lm aj 1261 b0 i quot 177 7 1M gig TEN U lSWs sDwrk Example airplane lift and drag during cruise 2g 0 blew a g 1 Mag 33 1 39P39D Qi k Pressure vs friction drag Edna vs 2 85 S 4 a Separation definition 9 W 3132012 3132012 Lift occurs parallel to the fluid flow direction A True 95 B False Separation why does it occur 5 o WJ W n AMJ39llk CP 397 Pr Alma cl ILAJL vlan39rvd 39 7 Muuu MSML anon quot DMOoM u s quot hunky Ln What bowlquott 13quot ms its Aw n chd39l n 04v39fMJLkueluw an cl Separation how to avoid it mum 1115 Kama Lt7awxlm M AM 3 W145 1w 1 MM slot allws I mum at I Mh Mama luw ilovug 43 4 SWAN a luau tabx SM Separation occurs when A The angle of attack is low B The flow is turbulent C he fluid cannot overcome the adverse D The slats and flaps are deployed 95 3132012 Today s Agenda 31312 Calculate lift and drag for objects ofvarious shapes Understand the physical basis for lift Explain how winglets increase lift and decrease drag Review common Project 3 questions AQNA 30 39 3 a 1W Z How lift occurs explanation 1 S39lwmlm awedw 5 a Palm L5 Mu f Du 1621 quot R ssW J J b ws Cody 0 wV M39h Pd quot PH at I 3132012 How lift occurs explanation 2 2 walm ye V me cm is delud m 53 1M MW 9 NW 3414 6013 415 RM am In o nh t It than ALLl eushs M Am Aw Flu and 0 push m Ll hf How lift occurs explanation 3 90904411 Mam Cr mmSu J MI pl we mums and M is Wuhh 439 quotp 939 41V ng 5113 ACLovJ viscous c wts 39 u in la w ml aimlmhg 39 oJMst Swv39L vi Viral Eq VHF L49L 3132012 3132012 Today s Agenda 31312 Calculate lift and drag for objects of various shapes Understand the physical basis for lift Explain how winglets increase lift and decrease drag Review common Project 3 questions hwwe Project problem 1 1 LiftDrag The Boeing 7878 is designed for a takeoff vertical acceleration of 1 89 at sea level maximum engine thrust and 190 mph horizontal velocity The aircra is designed for a cruise speed of 560 mph at 40000 it at 50 engine thrust If the maximum takeoff weight is 484000 lb and the engine thrust is 2 X 64000 lb determine the coef cient of li and coef cient of drag at both takeoff and cruise required to achieve this takeoff and cruise performance Total Lift Force 450002 lbf Total Drag Force 21598 lbf l0799lbf per engine AERODYNAMIC DRAG REPORT at MACH Altitude pressure 37000 feet AS 4 5 KEAS EO D KCAS 77 Project problem 1 1 LiftDrag The Boeing 7878 is designed for a takeoff vertical acceleration of 18g at level maximum engine thrust and 190 mph horizontal velocity The aircra is designed for a cruise speed of 560 mph at 40000 it at 50 engine thrust If the maximum takeoff weight is 484000 lb and the engine thrust is 2 X 64000 lb determine the coef cient of li and coeffcient of drag at both takeoff and cruise required to achieve this takeoff and cruise p rformance Wing Area definitioni Vm 39Piano Cioss defmmon apezoidal wingravea 39Wlmpvesi39 definitio Ecn area CDE area AEDC Project problem 1 5 V 1 LiftDrag The Boeing 7878 is desi a takeoff vertical acceleration of 18g at sea level maximum engine thrust anorizontal velocity The aircra is designed for a cruise speed of 560 mph at 40000 it a 0 engine thrust If the maximum takeoff weight is 484000 lb and the engine thrust is 2 x 64000 lb determine the coef cient of li an coeffcient of drag at both takeoff and cruise required to achieve this takeoff and cruise perform ance Gun FJ Ma Om ZR 0 28 O 3132012 Project problem 2 2 Airfoil design Using NASA s FoilSim so ware httpViMwgrcnasagovWWWk12airplanefoil3html design an airfoil that meets these needs Note that that you can change the angle of attack for takeoff and cruise ight Submit your solution as a screen shot of the NASA FoilSim page with the coef cient ofli and drag shown for both takeoff and cruise conditions w win awimww i W t i W Project problem 3 3 Engine t The required engine thrust T for aircraft dep ds on cruise velocit peed of sound 0 engine diamete air densit air viscosity u an gine rotational velocity w HF 1D m a A Develop a dimensionless relationship among the critical parameters B You build amscale model of your engine and test it in a sea level wind tunnel Can you achieve similarity for all your dimensionless parameters between your model an t e engine if the engine flies at 40000 ft at a cruise velocity of 560 mph a sz r l lls sl 3132012 Project problem 4 4 vuMuImn ulccu gtygtlclu cu tome uleeui air 39 compressor to supply the aircralt with air for cabin pressurizationair conditioning 39 39 quot or uulel 39 While the 39 quot usually large customer bleed system you still need some bleed air for the cabin You are modifying the existing CFM56 engine customer bleed system r the GEnx engine for this aircralt In this system air is extracted 39om the compressor 9 stage P 12 atm T 250 F m 01 slugs through a 2 inch diameter pipe routed through a high pressure valve and combined with air extracted 39om the compressor 5 stage P 4 atm T 140 F m 02 slugs A The high pressure valve reduces the pressure ofthe air coming 39om the compressor 9 stage to the same pressure as the air coming 39om the compressor 5 quot stage at point A What is the minor loss coef cient K for this valve B In your system you decide to omit the high pressure valve and mi h x igh pressure 9 stage air with low pressure 5 stage air at point A Identify three ways to decrease losses in the system for the new design Project problem 4 12 atm T 250 F m 01 slugs 4 atm T 140 F m 02 slugs r 2 inch diameter pipe Air pressure must be equal where 9 and 5 stage air mee Find KL for high pressure valve 01 14 HI 2 mumlusus Aquot KL39E39 at I39I 5 Luz m w 1 3132012 Today s Agenda 2712 1 Solve fluid flow problems using the Conservation of Momentum in CV form Midterm review Project solutions JON Conservation of mass l Steady ow DB 6 A 12 MN bV A Dr azCIVp 2L my O IpdVIpltV gtJA m CV CS Simpli cations 0 pi7fziA across each 0 at J pdV ZpVA Z pVA CV 014 m Constant properties inletoutlet 272012 272012 Conservation of momentum Dch a 4 FECJVpdeCjSpbV nyA Bml7 ZF IpI7dVJpI7V dA 7Z7 alLCV CS Simplifications Steady flow 2F Ipd 2 M4 Constant properties across each inletJoutIet What is the force on the flange bolts at 1 in x Water ows through the elbow shown and exits to the atmosphere The pipe diameter D1 is 10 cm and D2 is 3 cm At a weight ow of 150 MS and a pressure p1 of 23 atm gage what is the force on the ange bolts a 2 1mm Win atCV CS A 284 N B 1800 N C 2190 N D 46 272012 What is the force on the flange bolts at 1 iEIX if Water ows through the elbow shown and exits to the 7 atmosphere The pipe diameter D1 is 10 cm and D2 is 3 cm At a weight ow of 150 Ns and a pressure p1 of 23 atm gage what is the force on the ange bolts at 1 39 153 W p 4 9K vaals a s I A i z k quot can ml J I v 2 A Fw 3 3 WW 99 a39nquot 1 quot J zus N CoMm example Water jet w39v V A jet strikes a vane which moves to the right at constant velocity VD Find 1 Force to restrain the cart FX 2 Power delivered to the cart 3 Cart velocity for maximum FX or P Z ngSp WM ohs Mt k m 3 M 3959 arr R quotAw quotWA c nos J s 39 a J u ek a A MWU we tunav AN Irwingag 5 MWRJL0 m Ti 3 272012 What is the water jet velocity Ajet of water 3 cm in diameter strikes normal FE 3 5 to a plate as shown If the force required to hold the plate is 23 N what is the jet velocity V a a so F dv VV WIA 33 Z M4 0 n I A 285 ms 1 9quot VX VM 3m 75 081 ms D40 ms E23 ms 9 1 quot 0 1 ft x 63 by ex How high will the water jet rise m FE36 d hm 170cm 7 7 7d I1 cm A fire boat pump delivers water to a vertical nozzle with a a 31 diameter ratio If friction is neglected and the flow rate is 500 galmin how high will the waterjet rise Hall 0 0031 AM m 74v y ay ijamu B98m Eggs 0 i 56 D64m E98m CoMm example Satellite repositioning u IGquot Mn Sm my longu Ip39somrls i F rlwnvt39 A 4 KWXM 9000 N G o VM vii aSSmL MIJICMWM R 8m quot3 391 L W RIVA C 50h S Fat Lx V V M eksu vf 30 IS1 Today s Agenda 2712 1 Solve uid ow problems using the Conservation of Momentum in CV form Midterm review Project solutions JON 272012 Midterm review Fluid Properties Measures of uid mass and weight puSG Pressure absolute gage Viscosity shear stress Twig types of viscous fluids Ideal gas law P pRT use Pabs Cavitation keep Pliq gt Pvap at given T use Pabs Surface tension occurs at liquidgas or liquidsolidgas solve using force balance dropletbubble capillary tube as L Midterm review Hydrostatics Pressure in a uid increases with depth Pz R pgzl zz pgh Pressure is constant along lines parallel to free surface usually horizontal Manometer problems P as you go down P as you go up Pressure is same laterally in the same fluid 272012 Midterm review Hydrostatics Force on a submerged surface FR 2 PghcA 2M 0 hinge IXC 39 yR yc hys1n6 ycA Buoyancy density and volume are for displaced uid FB V if neutrally buoyant FE W Fluid moving with rigid body motion Solve Eqn of motion for fluid moving without shear stress dP 0 at the free surface a y dy azg VP pgl pa Midterm review Bernoulli mid2 pgzC 4 assumptions inviscid incompressible steady streamline Streamline coordinate system Total dynamic static stagnation pressures Types of problems Internal flow splitting pipe freejet open channel Bernoulli perpendicular to a streamline EZPVZ an ER 272012 272012 Midterm review Conservation of Mass Reynolds Transport Theorem takes equations for control mass and converts them to control volume form D5quot gcjypbdw 5prI7 iA Conservation of mass 05VpW 5pI7riiA If flow is steady 0 IpW yA If constant properties 0 LJVWEPVAEPVA If both steady and CP ZpVAZ pVA How to choose a control volume Today s Agenda 2712 Solve uid ow problems using the Conservation of Momentum in CV form Midterm review Project solutions wk Project Problem 1 incompressible The Macondo well drilled by the Deepwater Horizon rig was located 5000 ft below the ocean surface What is the local pressure at the ocean floor where the oil well was located Calculate this pressure first assuming that seawater is incompressible and then compare this to the calculation for compressible seawater For the compressible case you can assume that the bulk modulus Evis constant P g 3k m1 S HPhg2391 Lgbw 3103 ls Project Problem 1 compressible Forthe compressible case you can assume that the bulk modulus EV is constant A A a As AP Ev Ego PzalkCv Q a c T SJ 9quot 1 it 1 41 Absift hf Q k 399 Q 39 39P a in 039 3 1 31110 lhlg 272012 Project Problem 2 P varies with depth How did displacing 3300 feet of mud with seawater change the hydrostatic pressure on the cement job as well as on the well walls Plot pressure vs depth along the well walls from the well surface to the cement plug for cases where either mud or seawaterfilled the 3300 top feet ofthe well According to the report seawaterweighs 86 ppg and mud weighs 145 ppg where ppg are pounds per gallon A mk f S W 2 quot5 aka A P ll39I oml N A 3gsoSOIIZH1 Project Problem 3 ideal gas The primary cause ofthe explosion was that hydrocarbon gas leaked from the well into the pipe leading up to the rig These hydrocarbons expanded rapidly as they moved from deep within the well to the ocean surface Considering the hydrocarbon gas as methane and as an ideal gas how much would it expand as it rose to the surface Remember to consider both pressure and temperature effects I lul 3A4 Vk V 2 X2321 Hw gserS F r g39c 7 IL39L q lw39F as 272012 Project Problem 4 buoyancy Estimate the buoyant force on the dome as either all oil or all methane gas displaced the seawater How much applied force would have been needed to place the containment dome overthe well when it was full of either oil or gas 1 1 u R was 1 lib 1 11 my Fm n he u l is o MM g LW W IL Winn NutH31 272012 Today s Agenda 11712 Pressure variation with depth l Pressure measurement barometer manometer 3 Force on a submerged surface 4 Buoyancy 5 Rigid body motion Hydrostatics Pressure variation in a static incompressible fluid MIAquotM39s ZJ Law EPW b O 241 0 WSW ML F Pdw 7 339 W PVLSSIMA M Hall at 33 W3 Qua 5W Cm t39Q 1172012 Key concepts in hydrostatics 9M1 fasswu imam Mk ALfL Hm mssw s cmskd Aim31w Wm wJ qu m jkmmh A W m Hm msvlg w xvian I L k B Mr A r Lev M V3 Put the pressures in order from lowest to highest A PA PC PB PD PE B PA PB PC PD PE 0 PE PD PB PA PC D PE PD PB PA PC 98 D V W0 W VII 6 I A C o 1172012 Types of hydrostatics problems Pressure variation with depth l Pressure measurement barometer manometer 3 Force on a submerged surface 4 Buoyancy 5 Rigid body motion How deep is the pressure gage On a sealevel standard day a pressure gage moored below the surface of the ocean 8G 1025 reads an absolute pressure of 14 MPa How deep is the instrument If ranks9 A 4 m AK Wk B 129 m 25th 51 c 136 m k t D39 139 m IMP olm15quotquot E 2080 En Ww l mm o 1172012 Compressible pressure variation with depth it 41quot 1 J SW 2 RT 3 l h war Wm WM h RT E 1 2amp1 iquot 31 2 f 930 9 WWJM Compressible pressure variation with depth 51k WNW 32 Q A X Sad 5amp1 rm 9 umrriogd m um V 1172012 1172012 Types of hydrostatics problems Pressure variation with depth N Pressure measurement barometer manometer 3 Force on a submerged surface 4 Buoyancy 5 Rigid body motion 2 Manometers using hydrostatics for ressure measurement Huh MW 4J PM Ml Wt M bfl39n T Mm in a if L QmJS M39VMSaIw XML wMSW wuw Nosoiwh 51393quot KquotL SQKua m 3 39lt3L 1172012 How does pressure at 1 compare to pressure at 2 A P1 gt P2 B P1 P2 C P1 lt P2 D Not enough info 82 Fluid B How does pressure at 1 compare to pressure at 2 A P1 gt P2 B P1 P2 95 C P1 lt P2 D Not enough info 2 1 1 Fluid A Manometer problem 39 3 PA 1 ER my qt zn pg 135 01724 rsi What is the absolute pressure at point B The oil in region B has SG 08 and the Qulmhhu absolute pressure at point A is 1 atm A 5 6 kPa Wquot B 109 kPa c 1069 kPa D 1122 kPa E 1570 kPa Fig ma UI J lbcm 1 1172012 1172012 what is the absolute pressure at point B Ifthe oil in region B has SG 08 0 Fig FEE3 ram 361 and the absolute pressure at point A is 1 atm Manny Y ush or Glampk3 lu S MLQwgcc g39 93 Barometers measure atmospheric pressure L ALSo dl V d uquot tff loB 78m W 5 m k Ru l gtL h k k L7Lo WANK 39 D W Types of hydrostatics problems Pressure variation with depth l Pressure measurement barometer manometer 3 Force on a submerged surface 4 Buoyancy 5 Rigid body motion 3 Force on a submerged surface QH1 k01i31 ofm tstuMuad aim V EPA W w 2W0 milEel 1172012 1172012 How is a submerged surface problem solved A M o 96 azao QZ Z O D Both A and B Submerged surface define coordinate system 3 rs 4394 Subww gl W39nce In is ctrme LPN ku Suicu k Calculate force from fluid pressure FR R P 5 SMRMA PM cwsdux vmb quotAquot Tun 3k E a cs we k Q 3 Wm mm 4 4c Mug34 Mu Calculate location of FR center of pressure yR fussWL 4 399 SubWt u 1172012 Calculate location of FR center of pressure yR R1 1 We Saw y ltlh 5lgt l gum 3M 335149 guld m vb ln g R Wis m R Canal hams 73de AELPQA oIL X 3 at if Calculate location of FR center of pressure yR Zuni 5 mm In w leak GumLA lo Mom 0 ruJTa WA my would Mrs L Mr Iyln tin Ay S IJA Erl TX A f3 3 A Aj v AT AVA 3 l 1172012

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