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# Fluid Mechanics 057 020

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This 21 page Class Notes was uploaded by Rahsaan Hoppe on Friday October 23, 2015. The Class Notes belongs to 057 020 at University of Iowa taught by Staff in Fall. Since its upload, it has received 39 views. For similar materials see /class/228007/057-020-university-of-iowa in General Engineering at University of Iowa.

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Date Created: 10/23/15

11092008 Midterm 2 Review problem 057020 Fall 2008 l RTT Mass and momentum conservation The water in thisjet has a speed of 30 ms to the right and is de ected by a cone that is moving to the km with a speed of 13 m s The diameter of the jet is 10 cm Detennine the external horizontal force needed to move the cone Assume negligible friction between the wane and the vane 2 RTT Energy conservation The pump in Fig E120 delivers water 624 lbl39ll t at I5 lls to a machine at section 2 which is 20 ft higher than the reservoir surface The losses between I and 2 are given by h KV 2g when K 75 is a dimensionless loss coef cient see Sec 67 Take a Rs L07 Find the horsepower required for the pump it it is 80 percent ef cient p 147 lbfin2 abs D2 3 in z ft 1 10 1mm2 Water gt It 12320 11092008 Midterm 2 Review problem 057020 Fall 2008 3 Differential analysis Two immiscible incompressible viscous uids having the same densities but different viscosities are contained be tween two in nite horizontal parallel plates Fig P631 The bottom plate is xed and the upper plate moves with a constant velocity U Determine the velocity at the interface Express your answer in terms of U The motion of the uid is m and m caused entirely by the movement of the upper plate that is there is no pressure gradient in the x direction The uid veloc ity and shearing stress are continuous across the interface be tween the two uids Assume luminar ow U Q HFIGUFIE F631 4 Dimensional analysis A 149 scale model ofa proposed dam is used to predict prototype flow condi tions If the design ood discharge over the spillway is 15000 m3s what wa ter flow rate should be established in the model to simulate this ow If a velocity of 12 ms is measured at a point in the model what is the velocity at a corresponding point in the prototype 11092008 Midterm 2 Review problem 0572020 Fall 2008 Answer 1 R39l39l39 Mass and momentum conservation Information and assumptions provided in problem statenrent and xtemal horizontal force needed to mer cone Solution Select a control rolnnia surrounding the moving cone Select a reference frame xed to the cone Velocity analysis 111 e Vli43ms 112 43 ms zrlncnnentuln FL 7 when 1000 X 1r X 0052 X 43 X 2764 713 a 751st N FID 51 kX acting tole 11092008 Midterm 2 Review problem 0571020 Fall 2008 2 RIT Energy conservati n Asnmlpuans Steady aw negligibh viscous work1ug munoi v 0 Approath w V m the exit L r a Solution mps Us 86 unilsp 147144 2117 Infn1 map 1011 1440 bffrl i d v2 mm the known ow nu an m pipe diameter g Lsn ls A 1r432 m m39m 71 39 0and vo becomes I72 1V V z z I I 1 2 1 h n Kzg F1 39 Pl V1 1 or a K 23 Cumman Th pump musl balance four different effects me pressure changc me ele I range he exil jel kinelic tangyan he iclionl sscs For Ihe given dam we can cvalnal me requin pump head K 306flls1 7 2mm 7 u 20 124 I331 vnlim ch Firm solun39lm 1440 unnwm II 62A mun 20 439 I07 439 75 3 Iu ml unuin in Example 119 3 PM m 191 MI5133 m 024 n l H450 fl Ihfs 550 n e MK 7 hp 2 h 12450n 5 39h requimd P quot 2 29 2g hp ef ciency 08 11092008 Midterm 2 Review problem 0571020 Fall 2008 3 Differential analysis Ear 77k Vac6amp4 comAda 120 hr 284 and fa w 771A The x wmpontnl ol 7 71 MarinShall ina w39us 256171 Ar cVher 7 upper or anr lager redazzs t 32 4 L a I Im qmtiu a E II 95er 0 45 f 5 which gives 7 71 Vela7 5 xi rI39bkEa 1 2731 law In 711 In2y 70 az g24l u Uquot so nch 3 U Ath whrre 1hr turf2t I 392er 4a 771 up laym F 171 lower Myquot at 99 49 so Thai 51 a diver 751 subscript 1 MA 0 7h our layer Thus Ml A y le quotU a A1 5 M 314 on 61 Jo 7111 A 14D 244 13739 U A A z I a Can39t 11092008 Midterm 2 Review problem 0571020 Fall 2008 Suite 77 Velmlfq alulfrv39 ub 13 luien 4 Pack 44quot 7 71 Shearr17 n wss g I 4 77x7quot MW XV f3quot Is mnxfmi Wrouylmn Each layer 5r 1739 14 470 ZIEalJ and Ar 7 11 lunar by Tf Az 42E f M kn t t21 50 1711 ll LA or i A 3 lb Su rlkh n 0 Eg ly Min Eilzz axisAL x U 51 57quot 41 f1quot 6 04 A 1 H quot n 771115 lcu y of Thz M39 r ta 4 s L ul5 Azr Il u 11092008 Midterm 2 Review problem 0571020 Fall 2008 4 Dimensional analysis Solution The qude number criterion will be used Therefore Ftquot Fr Vm 1L emL m However gl 31V Thus we have 4 If We multiply both sides Df Lhis equation by the area ratio AmAP VmAlnAnl Lu vaP A L The lefthand side of lhe equation is the discharge ratio and AmAP LilL Hence we 0 min on Ln 52 3 39 2 Q Q G 5 000 089 mJs quot39 P 49 16800 39 From the fourth equation in this example vquot f Vp LP L Consequently p VHF 12 X 7 84 ms 7 At the given point in the prototype we would have a velocity of 84 ms 12142008 Final exam Review problem 0571020 Fall 2008 1 Head loss 0 6 g 0 2 Minor loss gt 39w oil 1 e 41gt39ltj139u 5mZs s 39 o j awsvfr om mg 39 pper lo the lower resalir vqir at a rate of 0028 nl s in 5113 15611 smogth pipe what is heeleva on of thie oil39surface in he upper reservoir f 39 39 1 Make use of Moody chart Fig 820 in Text Pp434 o 80 0136 020 h Kcvgng 090 006 010 KB KB Expansion DDz e 20quot 9 180 2 00 quotmg 0 20 n 30 87 0 40 025 0 7o 2 so 015 a 41 Ill Imam 030 010 015 Wl mlll K 11 39 vane 90 miter bend 1th Kb 02 39 vanes 111 5 and 1 K I135 15 o 2 a 19 90 smooth bend 4 m s 02 x v23 10 032 12142008 Final exam Review problem 057020 Fall 2008 3 Lift and drag coef cient m 0 kg and a wing planfmn E speed of 558 kmlh and the drag chamelerlsucs 012m t 11457 determine id l with and without extendr mm m Full l J n s w is a we mm we cu nmtms Fmtlllrnl minimum imamquot Mwmmm m 4 Boundary layer Adcllve lch curries Jvllg sign on lop as m Fig F7 56 If 39gn i very thin and quotIE veluclc moves al gt 39 ms me as mili a estimule u form on me Slgn With no t wiutl 12142008 Final exam Review problem 0571020 Fall 2008 5 R39I39I39 Momentum conservation Water at 20 C ows through the elbow in Fig P360 and exits to the atmosphere The pipe diameter is 1 10 cm while D2 3 cm Al 1 weight ow rate of 150 Ns the pressure pl 2 23 aim gage Neglect ing the weight of water and elbow estimate the force on the ange bolts at section 1 12142008 Final exam Review problem 057020 Fall 2008 113 msXOJS m 83 l val 2 l l 1 X 0 u s Since Lhn Reynolds lmlnbcr is less lhan 2000 the aw is laminar The loss par lOO m ix obluixicd from l l W 432le quotf f 7quot Here uy 4 nghellcg r azuLv l 39 r I 4 1 Thequot h 32610 V m A2000 1133 ms 983 m 93811113 015m The head loss is 985 m loo m unangm 12142008 Fina exam Review prob em 057020 F3 2008 2 Minor oss Sululibn39 Apply die E rgy aq a pxu39be ve 11 Iilc smfaccs of thq uppel I lower insarvous 3 r I 39 39 P1 V 72er i zv l h y Qgr J yzg 12 1 L r Mi 391 7 v2 07V0 zl 9450 13 0 1 111 D2 F2KPZSIIg2 Ku f Hanan 4 39739 quot quot uu quot lively These have values of 019 05 and I0 Table 103 To deferminc gstR ili orglmo cnt 39g 10x 39 39 103 3mm pipe climax 00357Twhen39 Tami L97 in zg o15 m 239oi9wofsf1 40 1C 138 05 139 L1361m V 12142002 Final exam Review problem 057020 Fall zone nd drag coerrieients The cruising conditions or a passenger plane anrl its Wing chars istics are given The minim m39safe landing and takeoff speeds the rut attack during cruising and the power required are to be determined lms 1 The drag and lift produced by parts of the plane other than such as the fuselage drag are not considers e wings are to he twordimellsional airfoil sections and the tip effects of the s a not considered 3 The lift and the drag characteristics of 2 Wings be approximated by NICA 23012 so that Fig 11 45 is apilicable 4 verage density of air on the gmund is 120 kgrn39i n iers The densities of air are LZO rri3 on the ground and kgm at cruising altitudei The maximum lift Coefficients CLm of the 348 and 152 wlth and without flaps respectively Fig 11445 Iysis a The weight and cruising speed uf the airplane are r lN n 1 r w W mg 7 l00kg98 ms211gm52 iX6700N v 558kmh ml 41 ssmr 36mm 39 5 he minimum velocities corresponding to the stall conditions without and h flaps respectively are obtained fmm 1 l 2w 7 739 ssm oii ilcgnisz709 I 39quot 39 pCLmHA39 lkgg l2150m1 iN f quotM incimsA LukgmXSARxlsomz 1N sins he the safe minimum velncities to avoid the stall reglnn are obtained by tililplying Hie values above by 12 W 12Vm l27n9 nis 35l ms 306 kinii LEV 12mm ms 562 ml 2 202 kinii V 12142008 Final exam Review problem 0572020 Fall 2008 4 Boundary layer Solution For air at 20 C take p 12 kgm l and p 18E 5 kgms Convert 65 milh 2906 ms a Ifjtherc is no crosswind We may estimate the drag force by appiale theory 39 03 070 Re L g o 1SSE7 tu l Int C l 397 l Re 15557139 1 2090 000291 1 5 vsztz sides 000291 29 0620 682 Sides 14 N A115 1 A Film LI 5 RW Momentum conservation Solution First from the weight How compute Q 150 Ns9790 NmJ 00153 rillS Then the velocitiescm 1 and 2 follow from the known 5 V QA7 0 01532gt13995m V2 Q 00153 221 5x Al pr4x01 5 A2 Ir4003 s The mass flow i in the x d39uection ts V 998 40t11l95 z 1525 kgs Then the balance of force 2Fx VFW pIAl i1u1 n391u rh VZ cos40 7V solve for PM 23gtltI013500 1 15252L7cos 40 l95 2100 N Ami Review problems for Exam 3 057020 Fall 2007 Friction factor and head loss Water at 40 F p 6242 Ibmi ft3 and p 1038 x 10quot3 Ibmft s is flowing through a 012in 0010 ft diameter 30ftIong horizontal pipe steadily at an average velocity of 30 fts Fig 8 18 Determine a the head loss lb the pressure drop and c the pumping power requirement to over come this pressure drop Minor loss 1091 Water is Pumped Lil a rate til 20 1113 s i rom iii rcscrmir and out through the pipe which has 21 diameter 0139 iii m What power must be supplied to the Wllit ili cl l cct this discharge Elevallon 140 In Elevation 135 m u it straighteners Dill user E level ion i Test Silencer soc Lion F i PROBLEM HWI ill Boundary layer mun 10 106 A small low speed wind tunnel Fig 10 106 is being designed for calibra tion of hot wires The air is at 19 C The test section of the wind tunnel is 30 cm in diameter and 30 cm in length The flow through the test section must be as uniform as possible The wind tunnel speed ranges from 1 to 8 ms and the design is to be optimized for an air speed of V 40 ms through the test section a For the case of nearly uniform flow at 40 ms at the test section inlet by how much will the centerline air speed acceler ate by the end of the test section bi Recommend a design that will lead to a more uniform test section flow Drag force A 22crn outerdiameter pipe is to span across a river at a 30mwide sec tion wh rle being completely Immersed in water Fig 11418 The average flow velocity of Water Is 4 ms and the we er temperature Is 15 C Deiermine the drag force exerted on the pipe by the river FIGURE 11733 DimensionalAnalysis A unesixteenth Scale model Iracmrrtraller truck IEwheeler is rested In a wmd tunnel as sketched In Fl 7 38 The model ruck Is 0991 1 Ian 0257 m ml and 0159 m wide Dunng ME tests the movmg ground belt speed 15 adjusted so as to alwa match the speed a the an muvrng mm g the test section Aercdynamlc drag force F is me wind tunnel speed the axpenmental vesuns are listed m Table 7 7 PICK the drag caemcrent 0 as a function of Me ey used h c r for car I uf CD IS romal area 0 me model truck the area S a ok at 9 model me upskream and an h s ale used iur calcu atlon of Re Is W WI h W Have we achieved dynamlc sun Ilarny7 we acmev d ynoid umber fndepe dense 1n n rwmd tunnel tesv Esl dyn x drag force an the prom e truck avelln m a my 2a the mghwa a 263 ms Assume that both the wmd tunnel an flowmg over the prototype car are at 25 C and standard atmospheric prasurev wmmm 1 mm mmdvumm drag H4 1an m MM m r a5 1 my Hun ofwmd mm s 1 Solution for review problems for Exam 3 057020Fall 2007 Friction factor and head loss SOLUTION The average ow veTocTty Th e pre Ts grven The head Toss the p n 3 u e 3 rm e o 5 re a o u 5 a pressure are Assumptions 1 The ow Ts steady and TncompressTbTe 2 The entrance effects are negTTgTeTe and thus the TTew Ts mTTy deveTnped 3 The pre vove n cm i a e e g g E e E a a a e g 2 2Tbrnrt3anua x s T rn 5Tespechey Anahsrs a FTrst we need to determme the TTow regTrne The Reynons hums her Ts Re IV 62421bmft 3 ft 00 t t e0 T TT s T 1803 w 033 x Ttr3 bunft thch Ts Tess than 2300 Therefore the TTow Ts TarnTnar Then the TnetTun factnr and the head Toss became 13 W5 1 1w s T a Netzng that the pipe 13 henzontat and rts marneter Ts constant the pres sure drop Tn the We Ts due entTreTy to the ThetTenaT Tosses and Ts equTvaTent to the pressure Tess 50 ft 6242 Ibm TU s1 Ibf 39 322 Ibm r fts c The voTume ow rate and the pumpmg puwer requTrements are T V A V Tani4T T3 msTTaTom mm 0000236 13 m 5 mg l W 7 Au 0737 lbf W5 x Therefore power input Tn he amuunt nf 030 W TS needed to nvercome the frictiuna Tasses in the ow due to stCnsTty WW v39 AP 0000236 3s 9291bfII3 v Minor 055 105 Lurommiuu and Anunnptiom A 397 U 16 mm menu mu uLw A 139 7 3 r Hquot mm prm39ulcd in pxoblzm smemm Find pump prmm Energy equation M77 VfZH 2 n 7 P 21 VEZw Wrziu 00100m 7 0 1521 130 V2 1UU3 fLDw V2 quot Q A 7 IlJ7r1gtl 7 110 1m m a 1152 strlmx10quot715x107 7 0 0000 Hum Fug 100 f 70010 nmn 2m Uis 0010 x nun15 gt34 m 0 7 1407100V 0 pm povcx s number Rem 5 x 105 and is even tower than Rs mt 1 x 106 and since the waHs 37 month and the new is etean we may assume that the boundary tayer on the wait remams tarnrnar thmughaut the Ieng39m at the test section As the haunda Fayer gmws atom the Walt of the wmd tunnet test section arr 2n the regmn of rrretattonat ow rh the centrat human of the test sectrdn aeeeterates as in F1 107105 n order to satr39sty cansewatiun of rnass We use E 1amp73 quot n to esttrnate the displacement tmckness at the end or the test secttoh Lm 72030 m LEISX loam 183mm I Two erossseehaha views at the test sec un are sketched m Frg 1amp107 at nhing and me at the end of the test seetrdn The etteeuye radius at the end nf the test section rs reduced b ti as eateutated by Eq 1 We appty cunservatton of mass to eatcutate the average an speed at the end of the test sectton vatt vteamsAems van at whrch yretds Vm 40 ms u IIIs a OJSm a 183 x10 15 Dra force SDLUTIDN A pipe is submerged in a river The drag force that acts on the ptpe is to be detevm39iried Assumptions 1 The outer surface of the pipe is srntmth so that Hg 11734 can he used to determine the drag coefficient 2 Water How in the nver is steady i The direchon of water flaw is normai to the pipe 4 Turbutence in river aw is not censittere Pmpemas The density and dynamic Viscosity of water at we are n 99 1 km and 1135 x 10 3k m Ahatysr39s Mating that D 0022 m the Reynoius number ls 7 VD 7 ND 7 9991 kgmlrm mm 077 m n tt 7 l sx m ngmrs Re 773 x 10 The drag memcient corresponding to this vatue 15 from Fig 11734 0 10 Aisn the fvontai area for ow past a cylinder is A LD Then the drag force acting on the pipe becomes 9991kgn r4mlst 1N 2 080 X 0022 ml 1 k N h g e n s t Vi m m mquot 10 IU 10 R mens nai Anaiy s SOLUTION We are to caicuiate and piot CD as a function oi R2 for a given se1ofwtndtunn i measurements and determine it oynamie simiiavity anoor heynnios nvrriher independence have been achieved Finaiiy we are to esti mate the aerodynamic drag force acting on the prototype truck Assiimitlans i The mooei truck is geometricaiiy simiiar to the prototype truck 2 The aerodynamic drag on the strutts hoioihg the modei melt is negirgibie Pipmules Fm air at atmospheric pressure anri at T 25 C i 1184 kgm3 and it 1849 x 105 in s Aliaiysis We eaicuiate CD and Re for the test data point iisteti m Tabie 777 at the fastest Wind tunnei speed 1 kg ms2 0139 mJl0257 in i N F C0 i t 1 mva lJMkymeml 0753 and vw imk 3 70 0 59 REMMW7IB gtlt 10gt I itquot 1849 x 10 5mins We repeat these caicuiatioiis 3907 an the data point5 in Tabie 777 and we oiot CD versus Rs in Fi 741 Ha chieveo dynamic stm39ilanty Wen we have geometric simiiarity between modei ano prototype but the Reyiioios number oi the prototype truck is E E n R lipquotin 1184 kyn iiz Sms160159 mi a 7 349 x 10 kgm where the width or the plomtype is speciiieo as 15 times that at the mudei Comparison oi Ens 39 more than six times iargev than tha of the modei since we cannot match the independent 1139s in the nrabiem dynamic simiiaiity has not Ileell achieved ave we achieve Reyna s number in eheh ence ram i we see that neynuius umnnei Illllelellllellbe has iiuieaii been aclllevekat Re g er n about 5 x 10 07 has ieveieo off to a vaiue of about 076 to since we have achieved Reynolds number independence we can extrape ate to the fuiiescaie prototype assuming that 00 remains constant as Re is im rsatsn in that nf the fiiHew aie nmimvns Pi39EdiL m39I oeiotivnmnic drug on HM pwloiyllr vaViAoCo n iii t 11347 1 258 13l630159 0257 076 7 71 Lglm it n s i inn unt i pm NW N

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