Fluid Mechanics PS1_sol

Fluid Mechanics PS1_sol - 1-3-3 A man is considering buying...

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Unformatted text preview: 1-3-3 A man is considering buying a IE—oz steak for 53.15.. or a EEG-g steak for $1.8G. The steak that is a better hn}r is to he determined. Assmuptt'eus The steaks are of identical quality. Arralrsl's To make a comparison possible: we need to express the cost ofeach steak on a common hasis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors: the unit cost of each steak is determined to he \ t's3.15“|-’1ssz'~.-” llhin .52 ounce steak: Unit Cost = II '-._12 oz ,.H1]hn13"|._D_-15339kg£ 320 gram steak: UnitCost=| $13” | 190% |=$B.T5tkg .NEEGg/H 1kg {, Therefore: the steak at the international market is a better bu}: 1-3-1 The thrust developed by the jet engine ofa Boeing 77? is given to he 35930-0 pounds. This thrust is to be expressed in N and kgf. Arralrsl's Noting that l lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed can he expressed in two other units as [-4.443 N" Thrust in I: Thrust = [83,030 lhf}: = 3.“! K105 H """l' r , -. TIJIuHinksf: 11111151431311? N}; “‘3‘ f teem; 4355:1114 kgf a a .r 1-? Sin automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must he bled ofl'to reduce the temperature to the original s-‘alue are to he determined Assmrrpn'ous 1 Est specified conditions. air behaves as an ideal gas. I The volume of the tire remains constant. Properties The gas constant of air is R = D2 37' kPa-ml-‘kg-K. Analysis Initially. the absolute pressure in the tire is Pi = P3+Em = 210+1|C|C|=31|3|kPa Treating air as an ideal gas and assuming the volume of the tire to remain constant. the final pressure in the tire can he determined fi'om Epl Ply: _ T3 _ 323K _ __, _ T; = T: _}P: _ T1 P1 _ 293K {:3 lePa] — sdePa 3:1: 23am Thus the pressure rise is nP=P3—P1 =33fi—31D=2l§ ItPa The amount of air that needs to he hled ofl'to restore pressure to its original salue is 1 1-; 3 m. =I—§_{J= {JlCIkPaHll'flssm } =0-Dgflfikg ' RT: {DIETkPa-m3:'l:g-K:II:293IQ H . ‘ 2s 3 a _ P‘tr' {leIltPaKflfl -m } = mmkg ‘ ' ar2 ={a_237kPa.m1..-kg.1t){323to am = m; —m1 moans-tosses =a.ooto kg l-lTE The inininiunipressure in a punip is given. It is to be determined ifthere is a danger (if-cavitation Properties The vapor pressure of water at TWP is 0.3 I532 psia. Analysis To amid cavitation, the pressure everywhere in die flew should remain abm‘e me vapor {er saturation ] pressure at the given temperature, 1which is P» = Punt-man ='i3'-3¢533PEifi The niininnnn pressure in the pump is [11 psia. which is less than the L'fipDI pressure. Therefore: there is danger of sanitation in the pump. Discussant Hate that the vapor pressure increases with increasing temperature: and the danger of eat-itatinn increases at higher fluid temperatures. 1-30 The percent increase in the density of an ideal gas is given for a moderate pressure. The percent increase in density ofthe gas when compressed at a higherpressure is to be determined. .«lssmrgln'ous The gas behat-‘es an ideal gas. Analysis For an ideal gas. P = cm" and (EFF 5,0).- = R1" = P ,o . and thus 1:- :iulgal =P . Therefore. the coefiicienc of compressibility of an ideal gas Is equal to its absolute pressure. and the coefficient of compressibility of the gas increases 1with increasing pressure. Substituting x= P into the definition of the coefficient of compressibility a' E — a cw -' u .o and rearranging gives *3 _ E o P Therefore. we percent increase of density of an ideal gas isothermal compression is equal to the percent increase inpressure. _ no a? 11—1[2l n. A 10 . _=_=_=1.:..... I dllIl. F P op 3P lCll—lfll:I _ a 1m ten: —=—= =15.) t a .o P mu Therefore. a pressure change of 1 arm causes a density change of 1091: at 1D arm and a density change of 1".-’.: at IUD atm. 1-30 The percent increase in the density of an ideal gas is given for a moderate pressure. The percent increase in density ofthe gas when compressed at a higherpressure is to be determined. .«lssmrgln'ous The gas behat-‘es an ideal gas. Analysis For an ideal gas. P = cm" and (EFF 5,0).- = R1" = P ,o . and thus 1:- :iulgal =P . Therefore. the coefiicienc of compressibility of an ideal gas Is equal to its absolute pressure. and the coefficient of compressibility of the gas increases 1with increasing pressure. Substituting x= P into the definition of the coefficient of compressibility a' E — a cw -' u .o and rearranging gives *3 _ E o P Therefore. we percent increase of density of an ideal gas isothermal compression is equal to the percent increase inpressure. _ no a? 11—1[2l n. A 10 . _=_=_=1.:..... I dllIl. F P op 3P lCll—lfll:I _ a 1m ten: —=—= =15.) t a .o P mu Therefore. a pressure change of 1 arm causes a density change of 1091: at 1D arm and a density change of 1".-’.: at IUD atm. 1-. T The density of seawater at the free surface and the bull: modulus ofelasticity are given. The density and pressure at a depm of 2500 in are to be determined. Assumptions 1 The temperature and the hull: modulus of elasticity of seawater is constant. 1 The gravitational acceleration remains constant. Pr'apai'n'as The density of seawater at free sttrface where the pressure is given to be 1030 kg-‘nid: and the bulls modulus ofelastjcity of seawater is given to he 2.3-4t-10g Mm]. Analysis The coefficient of compressibility or the bull: modulus ofelasticity offluids is expressed as I E: El 1 or 1:: pg {at constant T) .__cp .'1. tip The differential pressure change across a difi'erential fluid height ofa‘: is given as cl” = Jogpt': 2500 III. Combining the two relations above and rearranging: Fpfiflori _;. de=si= do do p‘ 1: Integrating from 3 = 0 where ,o = ,ofl =1030 ls:g-'n.t'1 to z = z where ,o = ,0 gives -.-=' aflo_£{5dT 1 1 g? | - E,- o. _> —_—=— -a p‘ r- Po :0 s Solving for ,0 gives the variation ofdensity with depth as 1 1.-,o.;.—g:.- Pt' Substituting into the pressure change relation :1? = pga‘: and integrating fiom s = 0 where P = P: = QEkPa to:=:whereP =Pgives .1“ JP: r 3:31: .o 1-"pU—gE.-'r I _v P = P: +a'l.t:t| 1 l w x1 — Po 33 1' ,- which is the desired relation for the variation of pressure in seawater with depth. At a = 2500 m. the values ofdensity and pressure are determined by substitution to be 1 = =1041 It rml 1:'[103-0]tg:'m’}—[9.31m-‘s'JCi00 m):'[2.34a10p tom-J g ,o NI P=tsaootrPaJ—{2.34x109u.-'m2)ml+3+t;rt—l txl—{10301cg-m )[9.Slm-‘s']{2300mj.-'{2.34><10 arm-J; =2.sjoa-mlsa =2s.5o MPa since 1 Pa = 1‘_‘~I.-'n1s = 1]tg:'m-s2 and lltPa =1000 Pa. Discussion Note that if we assumed o = oc. = constant at 1030 kgs'm3, The pressure at 2500 m would be P = P: hog": = 0.093 - 23.25 = 25.36 MPa. Then the density at 2300 n1 can be estimated to he as = pastP= (1030}[2340 It-fE‘aTIQilo h-IPa] =11.1]tg.-'m‘:' and thus p = 1041 lsg-"nii‘ 2-4? A clutch system is used to transmit torque through an oil film between two identical disks. For specified rotational speeds: the torque transmitted is to be determined. Assmuplt'ons l The thiclcness of the oil film is uniform. 2 The rotational speeds ofdisks remains constant. Properties The absolute viscosity of oil is given to be p. = 0.38 N-s.-'I:n2. Arralrsr's The dist-cs are rotting in the same direction at different angular speeds of cal and of mg . Therefore. we can assume one ofthe disks to be stationary and the other to be rotating at an angular speed of co; — co: . The velocity gradient anywhere in the oil of film thickness is is l’sil where i” = (to; — cs12 }r is the tangential velocity. Then the wall shear stress anywhere on the surfice of the faster disk at a distance r fl'om the axis of rotation can be expressed as do i” [all — or: jr' 1'“. = Ir:—= y— = Jr: o‘r l.‘ ll . Then the shear force acting on a differential area obi on the surface n and the torque generation associated with it can be expressed as 0,] [so] — £12 }J' o‘F = two‘s! = ,r: _— [Emira‘r 011 n — 'l 2 :2 . — 'u * dT = rdF = ,u in" f- 3" {Em'jdr = —"-“E“;1 “rd-j r’o‘r . I integrating, . ,I 4 D '2 L T = 2x;r(col —co]:| I'D Elgar = -mrrEro] —o:2} r = statics; —oo3 JD l': - r-D F: -—l 32!?! hi] Noting that co = Earl , the relative angular speed is . . , t _ 'r l min"; so; — co] = lrrfirrl — in) = (Est rad-'revj[|~145£| —1393,tre1.'.-'min] 6'3] P: 5.445 rad-'s ._ s 5 .- Substituting. the torque transmitted is determined to be _ gross N - 3me 315.445 5993.30 m}l 310.003 1.11} T =[l.55N-m Discussion Hote that the torque transmitted is proportional to the fourth power of disk diameter. and is inversely proportional to the thickness ofthe oil film. 2-49 _-'-‘L muEti-dislc Electro-rheological “ER” clutch 1arith a fluid in which shear stress is expressed as r = r3. —,1e|:dul.l"a‘v1-'} is considered. A relationship for the torque transmitted by the clutch is to be obtained. and the numerical T.‘alue ofthe torque is to be calculated. Aismuplfom l The thickness of the oil layer between the disks is constant. I The Eingham plastic model for shear stress expressed as r = r1. + Irtfldufldjr} is valid. Properties The constants in shear stress relation are given to be .u = 13.1 Pa-s and T:F2.5 era. F1=l.1n1m "T.— Clutput shaft Input shaft Plates mounted on shell Plates mounted on input shafi T-.-'ariable magnetic field Analyst's (a) The T.‘elocity gradient anywhere in the oil of film thickness to is 1’ it": 1a-‘here I’= on?“ is the tangential velocity relative to plates mounted on the shell. Then the wall shear stress anywhere on the surface of a plate mounted on the input shafl at a distance 1' fi'om the axis ofrotation can be expiessed as du F" for T--=r-+If—=T-+I—=T.'Lt'— ” -‘ 'a'r t “is J is Then the shear force acting on a differential area all on the surface of a disk and the torque generation associated with it can be expressed as a? = r.“ at =f «.3 +113 IlEZm'jwtr I_ if I. .« s r' x .3' d'T =m‘F=r‘i 1'1. +1:fl |[2arjdr=2:'r .71.?" +111 id: x ‘ ii 2 .l ' n t lutegrating. 3- 'z (or! I" r3 alt-1TH: T- - - a} - 1 T=23J' irrJ-wa— iii-=2: 1'}. —+’“ 2 =2: its? —Rfj+’“—[ag 42-4} r-R| k ' in _. 3 4h Hi“J12 3 ' 4h ' g This is the torque transmitted by one surface of a plate mounted on the input shaft. Then the torque transmitted by both surfaces of'lkT plates attached to input shaft in the clutch becomes l—T . T = test " L 3 {b} Noting that at = 2n] = Jrfiflflfl less-"mind = 15.030 rad-"min = 251.3- rad-"s and substituting: 3 3 #53 4 L Rs—R —— Rs—R { _ L] “if _ 1:1] {UlK-s-"mjjtlfiljs'sj trust] 12 m} l [[uromf —[o.oim}3]+ T = {Hm 1;][25410 :‘s'fm - action:ij4 {0.01 inf] = anal] N-m 2-51 The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be determined. Assmuplfaiis l The inner cylinder is completely submerged in oi]. I The viscous effects on the two ends of the inner cylinder are negligible. 3 The fluid is Newtonian. Analyst's Substituting the given values, the viscosity of the fluid is determined to be If {U.EN-mj{D.D012 m} Dl'sr'nrsi'sn This is the viscosity value at the temperature that existed during the experiment. T-.-'iscosity is a strong function of temperature, and the values can be significantly different at different temperatures. 2-61 A. glass tube is inserted into a liquid, and the capillary iise is measured. The sulface tension ofthe liquid is to be deteimtned. Assumptions 1 There are no impua'ities in the liquid. and no containination on the surfaces ofthe glass tube. 2 The liquid is open to the atmospheric air. Properties The denaitj-f' of The liquid is given to he 96E] kg-‘n13. The contact angle is given to be 15°. .-lirt:r{1'si5 Substituting the fl'llfllE'I'lC al values: the sniface tension is determined fi'onzt the capillar}r Iise relation to he 0",: o- .- 3 * .- 3 -".| ' 'I ' I' pgiil ={96EI kgm 3.3311113 Hilfifllg. -mj{D.DICI)n:L} 11s 1 Lungazmm Eco-s; licoslia} l 1kE-m.-'5‘ _.' Discussion The stuface tension depends on tempeiature. Therefore: the value determined is valid at the temperature of the liquid. 2-65 A steel ball flo ats on 1a-‘ater due to the surface tension efi‘ect. The maximum diameter of the hall is to be determined: and the calculations are to be repeated for aluminum. Assmuplt'ans 1 The svater is pure, and its temperature is constant. 2 The ball is dropped on water slots-'13; so that the inertial effects are negligible. 3 The contact angle is taken to be '3'” for maximum diameter. Properties The surface tension ofsvater at 20°C is o; = CLUB N-"m. The contact angle is_ talcen to he D“. The densities ofsteel and aluminum are given to be m = 73GB Ichm‘1 and log] = ETGU leg-"ml. Anaijsis The surface tension force and the weight of the hall can be expressed as F: =ail‘3hrrj and H’=mg =,ogV=pg.-ID3 .-"I5 When the ball floats. the net force acting on the hall in the vertical direction is zero. Therefore, setting F__. = H" and solving for diameter D gives '— 1}: IE 1' as Substititing the Imown quantities: the maximum diameters for the steel and aluminum halls become ,— .— _. K -2\' ‘50. | sun-33.- 1|c . . ‘- as t {TED-t}kg-"m’}l:9_31m.-g-:lk 1N /_ c -- " . 2 a fi - | ans-33.. 1b . ‘ D—‘--' = ii: Ifis—W.‘ g ms |=4.1x10_’ m=4.1mm 1| PE ‘t (limkgimJJEQSIm-‘sja 1N {J Discussion Note that the ball diameter is inversely proportional to the square root of density, and thus for a given material. the smaller halls are more likelyr to float. l-TDI. The minimum pressure on the suction side of a 1a‘ater pump is given. The maximum water temperature to avoid the danger of cavitation is to he determined. Properties The saturation temperature ofwater at 13.95 psia is llflUDF. A:ra{1.'st's To avoid cavitation at a specified pressm'e: the fluid temperature everywhere in the flow should remain below the saturation temperature at the git'eu pressure, 1which is Total: = T'arui IZISSpJfa = 1flfloF a Themfore. the 1'" <1 IGGDF to avoid cat‘itatiou. Discussion Kore that saturation temperature increases with pressure. and thus cavitation may occur at higher pressure locations at higher fluid temperatures. 2-?5 A circular disk immersed in oil is used as a damper. as shown in the figure. It is to be shown that the damphrg torque is Tmplg = Cm 1.t'here C = fi.5;‘ry[:lfrt+ lift: :LFE'l . Assumptions 1 The thickness of the oil layer on each side remains constant. I The velocity profiles are linear on both sides of the disk. 3 The tip effects are negligible. 4 The effect ofthe shafi is negligible. Damping oil Arrafi'sis The L'elo cit-3' gradient anywhere in the oil offilm thickness :1 is I-’.-'a u-‘here I’ = car is the tangential T.‘elocitp. Then the wall shear stress anywhere on the upper surface of the disk at a distanee r from the axis ofrotation can be expressed as ah: I" (or I“. = Irr—= ,u—= ,u— a‘i' a n Then the shear force acting on a differential area rid on the surface and the torque it generates can be expressed as are = was =r1£dd {I 1 n m. a"T=ra‘F=,u tit-l a! Noting that it = 211113?“ and integrating: the torque on the top surface is determined to be H 4 t =3 1 a I 1 - - are" Twp =fl [was =fl r' [ea-31m = "Wall Has- = “Tim’— = '” at at at - r-IZI a 4-: a 4 “U in The torque on the bottom surfice is obtained by replaping a! by b, sauna? 4 Thai-bunt = 2b The total torque acting on the disk is the sum ofthe torques acting on the top and bottom surfaces. .1 3 mmfl 1 l Tdampsng.tatal = Tatiana: +Ttop = I 2 |_\;+;Jl or. T —c 11 c—E-‘flfll f1 1‘ fllpmgltmal — 6:} “- EIE — 2 IHE+ This completes the proof. Discussion Note that the damping torque [and thus damping power} is inversely proportional to the thickness of oil films on either side, and it is propoitional to the 4-111 power o‘fthe radius ofthe damper disk. 2-HT. _-'-'1 glass Tube is inserted 111m meremjs'. The capillary drop of mercury in The Tube is To be determined. slismupn'am 1 There are no impunties in mercury: and no eonTaannaTion on The siufaees ofThe glass mhe. 2 The LuereLTTjr' is open To The atmospheiio air. Properties The surface Tension ofmeroury-glass ir- atmospherie air at 63”? [20”le 1s :3; = 0.44flx0fl633s = {$6015 lbfffi. The derisiT}r of LDEI'C'lT-l'jr' is p = 34'." lbm."ft’ at FTP. but we can also use This s‘alue at 63°F. The contact angle is given To be 140‘“. .-lnn!_1'51'5 3111:: sTiTuTiJJg The meerioal Values: The cap:l:illa:r1-r drop 1s deTermine-d To be ;. 35; €05 .d' 2mm 15 Tbs-fillings 140°) n" 32.2 1!:I111-fts's2 " I = — = —_ 1 I _ LE3 [34? 1h111-‘frj j|:32.2 fi-"s‘ ]I:I:I.-T-i L‘- frj '.\ 1 lbf j =—|;].|C|Dl:1:lfl m: = —|C|.|C| 17-5 in Discussion The uegaTis'e sign indicates capillar; drop i11sTead ofrise. The MEI-cm}. .—. ” drop is very small 111 This case heoause of The large diameter of The Tube. ...
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This note was uploaded on 10/15/2008 for the course MEEG 307 taught by Professor Smith during the Fall '08 term at Howard.

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Fluid Mechanics PS1_sol - 1-3-3 A man is considering buying...

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