hw2_sol - 1. MAE 101B — Summer Session II 2009 Homework...

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Unformatted text preview: 1. MAE 101B — Summer Session II 2009 Homework Assignment #2 Due Friday August 18 Ignoring frictional losses and assuming steady flow (tank is very large) and parallel streamlines at the exit, what happens to the exit velocity and volume flow rate when you add a diffuser? (a) Express exit velocity and volume discharge in terms of pressure head, H, and exit area A2. (13) Express exit velocity and volume discharge in terms of pressure head, H, and exit area A3. (c) Neglecting friction which is the correct statement: discharge flow rate at T(top) is greater, equal or less than B (bottom). Figure (a) Figure (b) Figure _(_c) affirm. Problem I Given: A tank of water with different exit conditions, but each with penile} sire-Mines at the exit. Figure (21) Figure (b) Find: (a) What is the exit velocity and volume flow mite of water of figure (:1), expressed in terms 02" {1113333113 head, H+ and diffuser area, A2. (1')) What happens to the exit val-wiry and volume flow raie (if water when a diffuser is addecig expressed in terms; Elf pressure head, 1-}, and diffuser area, A; (figure (13)). {e} Neglecting friction which is the correct statemem for figure (e): discharge flow rate at T(mp] is greater, :2un er Iess than B (bettom). Assumglions: {'1} No frictional bases (2} Steady flow, 32 {i 6%: Equations: (1] Energy equation {‘Bemeeiii since frictionless flow aieng a streamline) {3 V: — “l”? 'f" g: = mm! p ._ Sciation: (a) 1. Swing aieng a streamline from point 1 to 2 in figure (a). use equetien (I). i + 8H : & + p p 2 2. Note 1’. = $33 = atm. Then plug into new equation and find: V: = [23.8 )m 3‘ Since Q = IVA: Q = V2132 = A2 2(2ng {Evangeiista TofiicolE-i OMS-164‘?) — Student of Galiiieo, his early work on vacuums.- led to the invention of manometer, derived frictionless expression for discharge of water from tank. Know if frictionless that 1-1 = M g 12 and v = g1 so H = % vzlg or v = (2gH}'fl} (b) 4. Aéong the. same limes as for part (a), but with a new output area. m :>. Going along ti. streamline from point 1 to 3 in figure (13}, use equation ( l). p l9 2 {3. Note Pl = P3 = aim. Theo plug into new equation and find: V3 =(2gl‘1']”2 7. Since Q = VA. Q = v3.5.3 3 A; {ngm 30. with a diffuser the velocity does not change bul the voiume flow is increased by A3ng. Note that the velocity at 2 must be greater than 5113 by continuity (incompressible). Consequently the pressure at 2 111113le below ambient pressure. (0) 8. Same! This was a favorite questioo of a noted hyéro’log’st. Geoxge J. Pissing, and still is often asked of graduate students during oral exams. The flow rate only éepencls on d, the pressure head: v = ago)”. Another way of thinking about this is if the discharge flows were different. then if the spout was moved around to attach to the hole at B — if the flows were not exactly equal then one would have made a perpetual motion machine. 2. Prove that squeezing the end of your garden hose will only increase the exit flow speed (V; f V1 > 1) if which of the following conditions is satisfied: f L I D > 1; f L I D = I ; or f L f D < 1. Assume: turbulent flow; all losses are due to pipe frictiOn; V12 << V22 ; and “f” for piping is the same with and without the nozzle. IIIIIIIII I; / D f E ............. .. J . _ ’ . ........... .. _ ._ _._‘——'I-._ _ _.__ 3h- —“" 15— Problem 2 Given: Pinched garden hosed! when friction factor, “f”. is 00213221111 Find: Shaw fer which flow Ctmdititm the exit flow wif} incmaxe: fL I D > l; {L 1’ D = l; f E. I D < i: Assumgiinns: (i) No body fumes, g :0 (:2) Incumpressibie flow (3) Turbuiem' flaw, cc; =32 =1 Eiguations: (I) Bernoulli with 31836 i038: head I053=hfl ={finka £4» 321]“[p3 +er if; +gzz]-~Zh, +21% (2) Maéorhead loss (friction factor) :11, = f £5 Solulism: 3* I. First find ; m” {or the upper came (no nozzle) by plugging in equation (2} and p assumptitm (3) to equation (I) and mating Va = V; in this case because Ag = A3 and by {muse-wade!) of mass. poi/“AB 2 93192242. filly; 2: __ pram +IV2§ =pa-WXK I} 2 ’7 p go Do the same 31; {I} for the pipe and nozzle, but note Va :t V! a: V1. ’ w D 2 p 2 p 2 2 2 p . Combine the two equations just calculated and find: y ., V} V3 L L Divide by VZ'ND' on lhe Raft-hand side: 6;|:1"V—'2£1“ f —]:l = f E 0 " a» . Rearrange abcve and find: I. f 5 v3 v3 ' ' a 1-" +[amfi’1] V; 1‘3 . z, 7 W L . So 1f 2' <:<: V; , than V”: 2: . So far V; > V; (and mmemhenng V; :> V0 by {5 the physics of the probiem), {L i I} 3v 1. 3. Air at 15°C is flowing from a large clean room through pipe (a), shown below, with diameter 150mm. What is the improvement in flow rate, Q, when adding a diffuser, as shown by the modified pipe, (13)? The diameter of pipe (b) from 1-2 is the same as for pipe(a), and pipe (b) has a square-edged entrance as does pipe (a). (a) (b) Problem 3 Given: Flaw from dean mam out a pipe as shown halo-w“ $35 ....... WWW .... u‘ ....... .... .w = I . _ = f glflL ' I / x : _. KL)" M 1.3“ 3 . ,I 1.5— 3“ I 3' 1:31— ? 1.3w i i 1 13m WM ~ “3'39 J4 V2 . . . A?“ L3? L1” J". E ?__.i_ l . - a __L_ _ 0 w Q 623 QB 13} .15 2.43 SD 6.53 ifi 5.13 113.33 Find: improvement: Lxlflflgéz flew-amass sang»..st Assumgtians: (1) Vl = Figure 8.15 E O (E) (13 = a} (3.} 32 = 0 ('4) Uniform flow across each cross xeclion (,5) Neglect frictional losses Eguatinns: . . V3" 1 W (l) Tolai head loss: hi: =[fla§«a-I+é+gzl]-[p‘ +532 ' +g-81]=h.z +53 p p 2 i»: . . 173 (Z) MIHOI‘ head loss: hm = Km -—- + hf $.er .2 . ._ 1'73 A 2 (3} Head loss in diffuser: famffim 2 .3... I.“ __E MC” ’ " 2 A2 ’ Solution: 1. First solve for the flow rate in the origina} pipe (a). Cambine equatiens (i) and (2) with assumptions (E), (2), and (3} to fimf: 72 "9 _3 fl+a.L+gsa H p3+32V3 +33: =19,“ V2 )1) 2 P 2 ' 2 h.) Use the give-n information ta find: ' kg _ m. N p. — p2 = gyms-gm: = 999—. 9.81—2 (omzsm): 2&5 3 1 m _ s m 3. Leak up the entrance lures coefficient for this pipe in Table 8.2: szfléfi. 4. Lack up the density ()fair for 15°C in Tnbie AID: pm.r :- 1.21:3 In? Elk ., Hz . . — 2 . U ' 2 Ns'aw calculate the avezage velocny: V, M 25.211113” * lfikgfm‘ 0.5+: 6. Next solve the flaw rate for the modifietfi case. $323 33E of the equations and assumptions to simplify the tow} head 10331:}: I’ _2 —J —3 .—; 3 V 'l'jl 1 ‘l p_i+gl1€l_+gzi w fi+awi+gza =Kmflt_- +1“: 1— j};- "C? p 2 p * 2 - 2 0 A3 t I By consen'atitm of mass: VgAg = V3133 330 V3 : V3 (mffiu) 8. Again look up Km. it is the same as before. Km = 0.5 9. Since the pressure dmp is fixed we want it; find the maximum (213 to get the maximum velocity. Calculata NR = {3.45 f (0.152?) = 64 Using lhix look at Figure 8.15 and for NR = 6 find the maximum C3,. This is at abom: AR 2 2-"? and CF, = 0.62. If). Plug the-51: values into the latest equation and find: 2 *2 1 29:“191 =.Vl_ fl_5»§«]—~[A2} ~062 p L A: 3 A} __ 24.5me2 2 IZkgfm] 1.543552 133 J = 6.8m I s I I. New cnmpare 17}th rates; knowing Q = VgAl and Qumdmcd = V21 mdAz EMA: "1723»: 6.831%“w .Emm Q’" — Q xl{}(}% : —:—-_—‘x100% = —.‘t‘I [10% = 35.3% Q VBAZ 5.2»: I 3 So the impl'twement by mudifying the pipe output is 30.8% If! 4. Water flows down a 1.2 inch diameter pipe at 78°F with a flow rate of 0.04 ftsr'S. The pressure at the top and bottom are the same, as shown below. What is the slope of the pipe? Assume this is turbulent flow in a smooth pipe. . I." MW“ ' r I 'f’ “a. H.._ H... M“MV.\M\¥T : a“ my“. H- .. ..‘ Jr ’ am ~ ~ tttt my/ - , i’mhlem 4 Given: Smomh, turbuleni pipe flow: D = 1.2. inches? '2‘ = 78W, Q = 0.604 113)», p1 = pg Find: Slope of the pipe _- _! .1_ '*. f Assumgtions: (3) Fully developed, a, :53: ' . "'“w. _ ficw _ \\\\ «:92 95:, EquaLiuns: . 1?, V5 1?, VB (1) Handle-53:11}. = —+aJ—+g21 w —-+a, " +ng ‘ p 2 p ' 2 L 3'73 (2) Major head 105$ (friction factor) = h! = f (3} Fur smth turbuient pipe flow: f m :33: e . Seiut'ionz 1. Combine equations (3) and (2) and assumption {'1} to find {:zg-ZEVL‘ L172 V3 , "f ' f——=[fl+al#+g:i}-[p—‘+ag ‘ +gzgj=fla ~31) D 2 p 2 p 2 zlng __f 173 L W 2gb 3x — . f. 2. NowrearrangeQ:VAlu find: V zganw=5flQfih A {D " [mm 1f: T 21' — :r 2 '2 121'}: 3‘ Interpohte Tabie AK? [0 find the kinematic: viscosity at 73°F. VWF = 'E .86xk0'5_fi3 1’s and Vim”, 2 9.32.1r1lf}""‘(‘f72 {'3 80°F «70%? _ 78°F - ":00? 9.3le 0*“ ft: is — 2 mm" fr 1 3.: VW —1.06xm"5fi9 Ix rm = 9.58x10"" ftI 2's UI Caicuiate the Reynolds number: Re. : Q «t 1 _ (5.09fm)[1-2£n fl} _ 12in v 9.58xm'fifi2m = 53:30" 03:6 Find the friction Factm‘ usinv a nation (3). f = _;- = {1021 '3 q (5.3x10‘1“—-‘ Plug these into the first equalinn: 3- ‘ 3'? —3 53 _ I ‘ ‘va—20Jf12! ( wfim‘ =fi.169 L 25:}; Elm if? 282.213353} 2 l2??? Tm Em} {he slope (it) a iitlle geumetry: slope = sin'lflal-zflfL) = shit“). 16*?) = 9A?" 5.- 10 From White, assume Newtonian fluid and constant viscosity and incompressible: (1) please note other assumptions; (2) please list equations needed to solve problem (and only those equations); (3) please describe in words how you will solve the problem; if you like provide a numerical answer. 5 = 6.12 Given that flow is laminar and you know Ap, L Q, D so what is it? Further Assumptions: steady and fully developed. Equations: (1) hf = 32tt[V]Lf(szg) [from cons. of momentum and energy] (2) fish = AP (3) Re = pUDKu) Substitute (2) into (1) and solve for )1 Substitute it into (3) to check if flow is laminar (Re = 16, u = 0.292 kgf(mvs) 6 = 6.15 You know Ap, p, u, L, D and want to know if V is possible for vertical flow? Further Assumptions: steady and that straw is barely inserted into milkshake so pressure is atmosriheric (p1 = pm) and location (1) while just beneath the surface is far enough from the straw inlet so V1: 0. Equations: (1) pif(pg) + oerzf(2g) + 21 = p2f(pg) + OLVfKZg) + z; + hf [cons of energy] V, = 0 (note V1 is near surface but far enough from straw entrance that V] = 0), z] = 0, AZ = 0.3, Ap = -3000 Pa, V; = Vtube, OL = 2 since laminar hf = 0.255m — 0.3m — thfi’g < 0 which is impossible since hf can not be negative. If straw was reduced from 30 cm to 15 cm then possible for hf to be positive. If straw was horizontal there will always exist some V so that hf is positive regardless of the length of the straw. 7 = 6.25 For ethanol take 9 = 789 kgi’m3 and u = 0.0012 kg/(m-s); Assume steady and laminar then check. (I) pflpg) + oanKZg) + 21 = p2!(pg) + ochKZg) + 22 + hf [cons of energy] 13, = p2, V. = V; = 0; solve for h (= 0.9m) and substitute into (2) (2) hr = 32u[VlLJ(D2pg) = 128u[QJLf(nD4pg) [from cons. of momentum and energy] Solve for Q (= 0.00684 m3fh) Check Re number (3) Re = pUDKu) = 4pr(1tuD) = 795 LAMINAR 8 = 6.56 For galvanized pipe take a = 0.15 mm, covert D and Q into metric. Find in, use fl, = 0.0157 for your calculation of pipe length between pumping stations. Assume can use Moody Diagram (steady and fully developed); assume no minor losses (1) Re = pUDfu = p(QfA)UDr’p = 292,500 (2) 81D = 0.000123 fD ~ 0015? from Moody Diagram (3) Ap = f (L!D)(pV2)f2 know everything but L, solving for L get 188,000 m = 117 miles (4) Power = QApr’ Efficiency = 26.5 Mega Wats (35,500 hp) 9 = 6.63 For water at 20°C take p = 998 kg/m3 and p = 0.001 kg/(m-s) and 6 = 0.0015 mm for drawn tubing. Assume steady and no minor losses (1) p1f(pg) + (1V12K2g) + 21 = p2l(pg) + OLVfKZg) + 22 + h [cons of energy] p1: pg: atm.;V1= 0 h. = f(LfD)V2f(2g) = AZ — wag) [v2;(2g)][1 + (0.810.04)f] = 1.8m; v2 = 35.32;’(1 + 20f) Guess f= 0.015 then V = 5.21 mls and Re = 208000 But for this Re and ED Moody chart gives f = 0.0158 so use this f to calculate new V from V2 = 35.32;”(1 + 20f) and get V = 5.18 ms and new Re = 207000 which converges on Moody Diagram. Thus V = 5.18 ms and Q = V x Area = 0.00651 msfs 10 = 6.64 If liquid in problem 6.63 was viscous enough to produce laminar flow what should be used for hf? hr = 32uLW(pgD2) instead of hf = f(LlD)V2!(2g) Note if substitute f = 64I’Re then both equations are the same. #5 - 6.12 A 5-mm-diqmeter capillary tube is‘ used as a viscometer for oils. When the flow rate is 0.0?l mo'h. the measured pressure drop per unit length is 375 kPas’m. Estimate the viscosity of the fluid. 15 the flow laminar? Can you also estimate the density of the fluid? Solution: Assume laminar flow and use the pressure drop formula {6.13}: $pr = 8Qpfl n18}; 3?5000Pa.='m 2 310.0713 anometntonozsr}; u = 0.292 kgfrn-s Guess p = 900 kgt‘m3 [consistent with Table 1.4] Reg = pVDEu e 4pQ.-"[':I:uD] == 4(900}!.0?l.53600};'[{3.I415)[D.292'}(0.005i : if: 0K Laminar [t is not possible to find density from this data, laminar pipe flow is independent of density. Suppose did not guess laminar flow: pu'ipgi i a.'v'."'.='{2ga ' Z: = pflpgi + ozvglfitig] - 23 + [‘1p {no minor iosses] atm. pressure at both ends oftube and assume fuliy developed so .52 = ’rhiLr'D) st'flg} pgAz : on I pf},(i_.-"D} \5:.-"3: opr'L = (MED) V3122 V = DEA Solve for f}, and get about 4 Looking at IVloodj.r Chart {extrapolated for low Re} get Re about [6 11,. = mum Vl.-"{2g} {assume smooth pipe} { f: 4 is offchart, but ifextrapolated Laminar-flow line would find Re - 16!. #6 — 6.!5 (a: L = 30 cm py‘fpg) I HIVJJHZg'] t z; py‘lfipg} .' afl-"ffflg; - 23 - hr (no minor losses) {just beneath the shake surface with [3 ~ pm, and far enough from straw that V ~ 0] 0 I 0 - o = (-3000 Pa}:"[{1200 kgr'msflgjll "1.95331 : vim-lg + 0.3 +11.- hr: 0.255 m -0.3 m 49mm“? -: 0 which is impossible (b) L I [S cm pif'lpg) l aIV’;:.-"I[2g'] i z; = py'mg) i u3V;:.-"l2g) - z: - by {no minor losses} (inst beneath the shake surface with p ~ pat... and far enough from straw that V - 0] 0 + o - 0: (-3000 Pa]!‘[(l20{) kgr'm-‘MMI misgi] l vflmgg : 0.15 mt- h,—= 0.255 In -0-l S m —\*':,m.r'g -: 0 which is possible Assume laminar so hr = JZHLV-‘lpgfill = 32(6,0'}t’0.ISWI[(120{)}{9.81110.008)? = 3823‘! 3323va = 0.355 m — 0.15 m - Viubgg Solve for Vim. and get me = 0.002%? [not sensitive to whoihor a is l or 2 because Vmbcl is so small) #7 — 6.25 pE.-"{pg) I aIVJ‘fiiZg] + 2; 2 py'mg} l c13¥"3".-"t3g} ' 2; ‘ hr (no minor losses} {1 5 is at top rank surface [2) is at bottom tank office so V. ~ V3 ~0 and p; - p; ~ 0 hf: 32uLVJ‘ngD3) : [ammo-*{npgofl : IESIDBUIE)“.2)Q-‘[n(?89)(9_81){fl.002)"] : [2' 4.31: on (J 1.0.x 10'“ m worm m‘ h Rel, = pVDs'y = 4an’{ 11ij = 1'95 LAMINAR #8 — 6.56 (a: 70 x If)“ gallooSSday 1 3.07 m“:‘s: V QEA = 3.01% 311221354} 3 2.63 unis; Rm, 2 pVDflJ I (910212.63): l.22].-"{0.0[J]) = 292.5000 53D = 0.!5mmI‘1220mm = 0.000123 From Moody Diagram f1, ~ 0.015? {note instead ofusing the Moody Chan could use Eq- 6.49 or 6.48; o-‘gf 6.48 provides I], = ODISSS} f]; : {op-"URI E": {3 V1] solving for L gel 188.000 to - | [7' miles {by Powet = gap-"Emceency = 3.0? [113.55 (85: 10"‘Pa — 4x10-‘Pa1.-"o.38 = 2.65 x 10-T Watts #9 - 6.63 it'did not have Moody Plot could use Eq- 6.39 to solve for 11, Eq. 6.39 11, = (1.3 logIReua'691)‘: me energy equation have V = {35.32.50 '9' 2O?”-1 5 50 in principal can calculate Re in terms of f0, substitute into Eq. 6.38 or Eq‘ 6.39 and solve for 1]}. #ll] -— 6.64 Another way oflooking at if - can write hf: fr, {'LJ'D] Virugi regardless it'laminar or turbulent flow. If laminar know i}; = I34.-"Re so hr = [flip-'{pVDm L;"D]IV1.-'(2g) = 33pLW{'pDIgJ. so don‘t have To use Moody plot In find i}; vs RED relationship. 11 . The typical shape of small cumulous clouds is as indicated“ in the figure below (top of cloud further left than bottom of cloud). Based on boundary layer ideas, explain why it is clear that the wind is blowing from right to left as indicated. Given: Cioud formations Shawn to right Find: Why cloud is “leaning” Solution: Just like a bonndary layer on a plate, the veiocity in the annosphettic boundary layer increases with ai’titude (at least generaliy]. Thus the top portion of the cloud travels faster than its base - the clouds tend to tip towards the direction of the wind. That is, the wind is from fight to left. I -—————_—__.._—______________ 12. An atmospheric boundary layer is formed with the wind blows over the earth’s surface. Typically, such velocity profiles can be written as a power law: 11 = ay“, where the constants “a” and “n” depend on the roughness of the terrain. As is indicated in the figure below, typical values are n = 0.40 for urban areas, n =' 0.28 for woodland or suburban areas, and n = 0.16 for flat open country. (a) If the velocity is 20 ftfs at the bottom of the sail on your boat (y = 4 ft), what is the velocity at the top of the mast (y = 30 ft)? (b) If the average velocity is 10 mph on the tenth floor of an urban building, what is the average velocity on the sixtieth floor? Pmblem‘lz Given: Aimaspheric boundary layer wind u=ay Fimi: a) 130mm of 83% u =. 20.1% at y = 4ft, finé u at mp of mas; y = 39 ft b) 10"” Elm: of urban huildéng t3 = 10mph, what is average velncity on 6%?” flow”? Assumgzions: l) Ska-:15? flaw fluaticns: Given in gmfilem: 'u : 3y“ Solution: 8) 1. Create the ratin of veiocity at the top of the sail to the veiocity a: the bottom of the 5311 2. Plug in given equaii‘mas for u, and “value of .n in. open country. 3. Calculate u at £1}: tsp of the-$31}. b) 1. Again create: the ratia) 9f velocity at the 10"“ flfior'to lhai on the 60‘”. 2. Plug in given equation and value of n in urban areas. 3. Caiculaz'e u on me. :50“ floor. ' Numbe-rg: (a) a = C ya'm J Mam {Ti-a a mmfam‘ This} 5; - 0‘15 -. 9"” . .. - . H WH- __ - i=6?) 0" “3‘29? “225%: (In) a a E yaw!1 W/JBN 5 is a ems-kw} new, a - 0’” M 734%) w 13.A plate is oriented parallel to the free stream as indicated below. If the boundary layer is laminar, determine the ratio of drag for case (a) to case (b) using the boundary layer equations. “"1 .. I. is PLATE #2 .U r H m {a} (b) Given: Hates oriented parallel to the free stream as indicated below, each with a laminar boundary layer rim-g Find: Ratio of drag force. for case {a} to case (b) ..._ _..__.—-( Assumgtions: (I) No pressure gradient Egnations: w w} (l) 5‘ka friction mqfigigm = Cf = I 12.; 2 9.730 tier (2) fi‘fazfmr drangfl -_- LEW mfim deA Solution: .7 f} 1 0.730 1. Rearrange equation (1) tofind rm. z” = G - 3" ,3, {W :_pU2 I R3: 2 Re: 2 2. Plug this into equation (2} and integrate. 2 0.730 r. l Lami- mrfm'l' fwd” 2 l p—Ux V fl 3. Compare case {a} to case (b), where for case (a); b 2 4E, and L 2 .E' ; for case (b): a, = bdx £0,732: new} E1226, and Lad-1?. FD (case a) 3.7%4f)1ll,tt(f}pU3 _ 2 Fatwa b) _ 9..73{§}‘llfl{4f)pUJ “ 14.1f the drag on one side of a flat plate parallel to the upstream flow is FD(U) when the upstream velocity is U, what will be the drag when the upstream velocity is 2U; or UIZ? Assume laminar flow. Given: Laminar flow ever plate parallel to flew Find: New drag force if: (a) U m 21}; or (b) D = U12 Assumgtfims: (1) 1% pressure gmdient Emions: (a) Skin fffcrian coqufiefem = C! = I f” = 0330 _ Re; {2) fiicn’on drag-APE, : J TFM palm acme-r solution: I- Rearrange equatienflflefiad run I r... = 0330 m; f :1 pg: 0'30 _pU2 “Rex * 2 2 2. Plug this into equation (2-) and integrate. {173% II ,on 1-! .3. (a) Cmnpare drag force fer U = 2131:; original. F990):ng— m} FDCZU}=2«83F3(U) my) 033a: MW 4. {13) Compare drag force fer U = U1? :0 efiginal. 3 a 04735 pr[—} 0’ F . 2 D(UI2)2____:‘/E a F5(2a')=e.354Fp{U) PU 3 FECU} I snake 3“ _ _ 1 2 3 F” " Imam waft“ " 39” Mr —0.?3b .uLpU afl—rqh ...
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hw2_sol - 1. MAE 101B — Summer Session II 2009 Homework...

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