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308_mt2_sample_solution

308_mt2_sample_solution - Mechanics of Fluids Aerospace...

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Unformatted text preview: Mechanics of Fluids: Aerospace 308/308H Sample Second Midterm Examination You have 75 minutes to complete the exam Answer 4 out of 5 questions only Question Points Check 1 r 25 2 25 3 25 4 25 5 25 1. Two point sources of equal strength m are located at z = c and z = ’iC, with c > 1 and real. Find the complex potential for the flow repre— sented by these two sources outside a circle of radius a = 1 centered at the origin. Describe the resulting complex potential in terms of a distribution of point sources. Where are they located. and what are their strengths? fl». 610344;! fie» Jo Medial 5 fake W 4‘9, i=6 ml é= Ca. 71405;, . ’/(£) = ”lg!" (3'6) 4. «(a (€‘C<L)} 7?" "“25; J» gm.“ ’7 #6172) rd" “’ ”/9“ #99 = mflifa‘lc)+J«(3?"+5¢)l L #‘fl/t) = mfifi (£2 "3) 4"" {VH9} 1., __ 1 _ " - _ “- W6 C_E_(Iéc)- 5(6 2") M :46: =é<l+€c€>= ?(é_22) J; .& W fled far 24"“ ‘3 Me): fife)4 7571A) / I) = m {&(£-c)+ 4.4x (an-6c.) +46?) 'J“é)+ A $.25 +£(fc)-1\{£)+J"(é-é)} ’1 A; M62) , m/Vg\(Z-¢) +’(«(é‘~6¢) -1JA(§) +4 (a-a()*.(,‘(£-%) 4.462)} 52% fa» a (aqua .558 ”5,24 74m .90qu ysfififiim 4...! (7') My yacylfi m 4,91 gag/c (17/ 77‘ (Crux £62923 duo/ a Maui “a! Mam/VIM 74% (14,1069- flux. 2. The curl of the velocity, or vorticity, is given by Z .1 E g:VXy:= (9/(9x (9/03; (9/82 u v w Consider developed laminar flow in a channel of height 2h where the axial velocity is given by where U0 is the centerline velocity. (21) What is variation of the vorticity with y? (b) If a streamfunction can be determined for this flow find w Cay). If not, explain why a streamfunction can’t be determined. (c) If a velocity potential can be determined for this flow find gb (1‘, g). If not, explain why a velocity potential can’t be determined. (0)414 aid/Lea, «9/93=o 4.4:! “no, kztmj Manama 7:) 4‘” 06 = 23‘3: ex. 33 gm “Ct; fie“) (‘5 VOCLJb‘o/zeol” ‘9/31.()EO H -99. 2:03 LJg. = - U0. “’1? '- I; 3 . . o z} (5) 71*- f’ku 75 hm-obmwmwfl,&&:wfj moo MOWM’ :23 M'%~e,( made. Ugo-J, from (Sm/1;?) (944+9V*O OJ 9‘“.0 M,b&fle‘JisaZu—Léofuo. ex. 53 3-» C) gnu. «06. w (J) WM'MuQI fr,“ MCQ:)IM cM'LuA—L A W/‘aaéi path/4'1 Mg} ”:9 4”,”, {7, .‘rrohhmd 00am; L.______m , 3. Consider steady, two-dimensional viscous flow between two parallel plates at y = :l:h. The upper and lower walls move in the w-direction with velocities 2U and U rcspcctively.The flow is independent of :L' and 2. There is no pressure gradient imposed in the redirection, but there is a body force due to gravitational acceleration given by, gy : —gi From the equations of motion determine, (a) The velocity component in the y‘dircetion, v (9:, y, t) (b) The pressure gradient in the y-direction, 823/03; (1‘. y, t) (e) The velocity component in the x-direetion, u ($,1 ,t). Justify your reasons for your simplification of the two-dimensional equa tions of motion given on the crib sheet. a» Amiga Mtge «WA-m 6W W DP =—@ 3 {9&33) a l’flo") .. 635(54-‘0 mag 9,! :dflzo a» 01x— “we a. : Clj‘f' Ca. A/o—u K(A)=02U 4-" 18‘:c,A+CL “(‘A) = ‘0— 0- :---c,/\ + Ca. 2 ’ 3U " (9:1 4. A laminar boundary layer on a flat plate (zero pressure gradient) is approximated by _ Uoosin(7ry/26) 1/ch “ ’ U00 3/ > 5 The momentum thickness is found to be, 0? u 9=/— (1——)d 20.13666 U00 U00 1/ (1 ) 0 If the skin friction coefficient is related to the momentum thickness by (if) . : 2” Cf drli Show that 6 ~ — 4 795 RC”? rc where, Reg, = onom/u. Moo 6,: = {123/ ml fa =/4:_;‘/=/”‘ wrm( SN” 2% Cf: ///‘U/a°". -20}? 2 08660545 emf/.8 dz. 41 So 263‘8 ‘ of}: a T 1) x‘ a“ 043% 3,. All: 1]; v1 4:. 5M'It ago ‘1‘! 8(0)=O o Izéé 000 u/ an m <5 T (a; L a 4 7"“: (at, 5. Consider a steady, laminar boundary layer where the freestreani veloc- ity is given by U00 = Vat/L where {IS/L is the non-dimensional distance from the leading edge and V is a constant velocity. The velocity profile is assumed to be of poly~ noinial fOI'Il'l , , U00 (do + dry/5 + LIME/52 + (MUS/53) y S 5 u _ U00 3/ > 6 Show that to satisfy the appropriate boundary conditions, the velocity profile takes thc form u 3 13 V62 2 3 #2 _ __ _ _ < U00 (277 3n>+4uL(77 277 +">’y*6 where, 7/ : 11/6. State what boundary conditions you ehosc. 0 “(0) = O U. 0 ‘ ‘ i {p .+ 1] Big,- 1 (—0 d L a U I a usage =-12l.e6 i v3}! =X‘x dz fax L. L. L.“ @/E b: «L; 2. a. V93L9)= AP =.._\{__x. @a. 7335-27“ f 7. ® ‘33? 2—4-10” .4; 42. a - V’.‘ z ——". W- 5 5: vaz'U; i L“ 4. +343 = ~20”. = Vansl r 2371-.- a. +a3;l~a; ‘ l4, Viz-s; 12:2sz - a'u; =-' + mag: 43 g -1 + ms: qozv-U 7- €4vL‘ 014, = 3 '1‘ VI-x' St 1. 1 q‘Z’VLI- 4‘ - i‘v—is‘ z UfiyL 5 a 3 v1 S‘ 2 3 ‘0' __+ zt§_\/x£ +(~i+’v‘z$‘)§_‘ no 2 any». 8 gzvué* 2- 1541: L‘ b“ . q 3 -.'_ 3 V1 31 1'. 3 '4. _—~ ‘ In in)+ x. ‘(n‘2fl*n) 0.. ‘1va M 3:6,. So 1 I. I" 2‘. . (inn-'- 3) + V. S (”7'2” +4) ,0.” 2. 2. 4-" w. ...
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