{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam1Solution

# exam1Solution - Fluid Mechanics II Examl Name Score 1 A...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Fluid Mechanics II Examl Name: Score 1 A particle moves along the horizontal centerline of a converging channel. The velocity at the centerline of the channel is given by 17 = V1(1 +f)t, where VI and L are constant. Find an expression for the acceleration of the particle. (5 points) .3 d? )Tf —-* 4‘ all? 31? a?! (3?? or r (at + 'v-VV- n + an War“??? r‘ n _‘ Y5” :r ﬂighwmjmt— Loni 2. The Reynolds transport theorem can be written as DEW—6f de+I bV‘dA or "at p p '71 Cl? CS Where bis the intensive property and B is the extensive property. Explain the physical meaning of each term and show which term employs Lagrangian description and which term uses Eulerian description. (7 points) ' a " " " . -. .. L-. , 4b“. {We “(aft 0% char]? b 0900:0th wqh S‘then, Lﬁ3wajmq 'D't re r I _ I H - (Ir-0(- . I I- . In,“ . e ‘ {.9}er {he tune (at: 0-? Chm}? 0'; {'9 G j waifd w H\ CV“ in: We .9 C; {V l ___,,..—- a _ TEL-delta n I "" ‘dA. Mfr-1% b on the thrat ST-qw due h {M ‘ l 1 Name EM‘f’Y‘t-bn 3. One velocity component of a two dimensional incompressible ﬂow ﬁeld is given by u = 2xy, ﬁnd the expression of another velocity component. (7 points) "‘ a [t if 2D Tﬂfmptrm'b‘e silent :0 at w ’ )l1+-E:j':o ﬂ— w. my"? 4. The two components of velocity in a ﬂow ﬁeld are given by u = 3’2 —x(1 +x).v = y(2x + 1), ‘Wo’cahunat is this ﬂow irrational? Show your work. (6 points) a "3’ 3 -;(§L§ml_ al'tz—vnmij) “1 ‘2'" a; an 2- 31 “yr—— 2 Name S. A flow ﬁeld is described by the equation V = (3x2 + 1)? - 6xyj‘, determine the following (i) the volume dilatation rate; (ii) rate of shear strain (or angular deformation), and (iii) rotation. ('12 points) . r _ - W F .,_ ".- \f I _ - — v.1. L ..r_ g f ‘1' l ( {It} _ r If r l 3‘ E -r\':" "' h if! _|‘{ i'fl' -\ I K- - '_ 3.! l _: .. ' Fr. “'1‘! ' ' ' . F ._ - - .. ." ‘3 {if}. . r I '--I {J “r h I} ) h- ;4 'l ‘ ‘ II J .-._.\ - 2 F '=' "J k 6 Flow of a viscous ﬂuid over a ﬂat plate surface results in the development of a region of reduced velocity adjacent to the wetter surface as shown here. The region of reduced ﬂow is called boundary layer. At the leading edge of the place, the velocity proﬁle may be considered uniformly distributed with a value U. All along the outer edge of the boundary layer, the ﬂuid velocity component parallel to the plate surface is also U. If the x direction velocity proﬁle at section (2) is a - [3] Develop an expression for the volume flow rate through the edge of the boundary layer from the leading edge to the location downstream at x where the boundary layer thickness is 6. (20 points) .' . W _I_.4 _. . W. . .. W ( ; II section (1), U .. IS?°ti°“(2) L . n l_ 'U _ . f“ If l l I l l l I t l I it ' J r. 7 A liquid ﬂows down an inciined plane surface in a steady, fully developed laminar ﬁlm of thickness of h. Assume flow is two-dimensional 1. Give your assumptions (4 points); 2. Simplify the continuity and Navier—Stokes equations to model this ﬂow ﬁeld (8 points); 3. Find expressions for a. the liquid velocity proﬁle (10 points); 1). the shear stress distribution (4 points); c. the volume ﬂow rate (10 points); d. and the average velocity (3 points). 4. If the film thickness h=1 mm and the plane inclined at 9:15”, ﬁnd the average velocity (4 points). If I ﬁf’ﬁﬁ‘fj'i -- . I _ a.” _ A V II: 1 d devede 307‘ 7.7 e o 69) mcmpmmte ® iwo dr-m w: 0 ’ ® L0 If“ "LIA {1T LEW,— ‘ ' h" C"? a? 1;": M,- up be V30 6,}: - r r» -‘ n - — — v —- a»: . Q 8,10,, 6 .7. b "7 0 g) : ‘ l . l. ' ermﬁ‘lbm 5""? Page“? 4 M V 4 \76% : —%+/6'{ ‘a " ’ <12 6 7 a a ."l :' 4 ' ‘- —-—--------—~ 1/1 ‘4V f3L3)” 5:] “P/Ml 2-+ 5 Name CINZ'W‘Wr 91mm“ “'9 hm" . ﬂ— y = \ -3 VI’ “ﬁgro T“ #uﬂa V': But V30 {0! V73”ng 0'1 JD“? N‘H- V: 0. iv Ffum {he 1’0“ng e‘iuab‘m 1% T—oﬁrpcffm W? knw-v 3” a’u ) -. 'W‘F/“W ~+ [933:0 => % z/Lq—ﬁg.+ {>31 TM» left We is a fwcbm «F “r and the 193% Rob 11‘ afwbh b “f 2f.- Uggtzo), 5‘0 ﬁt: (:va Becaﬁfe {*9 Wm”? 9+“ mfg” ﬂ “WW (Batu) WP {Tet 2.13 = 0 \$0 the Muménkvm 919‘“th Can he fﬁanl.‘ed a} at“ am (MW/a m1” {ﬂwﬂﬁnm --® ') ' aP —%+%1:O *W‘ pﬁawzo ——[email protected] ...
View Full Document

{[ snackBarMessage ]}