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mae_101a_midterm1s

mae_101a_midterm1s - Pr.1 Pt2 2009 1st Midlerm— Sultions...

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Unformatted text preview: Pr.1 ) Pt2 ) 2009 1st Midlerm— Sultions Both statements a and b are False. Given the steady two-dimensional velocity distribution u=Kx v=iKy w=0 (I) where K is a positive constant, compute and plot the streamlines of the flow, including direc- tions, and give some possible interpretations of the pattern. Solution Since time does not appear explicitly in Eq. (1 ), the motion is steady, so that streamlines, path-.ines, and streaklines will coincide. Since w = O everywhere, the motion is two dimensional, in the xy plane. The streamlines can be computed by substituting the expressions for u and v into E : ,Ex KI Ky Q: _ , £12 or J x — i y Integrating, we obtain In x = ’11] y + In C, or xy = C Am. (2) This is the general expression for the streamlines, which are hyperbolas. The complete pat- tern is plotted in Fig. by assigning various values to the constant C. The arrowheads can be determined only by returning to Eqs. (1) to ascertain the velocity component direc- tions, assuming K is positive. For example, in the upper right quadrant (x > 0, y > 0), u is positive and v is negative; hence the flow moves down and to the right, establishing the ar- rowheads as shown. Note that the streamline pattern is entirely independent of constant K. It could represent the impingement of two opposing streams. or the upper half could simulate the flow of a single down- ward stream against a flat wall. Taken in isolation, the upper right quadrant is similar to the flow in a 90° comer. Finally note the peculiarity that the two streamlines (C = 0) have opposite directions and in- tersect. This is possible only at a point where u = v = w = 0, which occurs at the origin in this case. Such a point of zero velocity is called a stagnation point. Problem 3. Giuesx. '. ‘05}mG mommnérx‘ ween M5522; c295 $.31€<efi r25?» ég‘xoxxéiir Ofiéfium K‘Cky‘ioéxfi,‘ Oi; Quads“. \T\v\c&“.x€‘e \S‘xg éuQmjfucr R \ (CF wt»: :50 «A «x , E J MW; vvq)*g\v\c£\ 1 g; N - '2'“ E nguxxngkxehi. ‘ U‘x é‘gmmg MCUJLLcE. Rh \hmvxgyw s£§b£ E :qumcnx \c‘vq st-w» \DQSK a: fjjkcfiuxcf {vgaM (fffi‘itt‘xll S\C:\:e; a} 80 *0 E gmmcckc‘a Rafi; OE: 3C (2sz E \\ n (“1”:3‘! E . ?%C§j§ kWCFQ E __ _ , (" "\vowx Wm \a‘fi; ®aafmm '. Nfi‘fihfiicxi‘ng = 3w %*4. x L“ QC, {(1 {in _ , fl _ A U E'mw RQVQ. ‘3‘ VCR aka a" mfx “Q ‘- (r J? h E K «E (:3. " 6:) ‘32: K : 43 I? 1‘ K A!” I? ”9 C 00 M ‘c . ‘ {“x 4:2, “‘ QGRm 2 Emcéfiqu‘ 4' $00 XL —» ~ 42>: . *7. t ~64 . " Skim och,“ 5‘3 “EEO h (T‘QV‘WK Ego.“ gt" :5 *‘efi‘v awxg‘ncg, Q‘mfix ‘33.} E cunxia {R \x a 1:1: "j (:3 1 (3’9: EwCSK 7:ka (93%" RQEQS: ‘LuCéE‘CSKRJ‘gl‘wQEE ) '1 ; fie <xm¢ {flag ‘~ “L 24:) , Rake. ELK: \‘\ . “Rama“ ‘3“? 2: ngim4 1' ‘i ‘ ' “‘ \kc, can <‘Q\0E\>¢ QM :kxc: “\‘x \ié Vfl:n€: $fig \ng \; Ska/wave, of ”£32,?“ ‘fiubx \xz Qefiagvued . (5 Q vii?- ’ 3‘}? m SK¢KYJSZJKVA \Wq \V\\.o \C QM \ W W _. . A? ?w%\) CW?“ \\\%&L\l 3 3:“ \NM ., §0\Nw~xc\ {Z \\'\\ W C\ \\)‘Q S =Q j, I! {7 ' ( k l P) g"; j»??? 49 5?)? 05: F‘ fl"? L" r: i; , q 7;: fi/Eg” “V E 7L 9E6 5 3 Problem ;4 \.>- "ann we. QC EEK; \s \\ SKKL G‘ézés \E- \n waxzvxon Racy; Look»: aQQSK <5»- , aw $0.2M. ‘ LWLoXAm ‘\ Sin" §g§ CQNQ-“M‘g WW“ ‘- KER“ 3:3- ga fim= 11. \stW'. \\\ “$655.“. MW kt) e: mum {3% ”9“,“ (LL 2: K ‘CT‘L smrQMn. omA cfi W‘a\&fl_ o-C 8503a . ...
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