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Exam2Solution - Exam 2 CE 3700 Fluid Mechanics Fall 2005...

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Unformatted text preview: Exam 2 CE 3700 Fluid Mechanics Fall 2005 Show all work. Show all units. NAME: 3 ‘4me (20 pts) 1. For the open tank below, find the value of h that will produce a steady water surface elevation (that is. a water surface elevation that does not change in time). CV CS ® 1-ft diameter —Ql + Wfifldz> :O Vout:@fi/$ 4— L 2% = f j 3 (1m Q 7. t 2 2L w L 3 £74”) I Z6 “3/5 L _ _,_._— f 4321?) [141005;] PW 0.188341 (20 pts) 2. Water flows in a 0.5—m-diameter pipeline at 1.0 m3/s. The pipeline is 4 km long and is connected to a reservoir at the upstream end. a) Calculate the critical closure time. b) Find the pressure rise that would occur if the valve were completely closed before the end of the critical closure time. s Q 1 M/s b) A 'm : A : —— 2 .0 __ V A ':%:(OSM)L b 3 S 1 0.1%“?- E‘V "' 2.7. (EVA 2 2.7.xlo'1 Va. (Fun—Wide M ”Wale AS) \_.____‘.\ Q, - ti)” s /Z.Z)<Io1 N/M‘L 4 «m Q F5 ‘fiv’lAiL l 83 s looc- x __._r J< MS ““3 \ Qrflq’coi closure el—nMe 2 4mm»; {Sew pressure waw +9 MW- rme valve lo reservu'w- +3 wilt/‘8. med ‘Hme Mm valve h VQRNMY Crel’l‘ca,‘ Unsure l—me -: 1(23 5) : Q4 5 = 7,563,000 FL (20 pts) 2. Water flows in a 0.6-m-diameter pipeline at 1.2 m3/s. The pipeline is 5 km long and has a valve at the downstream end. The pipeline is connected to a reservoir at the upstream end. a) Calculate the critical valve closure time. b) Find the pressure rise that would occur if the valve were completely closed before the end of the critical closure time. If; "‘73 V 2 £3: 1: 2 4.14, "4/: TT‘ 7. 7; (OW) L—V—F“ 0.293%— Ev = 2.; C779. = 2‘?— Ktbq la. Offline“ M T034“ A‘s) E - t, = ,2;— ' : l483 "Y: Q) Cfl'h’uJ Closure Hue '; Hm! Rm Tressuv-Q wouelo Havel ”9mm v«\ve 4o Hitch/off h valve. Travel time. (EX/WA valve in vetervsu‘» - L- goQQM C ”(53 ”(t : 3.375 [HA-{Lad doswe lime : 1(337 s): (2.75 = c.1825, 000 7"; (20 pts) 3. The siphon discharges into the atmosphere at point (2). The surface of the reservoir at point (1) is open to the atmosphere. Ignore energy losses in the siphon. 21. Find the discharge Q in the siphon. b. Find the pressure at point A. Reservoir 5 ft 12. 2 6H 13 -i p, V; ~ (WXZWLL 9) 2 {7.1 "21“ ngx 2 m %)[% ”fl W 0.18”»? = 14—.\ 333 E) f 5. ° ‘ . I A n 4; w 9 3“: /‘\ L9 3, \f/ (20 pts) 4. Determine the forces in the x and y directions exerted by the oil jet on the vane. The vane is moving to the right at 5 m/s. The vane turns the oil jet in a horizontal plane. V. 2 V3 : 16 m/s. Q = 0.13 m"/s. Ignore energy losses. y l Wul’ln wovmg camlwl volume: on (s = 0.90) V” = - 4.63 “/5 WYSL: - S-SOWYS 30° N "Y; ————— __.x . '3 v“/ A z: {3‘ 2 (1:11,: = 0 003‘ m Qlflifl) — v A = \\W/,Xo.oorlm1) Vol 3 = 0.0%‘t4 “/5 EFX : (Q(T¥v’ W1) ELSL 3 w h = @‘l x “300%: x L“; mostlng’iSle - n 4) X W‘ . l Kj : — \USL N Vane ctr/1mg on gel > lLaSL (\l 3d- chl-m3 m \rqwe CESZN—ez 2F : (Q (martial) - U N’S'L w} F = Q51 X(DOo [513 x LI>(0.0314"§>(”5573M6 ~ O> : " 4'43 N vane M/HVL] W 32A» 7 443'N gel (LL-{rm} w Vanna £44’3N H (20 pts) 4. Determine the forces in the x and y directions exerted by the oil jeton the vane. The vane is moving to the right at 7 m/s. The vane turns the oil jet in a horizontal plane. V] = V; = 22 m/s. Q] : 0.18 m"/s. lgnore energy losses. ly Oil (S = 0.90) 30° l N—SL k3 lit/{W WDVYVK? oowlvm Volmmt‘. vii: 42.4% ml: law—7.50% \s'vg ““““ , - g _ mew/s = 7, A—_ V - 22% oooxlim QGfi‘i’l‘) VA : i5flX0.00?l8mL) VanMC " = 012?) ““7; 3 m 1/0423 m§)(~ll.q1’§ — 15 £35 ‘30‘13 {\l \(vae £1:th on Set 3018 H ‘Je-l (Lei-M3 on vawe, (BOTBN a, M R X ZFY : g Q (my : K Q3 x loco f; 2n, = M (Tar-Va.) F3 : (0“! x [000 —830 N 8'50 N l l 39* Adams on wmc N—sL 'YVI I: M ho \féme “.ij on Eel— 82w T (5 pts) 5. Start with the general control volume equation shown below and derive the general continuity equation. stys (if J _ _ + v I it it bpdV bp dA Lei l5,“ : mas: a? lib» Sysl-e/m b a mus ” ‘ — ms: g4}; (MASS-oi: “HM. TLLS'i-e’w‘) :c 0 Hum Jai‘nM\i‘l:au a? ‘1 “(Sin)"; Wm.“ ”p ' Sys—l-em is MSW (8 pts) 6. Write the type of flow represented by each of the equations below. 5km! ‘3 yum/mum ( «(mm/M uuerUMH Wflom ...
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