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327exam02-fall06

# 327exam02-fall06 - 530.327 Intro to Fluid Mechanics Su Exam...

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Unformatted text preview: 530.327 - Intro. to Fluid Mechanics - Su Exam 2 — 20 November 2006 Name Swamps 50 points Problem 1 - Short answer. (5 pts) In class we looked at pressure variations in a liquidwﬁﬂed tank that was aeceierating horizontally“ Consider here a tank that contains water (density 1000 kg/m3) to a depth d = 2 m. The top of the tank is open to the atmosphere. What’s the pressure at the bottom of the tank if the tank is accelerated upward (the positive z—djrection) with acceleration a: = g = 9.8 111/32? ’ P“: w “’42 was owners: 0% m4}: 56mm, 'fmsme m ”we £~-‘Dt§2ﬂ;\’m~1 ~— /1 use my? Cmﬁuﬁcxﬁﬁ) => P(gj 22913—13 + com: MM c, G) a a Hammer Wm So WZMI: Rem-r 2‘ “000535 my“; 01m) 2. " Wm + 51,200 Par,‘ 0 Problem 2. (Total 25 pts) Consider the steady flow of an incompressible, inviscid liquid, whose density is p, through the horizontal Venturi section shown: ml?“ all “—11,.” Q A; A The entrance and exit to the Venturi have diameter at}, and the throat has diameter d2» The volumetric ﬂow rate through the Venturi is Q, and the outlet pressure (the pressure at the right side of the Venturi) is 190. a.) (4 pts) Suppose that the only limitation on the ﬂow rate is that the pressure at the throat must exceed the boiling pressure, 191,, of the liquid. Does this condition establish a maximum or minimum on the allowable ﬂow rate? Explain very brieﬂy. b) (9 pts) Denote the extreme (11133: or min) allowable ﬂow rate from part (a) by Q = an Using dimensional analysis, determine how Qm depends on the other variables in the problem (all of which are mentioned above) in terms of nonwdimensional parameters. c) (9 pts) Solve for Qm exactly, using whatever equations are relevant" d) (3 pts) Express your answer from (c) in terms of the non-dimensional parameters from part (b) E The vewom’ \s A mum A1 1% Tatum; so 8" BEFMULM Tee Wes-ate 15 (9 Newest ram“ Teas me commas Em,“ >Pb weasenosn A MAXIMUM vtwcm’ entities) A MAXMM Fund {men 2' El qty-rte QM:¥“ l‘lx,Az;P,Vo,?b) “=9 3 messages =3 Wu?) :3 WM: 3 now—Dwmsmum sweetness ° 19°F- WWM‘W YWMEFQ CW} 9527 GM, oko’r use ear“ at Asp Ola. 6?. \$01“ ?0 W50 H: 1.2;) U59 A‘} e) ”900 O ‘ I olwoeFTue '61; Age mm: T? = vim ,1 P0 l T1“2," “h. 2W MW (Extra workspace for problem 2) ¥0¥~V5=qu€WW1o€HLYl 3 g .3 so «a- = My”): GD All ‘?0 mg raw, \1 : _....._ \J _ J: ‘ “if/it «ND 2' WT Az‘ﬁt . ‘Yo - 9p!» (9 \$3 OW E . : EL \1 2' _ 2' _ . Y'NO Q (9 4. 'J£*QZI " %*%*Q%z (%|'%z\ "gamma 3, E: "K SGE5ETUT6 @: VLZ'Vﬁ: 7.. Qua" (\ , 3 v.13 Q 2. u "H“ "3-7.. 3...}: : .31. J... A _ (“W ”‘3‘ 13‘ 65 ma" «41'1“: ‘3 So we HAW. ~. \thﬂ (1‘1 (4J1: ~\Tz(?°'Pb)|/1\ ® > \T w( A ”I: T ”1 JSWbTH'Lﬁ'" : 3: “‘" Tl“ -\ \"\1 \ 5’| m ‘ H 3 F gomfﬂ Problem 3. (Total 20 pts) Water (whose density is p m 1000 kg/m3, and whose dynamic viscosity is ,u m 1 -- 10—3 kg/(m~s)) ﬂows leftwto—right through the pipe system shown in the ﬁgure. The ﬂow enters a pipe with diameter d; = 10 cm, passes through a square—edged contraction into a. pipe with diameter dg u 4cm, then passes through another equate—edged junction into a. pipe with diameteI d3 m 12 em. The volumetric ﬂow rate is Q = 3 liters per second. You can measure the ﬂuid pressures 131, p2 and 393 at the locations shown" The indicated lengths are 11 = 15 cm, lg n 20cm, and £3 w 15 cm. Suppose that p; = L1 - 105 N/mgw Q “"3 l 4: 3 W 1: +---—--+ [2 (a) (10 pts) If you ignore major losses, but account for the minor losses at the contraction and expansion, what is 393? (b) (10 pts) What’s 133 if you include major losses? Treat the pipes as smooth. WI ﬂ EM?“ 60‘" [EVdKY-L +a%:\dlfi+é}g 4:323 : 2M9 ’21 mmﬁssec, 2 P CD “(15.62 NUT?» wt: {1.3 :\ (Assumes); 2323 f, \l \ :2 3'- ﬁozzﬂ‘ FWM Jab. 8.4-1 COMTVACVION/ Nita“ smog = 0.110, K02: on L6 Ethst, A? :- OJX, K9,: 0.15 {)3 l" \f All 1.9."2' m... + l 5 _, .._...\I3 6’ I). Z. 2. (twee) (Extra workspace for problem 3) ... ‘ ~5wn/ m. Van, no *3 5 \M M = 0.537175 V; = Lima/5 : “405:? + \ocng 3331 ::]:\ \Mog (NO—«z 3173‘ :5 \dm Mao? b05505 £166?“ ”momma 96mm :91 -_- “.4, 2‘4“. ... 2h! P P 2 \L _.A SAM/«1mg HMO?- Lossie', w. L m z ..._ 1 Zkv gig—.31. *ssziaihxsgﬂi (a) ‘ 2‘ Dz. 3-" D5 2. - To \$pr me £12. a web wfsews #3, \< _ : ‘00“ 9k? . “‘ = «15" = no“f lo” (33‘ “\$3521 (Jim 05:24, 3.6 0/ W2 ‘1 b1 1 l 3 3 H5 M D‘ V l ”3 \$9M M000? BHG‘MM (swam W15) 2 £30021), 431’! 0-01‘3/ 4:3 ; 0.01.; 3 2. m " So 21% = 0.0211.- ‘393 --(°_I3”'"‘B5 4. com; 259% iﬂ/ﬁ o 02.3 {5916-1995) Mom 1 ‘fCM '1. HIGH 2. W =9 P = Vatmn1matooo§a~om =\l0fam 550w- gl Q) ...
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327exam02-fall06 - 530.327 Intro to Fluid Mechanics Su Exam...

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